Class 11 Physics

Hydrostatics Long Answer Type Question And Answers

Hydrostatic Pressure Questions Explained

Question 1. How can a body be cut easily with the sharp edge of a knife, but not with its blunt edge?
Answer:

A body can be cut easily with the sharp edge of a knife, but not with its blunt edge

The working of a knife depends on the pressure, i. e., the force per unit area. We know that pressure (p) = \(\frac{\text { force }(F)}{\text { area }(A)}\); so keeping F constant, if we decrease A,p  will increase.

Now, the surface area of the blunt edge of a knife is more than that of the sharp edge. As a result, for the same force, the blunt edge produces less pressure, but the sharp edge creates a much higher pressure. For this reason, a body can be cut easily with the sharp edge of a knife.

Read And Learn More WBCHSE Solutions for Class 11 Physics

Question 2. During the construction of a dam, why is the bottom of the dam wall made thicker than the top?
Answer:

During the construction of a dam the bottom of the dam wall made thicker than the top because

The lateral pressure of water depends on the depth of water. With increase in the depth, the lateral pressure increases. The depth of water at the top of a dam is less and hence the lateral pressure of water is also less.

But due to the maximum depth of water at the bottom of a dam, the lateral pressure of water is the maximum there. To withstand such enormous pressure, the bottom of the dam wall is made thicker.

Question 3. Some liquid is kept in a container inside an artificial satellite revolving in a circular orbit around the Earth. What will be the pressure at a point inside the liquid?
Answer:

Some liquid is kept in a container inside an artificial satellite revolving in a circular orbit around the Earth.

We know that a body becomes weightless when placed inside an artificial satellite revolving around the earth. For this reason, the liquid inside the satellite has no weight, and hence no pressure develops at any point within the liquid. However, there is the gravitational attraction of the satellite itself. This force on the liquid is so low that the pressure due to this is negligible.

Step-by-Step Solutions to Hydrostatics Problems

Question 3. A cylindrical vessel is filled with a liquid such that the thrust on the bottom of the vessel and that on Its side wall become equal. Prove that the height of the liquid column in the cylinder is numerically equal to the radius of the vessel.
Answer:

Given

A cylindrical vessel is filled with a liquid such that the thrust on the bottom of the vessel and that on Its side wall become equal.

Let the height of the liquid column in the vessel = h, the radius of the cylindrical vessel = r, and the density of the liquid = ρ.

∴ The weight of the liquid in the cylinder = πr²h x ρ x g

∴ Thrust on the base of the cylinder = weight of the liquid in the cylinder = πr²hρg ……(1)

Now, the average pressure on the wall of the cylinder
=1/2 x hρg, area of the curved surface of the cylinder = 2 πrh.

Total thrust on the wall of the cylinder

= \(\frac{1}{2} h \rho g \times 2 \pi r h=\pi r h^2 \rho g\)

According to the problem, from equations (1) and (2), we get

∴ \(\pi r^2 h \rho g=\pi r h^2 \rho g \quad \text { or, } r=h\)

i. e., the radius of the cylinder will be equal to the height of the liquid column.

Real-Life Applications of Hydrostatics Questions

Question 5. A large shallow wooden container is filled with water. But the container does not crack. A hole is then made on the surface of the container and a long narrow tube is inserted vertically into the container through this hole. The tube Is now filled with water. It is seen that the container now cracks. Explain why.
Answer:

Given

A large shallow wooden container is filled with water. But the container does not crack. A hole is then made on the surface of the container and a long narrow tube is inserted vertically into the container through this hole. The tube Is now filled with water.

The thrust exerted by the liquid on the bottom of the container depends on the base area and the depth of the liquid in it, but not on the quantity of liquid kept in that vessel.

• In the present problem, initially, the depth (h) of water in the container was not large before the inclusion of the tube, although the amount of water in it was large enough. It produced less thrust on the bottom of the container hence the container did not crack.
• When the narrow tube is filled with water, up to the height (h + H), the water column increases considerably. It exerts a large thrust on the base of the container, and so the container cracks.

Question 6. A vessel full of liquid is descending with an acceleration a [a < g]. Find the relation between the liquid pressure with the depth of the liquid in the vessel.
Answer:

Given

A vessel full of liquid is descending with an acceleration a [a < g].

Let the density of the liquid = ρ, area of crosssection of the vessel = α, mass of the liquid =m, and pressure at a depth h inside the liquid = p.

The upward reaction force exerted on the liquid in part A by the liquid in part B = pα.

So, considering the motion of the liquid in part A, we get mg- pa = ma

or, pa = mg-ma = m(g-a) = ahp(g-a)

or, p = hpρ(g- a).

Question 7. A cylindrical vessel contains a liquid up to a height h. A hole is made on the wall of the vessel. Water I comes out through the hole and falls at a distance x from the base of the vessel. At what depth should the hole be created so that x becomes maximum?
Answer:

Given

A cylindrical vessel contains a liquid up to a height h. A hole is made on the wall of the vessel. Water I comes out through the hole and falls at a distance x from the base of the vessel.

Suppose a hole is made at a depth y from the free surface of the liquid and the horizontal velocity of efflux of water through the hole is v.

∴ ν = √2gy

Suppose the liquid takes r second to fall at a distance x from the base of the vessel. Considering the vertical motion of the ejected liquid, we get

h-y = \(\frac{1}{2} g t^2 \text { or, } t^2=\frac{2(h-y)}{g}\)

Considering the horizontal motion of the ejected liquid, we get

x = νt or, x = √2gyt

or, \(x^2=2 g y t^2=2 g y \cdot \frac{2(h-y)}{g}=4 y(h-y)\)…….(1)

Now, for the maximum value of x, x² is also maximum, and hence \(\frac{d}{d y}\left(x^2\right)=0\).

So, differentiating equation (1) with respect to y. we get

⇒ \(\frac{d}{d y}\left(x^2\right)=4(h-y)+4 y(-1)=4(h-2 y)\)

∴ \(h-2 y=0\) or, \(y=\frac{h}{2}\)

Hence, for the maximum range of the water coming out of the hole, the hole should be made at a depth of h/2.

Question 8. Prove that In static equilibrium the pressure exerted by a fluid decreases with its height.
Answer:

Let us assume two layers at heights h and h + dh from the base inside a fluid. Let the density of the fluid = ρ. A part having unit cross-sectional area and thickness dh is considered.

So, the mass of the part = ρdh.

Let the upward pressure on this part be p and the downward pressure = p + dp.

∴ In equilibrium, (p+dp)-p + (ρdh·g) = 0 or, \(\frac{d p}{d h}=-\rho g\)

or, \(\frac{d p}{d h}\) is negative, which indicates that with increase in h, the dh value of p decreases.

So, it is seen that the pressure exerted by the fluid decreases with its height from the base.

Question 9. Two holes are made on the wall and at the bottom of a cylindrical vessel. Now the two holes are dosed with two corks and the vessel is filled with water. If the vessel is now allowed to fall under gravity and the two holes are opened during its flight, then state what will happen.
Answer:

Given

Two holes are made on the wall and at the bottom of a cylindrical vessel. Now the two holes are dosed with two corks and the vessel is filled with water. If the vessel is now allowed to fall under gravity and the two holes are opened during its flight,

We know that a body falling freely under gravity experiences weightlessness. So, the vessel filled with water remains weightless while it is falling under gravity. Then, the upper layers of the liquid exert no pressure on the lower layers and hence there is no pressure difference in the vessel. As a result, no water will come out through the two holes.

Question 10. State whether pressure applied on any part of a confined liquid is transmitted instantaneously to different parts of the liquid. Also, state whether Pascal’s law is applicable.
Answer:

The pressure applied on any part of a confined liq¬uid does not reach different points of the liquid at once. The applied pressure is transmitted through compression and rarefaction within the liquid, and hence, with the velocity of sound. During the transmission of pressure, Pascal’s law is inapplicable. Pascal’s law is valid once the system comes to an equilibrium.

Question 11. Does a hydraulic press work, if the liquid used in it is replaced by a gas?
Answer:

The thrust applied will no doubt be multiplied if the liquid used in a hydraulic press is replaced by a gas. But, due to the greater compressibility of a gas, the thrust developed on the larger piston will not be large enough. For this reason, the press cannot be used for efficient pressing.

Question 12. Does the density have any significance in the weight-less state?
Answer:

Density = \(\frac{\text { mass }}{\text { volume }} \); the quantity of matter contained in a body is called its mass. This mass remains the same even in the weightless state of the body and hence the density of the substance remains unchanged.

Question 13. A long cylinder is fitted with a tap on its wall at its lower end. Keeping the tap closed, the cylinder is filled with water. Now the cylinder is made to float by placing it over a cork and then the tap is opened. State what will happen.
Answer:

Given

A long cylinder is fitted with a tap on its wall at its lower end. Keeping the tap closed, the cylinder is filled with water. Now the cylinder is made to float by placing it over a cork and then the tap is opened.

Due to the lateral pressure of water, it will be seen that a jet of water comes out through the opening of the tap. If the opening of the tap is in the horizontal direction, then the water jet will come out horizontally. Due to an equal but opposite reaction of this thrust, the cylinder (floating on the cork) will move in the opposite direction.

Question 14. A cubical box is completely filled with water. Prove that the total thrust exerted by water on one of the vertical walls is equal to half the weight of water kept in the box.
Answer:

Given

A cubical box is completely filled with water.

Let each side of the cubical box be of length x. Mass of water kept in the box, m = x³ x 1 = x³

∴ Thrust exerted by water on any one of the vertical walls = lateral pressure x area

= \(\frac{x}{2} \times 1 \times g \times x^2=\frac{1}{2} x^3 g=\frac{1}{2} m g\)

= 1/2 x weight of water kept in the box.

Question 15. 1. Prove that the density of the mixture of two substances with densities ρ₁ and ρ₂ of equal mass will be \(\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}\)

2. Prove that if the two substance with densities ρ₁ and ρ₂ are mixed in equal volumes, then the density of the mixture thus formed will be \(\frac{1}{2}\left(\rho_1+\rho_2\right)\)
Answer:

1. Let the mass of each of the substances be m.

∴ Mass of the mixture = m + m = 2 m

Volume of the mixture = \(\left(\frac{m}{\rho_1}+\frac{m}{\rho_2}\right)\)

∴ Density of the mixture = \(\frac{2 m}{\frac{m}{\rho_1}+\frac{m}{\rho_2}}=\frac{2 \rho_1 \rho_2}{\rho_1+\rho_2}\)

2. Let the volume of each of the substances be V.

∴ Total mass of the mixture = (Vρ₁ + Vρ₂)

Total volume of the mixture = (V+ V) = 2 V

∴ Density of the mature = \(\frac{V \rho_1+V \rho_2}{2 V}=\frac{1}{2}\left(\rho_1+\rho_2\right)\)

Question 16. State with reason whether Pascal’s law is applicable to the water in a pond.
Answer:

Pascal’s law is not applicable to the water in a pond. The law is applicable only in the case of confined fluids. As a pond’s water is not confined, the law cannot be applied here.

Question 17. A piece of iron (specific gravity = 7.8) sinks in water, but floats on mercury (specific gravity = 13.6)— explain why.
Answer:

Given

A piece of iron (specific gravity = 7.8) sinks in water, but floats on mercury (specific gravity = 13.6)

The density of iron is 7.8 g · cm^-3 and the density of water is 1 g · cm^-3. The density of iron is greater than that of water and hence the weight of a piece of iron is greater than the upthrust of water exerted on it.

• As a result, the piece of iron sinks in water due to the action of the resultant downward force. But the density of mercury is 13.6 g • cm^-3 which is greater than the density of iron.
• Hence, the weight of the piece of iron is less than the upthrust exerted on it by mercury. As a result, the piece of iron floats on mercury due to the action of the resultant upward force.

Question 18. Why is it easier to swim in sea water than in a river?
Answer:

Due to the presence of salt dissolved in sea water, its density is more than that of river water. Due to the greater density of sea water it exerts a stronger buoyant force on a swimmer than what is exerted by river water. Hence, it is easier to swim in sea water than in a river.

Question 19. Explain whether Archimedes’ principle is applicable in the case of a freely falling body under gravity.
Answer:

A freely falling body has no weight. Hence, the weight of a freely falling body and the weight of the liquid displaced by it—both will be zero so the body will not experience any buoyant force. It means that the application of Archimedes’ principle in this case is unnecessary.

Question 20. An empty balloon (or a soft plastic bag) weighs the same as what it does when filled with air at atmospheric pressure—explain why.
Answer:

An empty balloon (or a soft plastic bag) weighs the same as what it does when filled with air at atmospheric pressure

An empty balloon or a soft plastic bag displaces a very small quantity of air. The balloon gets inflated when filled with air at atmospheric pressure and displaces a larger amount of air. More the amount of air inserted into the balloon, the larger the volume of air displaced by it.

1. Hence, the increase in weight of the inflated balloon due to the inclusion of air into it becomes just equal to the apparent decrease in the weight of the balloon due to the displaced air.
2. For this reason, an empty balloon (or a plastic bag) weighs the same as what it does when filled with air at atmospheric pressure.

Question 21. A wooden block is floating on water in a closed vessel. What will happen

1. if the air above the water is compressed,
2. if all the air above the water is removed from the closed vessel?

Answer:

1. If the air above the water in the closed vessel is compressed, the block will float upwards some more.

Here, weight of the wooden block = weight of displaced water + weight of displaced air.

If the air is compressed, then its density increases, and hence the weight of the displaced air increases while the weight of the displaced water decreases. Hence, the block will float upwards some more.

2. If all the air above the water is removed from the closed vessel, then the wooden block sinks more. This happens because when the air is removed, the buoyancy due to air is absent. Consequently, the apparent weight of the body increases, and the block sinks some more.

Question 22. Explain why it is sometimes safer for a ship while floating on water to load more goods than to get rid of them.
Answer:

For stable equilibrium of a floating body, its centre of gravity should lie below its metacentre. So, if the metacentre lies below the centre of gravity of a floating ship, then, to lower the position of its centre of gravity below the metacenter, more goods should be loaded in the hold of the ship.

In that case, the ship will float under stable equilibrium and hence it is safer for a ship floating on water to be loaded with more goods.

Question 23. A piece of ice is floating on a liquid of density 1.5 g · cm^-3 kept in a beaker. Will there be any change in the level of the liquid in the beaker as the piece of ice melts completely?
Answer:

Given

A piece of ice is floating on a liquid of density 1.5 g · cm^-3 kept in a beaker.

If the piece of ice melts completely, then the level of the liquid in the beaker will rise.

Let the mass of the piece of ice be m. So, the mass of the liquid displaced is also m.

∴ Volume of displaced liquid, \(V_1=\frac{m}{\rho_l}\left[\rho_l=\text { density of the liquid }\right]\)

As the piece of ice melts completely, the mass of water formed = m.

But the volume of that water, \(V_2=\frac{m}{\rho_w}=m\left[\rho_w=\text { density of water }\right]\)

According to question, \(\rho_l>\rho_w \text {, so } V_2>V_1 \text {. }\)

Hence, the level of liquid in the beaker will rise. If the liquid is immiscible in water, then water will float above the liquid.

Question 24. A wooden block is floating on water at 0°C keeping a portion V of its volume outside water. If the temperature of the water is increased from 0°C to 20°C, then what change of V will be observed?
Answer:

Given

A wooden block is floating on water at 0°C keeping a portion V of its volume outside water. If the temperature of the water is increased from 0°C to 20°C,

The density of water increases with the increase in temperature from 0°C to 4°C. With the increase in density of water, the upthrust on the block will increase. As a result, the block will gradually move up in water, i.e., the value of V will increase.

During the rise in temperature from 4°C to 20°C, the density of water decreases and hence the upthrust on the block will decrease. As a result, the wooden block will gradually sink in water, i.e., the value of V will decrease.

Question 25. A piece of ice with a cork in it is floating on water in a beaker. Will the level of water in the beaker change after complete melting of the ice?
Answer:

Given

A piece of ice with a cork in it is floating on water in a beaker.

The level of water in the beaker will not change.

Let the density of water be ρ, the weight of the piece of ice be W and the weight of the cork inside the piece of ice be w.

Total weight = (W+ w) = weight of displaced water

∴ Volume of displaced water  = \(\frac{W+w}{\rho g}\)

Suppose the ice melts completely. In this case, the weight of water thus formed = W and the volume of that amount of water = \(\frac{W}{\rho g}\).

In this situation, only the cork will float on water. Weight of water displaced by the cork = weight of the cork = w.

∴ Volume of water displaced by the cork = \(\frac{W}{\rho g}\)

Total volume of water formed due to melting of ice and also due to the water displaced by the cork = \(\frac{W + W}{\rho g}\)

Therefore, even after the ice melts completely, the level of water in the beaker does not change.

Question 26. A piece of ice with a stone in it is floating on water in a beaker. Will the level of water in the beaker change after complete melting of the ice?
Answer:

Given

A piece of ice with a stone in it is floating on water in a beaker.

If ice melts completely, the level of water in the beaker goes down.

Let the density of water be ρ, the weight of ice be W and the weight of stone be w.

∴ Weight of displaced water = weight of ice + weight of stone = (W+ w)

∴ Volume of displaced water = V = \(\frac{W+w}{\rho g}\)

Let the density of the stone be ρ_s.

Suppose the ice melts completely. In this case, mass of water thus formed due to melting of ice = W and volume of this water = \(\frac{W}{\rho g}\)

In this situation, the piece of stone will sink into water. Volume of water displaced by the stone = \(\frac{w}{\rho_s g}\)

∴ Total volume of water formed due to melting of ice and also due to displacement by stone

= \(V_2=\frac{W}{\rho g}+\frac{w}{\rho_s g}\)

∴ \(\rho_s>1 \quad therefore V_2<V_1\)

So, when all the ice melts, the level of water in the beaker goes down.

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Question 27. A man carries a bucket containing water in one hand 1 and a live fish in the other. If he releases the fish into the bucket of water, will he carry less weight?
Answer:

A man carries a bucket containing water in one hand 1 and a live fish in the other. If he releases the fish into the bucket of water,

The man will carry the same weight when he releases the live fish into the bucket of water. When the fish is released into the water of the bucket, the displaced water exerts an upthrust and hence the apparent weight of the fish will decrease.

• According to Newton’s third law of motion, an equal reaction force will act downwards at the bottom of the bucket. So, the decrease in the weight of the fish becomes equal to the increase in the thrust at the bottom of the bucket.
• As a result, the net weight of the bucket will remain the same and hence the man will carry the same weight as before.

Question 28. A bird is sitting at the bottom of a cage and, in this condition, the cage is weighed with the help of a spring balance. Now, the bird starts flying inside the cage. Will the reading of the balance change?
Answer:

A bird is sitting at the bottom of a cage and, in this condition, the cage is weighed with the help of a spring balance. Now, the bird starts flying inside the cage.

The reading of the spring balance depends on the design of the bottom of the cage. Suppose the bottom of the cage is totally covered with a metal sheet. As soon as the bird starts flying, it flaps its wings.

• Hence, due to an extra thrust developed in each flapping acting on the bottom of the cage, the reading of the spring balance fluctuates.
• But, if the bottom of the cage is covered with a net, then the reading of the spring balance will decrease. Because the spring balance will then give the reading of the weight of the cage only.

Question 29. Two balloons of the same volume are filled with two gases at the same pressure, one with hydrogen and the other with helium. Which of the two experiences a greater upward force?
Answer:

Two balloons of the same volume are filled with two gases at the same pressure, one with hydrogen and the other with helium.

The upward force for the hydrogen-filled balloon will be greater.

Here, resultant upward force = upthrust – the weight of the gas-filled balloon.

• Since the volumes of both balloons are the same, the total weight of the air displaced by them is also the same. So the upthrust exerted by air on both of lime is equal.
• But, under the same conditions of temperature and pressure, the density of helium is more than that of hydrogen. So, due to the lower weight of the hydrogen-filled balloon, the upward force on it will be greater.

Question 30. An egg sinks in freshwater, but it floats when a suitable quantity of salt is mixed in that water. Why?
Answer:

An egg sinks in freshwater, but it floats when a suitable quantity of salt is mixed in that water.

An egg is heavier than the weight of an equal vol¬ume of water and hence it sinks in water. But when a suitable quantity of salt is mixed in water, the density of this saline water becomes greater than the density of the egg and so it floats.

Question 31. A solid sphere and a hollow sphere having the same mass and external radius are immersed in the same liquid. Which one will feel heavier?
Answer:

A solid sphere and a hollow sphere having the same mass and external radius are immersed in the same liquid.

Since the radii of both spheres are the same, their external volumes are equal. So, both of them will displace an equal weight of the liquid.

According to Archimedes’ principle, the apparent loss of weight of a body in a liquid = weight of liquid displaced by the body. So, loss of weight suffered by both spheres will be the same and, hence, their apparent weight of them will be the same.

Question 32. A glass of water is placed on one of the pans of a balance. On the other pan, another similar glass of water is placed with a piece of wood floating on it Water is at the same level in both the glasses. Which of the two glasses is heavier? Explain your answer.
Answer:

A glass of water is placed on one of the pans of a balance. On the other pan, another similar glass of water is placed with a piece of wood floating on it Water is at the same level in both the glasses.

The masses of both the glasses will be the same.The amount of water contained in the second glass is less than that in the first glass because the submerged part of the floating piece of wood has displaced an equal volume of water.

But we know that the mass of this displaced water is the same as the mass of the floating piece of wood and hence the masses of the two glasses will be the same.

Question 33. A flat disc, a solid cube, and a solid sphere of equal mass made of the same material are completely Immersed in water. Which one of them will experience the minimum and which one will experience the maximum buoyant force?
Answer:

A flat disc, a solid cube, and a solid sphere of equal mass made of the same material are completely Immersed in water.

When a body is immersed completely in water, then the buoyant force becomes equal to the weight of the water displaced by the body. Here, the masses of the three bodies are equal and they are made of the same material (i.e., of the same density) hence their volumes are equal. Hence, in each case, the weight of the displaced water becomes equal and the same buoyant force is experienced.

Question 34. A hollow glass sphere is balanced by counterpoising weights made of brass in a common balance. The whole system is then covered with an air-tight bell jar and the jar is evacuated. What will be the result?
Answer:

A hollow glass sphere is balanced by counterpoising weights made of brass in a common balance. The whole system is then covered with an air-tight bell jar and the jar is evacuated.

Since the density of glass is less than the density of brass, the volume of the glass sphere is greater than the vol¬ume of the same mass of brass weights. So, in air, the glass sphere experiences a larger buoyant force than the brass weights.

Now, when the bell jar is evacuated, this buoyant force ceases to act and hence the weight of the glass sphere becomes more than that of the brass weights. Hence, the end of the balance beam containing the glass sphere will move downwards.

Question 35. A common balance has a beaker of water and a piece of stone in one of the pans and is balanced. Now, the stone is dipped into the beaker of water. Is any change in reading of the balance observed?
Answer:

A common balance has a beaker of water and a piece of stone in one of the pans and is balanced. Now, the stone is dipped into the beaker of water.

When the piece of stone is immersed in the water of the beaker, the balance beam remains in its equilibrium condition. Due to the upthrust exerted by water, the piece of stone loses a part of its weight.

According to Newton’s third law of motion, the reaction of the upthrust of water will act vertically downwards on the pan of the common balance. As the apparent loss in weight of the piece of stone becomes equal to the downward reaction force on the balance pan, the net weight remains the same and hence the balance beam remains in its equilibrium condition.

Question 36. What will happen if the upper part of a ship Is made heavier than its lower part?
Answer:

If the upper part of a ship is made heavier than its lower part, then the centre of gravity of the ship lies above the centre of gravity of the water displaced by the ship. As a result, if the ship is tilted a little, then, due to the couple that develops due to the weight of the ship and the buoyant force, the ship tilts more, and hence it may turn over.

Question 37. Determine the change in potential energy of a body when it is raised through a height h, inside water. Volume of the body is V, its density is ρ and the density of water is ρ₀. Give your answer for the situations when

1. ρ > ρ₀ and
2. ρ < ρ₀.

Solution:

Two forces act on a body when it is immersed in water. One is its weight (acting downwards) and another is the buoyant force (acting upwards). Force due to weight = Vρg, and upthrust due to buoyancy = Vρ₀g

1. If ρ > ρ₀, the weight of the body becomes greater than the upthrust.

Hence, resultant downward force = \(V \rho g-V \rho_0 g=V g\left(\rho-\rho_0\right) .\)

If the body is raised to a height h against this force, then work done = \(V g\left(\rho-\rho_0\right) h\) = increase in potential energy.

2. When ρ > ρ₀, the upthrust becomes greater than the weight. Then resultant upward force = \(V g\left(\rho-\rho_0\right)\). To raise the body through a height h, work done = \(V g\left(\rho-\rho_0\right) h\) = decrease in potential energy.

 

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Useful Relations For Solving Numerical Problems

If the mass of volume V of a substance is M, then the density of that substance, D = M/V.

• Specific gravity, S = \(\frac{\text { density of the substance }}{\text { denisity of water at } 4^{\circ} \mathrm{C}}\)
• In the CGS system, the numerical value of the density of a substance = its specific gravity.
• In SI, the numerical value of the density of a substance = 1000 x its specific gravity.
• Pressure (p) = \(\frac{\text { normal applied force }(F)}{{area}(A)} \text {. }\)
• Thrust exerted by a liquid (F) = pressure of the liquid (p) x area (A).

1. The pressure of a liquid of density p at a depth h, or gauge pressure, p = hρg.
2. The pressure at a point in a liquid at a depth h, when the liquid is exposed to free air, or absolute pressure, p = B + hρg

[where B = atmospheric pressure].

The lateral thrust exerted on a rectangular lamina dipped vertically in a liquid

= average lateral pressure x area = \(\left(h+\frac{1}{2} b\right) \rho g \times a b\)

[where a = length of the lamina, b = breadth of the lamina, h = depth of the upper edge of the lamina from the free surface of the liquid].

In a hydraulic press, if the cross-sectional area of the smaller piston is a, the cross-sectional area of the larger piston is b, and the force applied on the smaller piston  is F₁, then the thrust developed on the larger piston, \(F_2=F_1 \times \frac{b}{a}\)

Apparent weight of a body immersed in a liquid = real weight of the body – buoyant force.

Apparent loss in weight of a body immersed in a liquid = weight of liquid displaced by the body = weight of an equal volume of liquid = buoyant force.

In the case of floatation of a body partly immersed in a liquid,

⇒ \(\frac{\text { volume immersed part of the body in the liquid }}{\text { total volume of the body }}\)

= \(\frac{\text { density of the body }}{\text { density of the liquid }}\)

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, statement 2 is true.

Question 1.

Statement 1: If a barometer is accelerated upwards, the level of mercury in the tube of the barometer will decrease.

Statement 2: The effective value of g will increase, so upthrust will increase.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 2.

Statement 1: A hydrogen-filled balloon stops rising after it has attained a certain height in the sky.

Statement 2: The atmospheric pressure decreases with height and becomes zero when maximum height is attained.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 3.

Statement 1: A solid sphere and a hollow sphere ot same material are floating in a liquid. Radius of both spheres are same. The percentage of volume immersed in both the spheres will be same.

Statement 2: Upthrust acts on volume of liquid displaced. It has nothing to do with whether the body is solid or hollow.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 4.

Statement 1: For a floating body to be in stable equilibrium, its centre of buoyancy must be located above the centre of gravity.

Statement 2: The torque developed by the weight of the body and the upthrust will restore the body back to its normal position, after the body is disturbed.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 5.

Statement 1: The blood pressure in humans is greater at the feet than that at the brain.

Statement 2: Pressure of liquid at any point is proportional to height, density of liquid, and acceleration due to gravity.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 6.

Statement 1: A cylinder fitted with a movable piston contains a certain amount of liquid in equilibrium with its vapour. The temperature of the system is kept constant with the help of a thermostat. When the volume of the vapour is decreased by moving the piston inwards, the vapour pressure does not increase.

Statement 2: Vapour in equilibrium with its liquid, at a constant temperature, does not obey Boyle’s law.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 7.

Statement 1: A body floats in a liquid with a fraction n of its volume above the surface of the liquid. If the system is taken to a planet where the acceleration due to gravity is greater than that on earth, the fraction n will decrease.

Statement 2: For flotation, the weight of the body is equal to the weight of the liquid displaced.

Answer: 4. Statement 1 is false, statement 2 is true.

 

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Very Short Answer Type Questions

Question 1. On what other factor does the pressure at a point in a liquid depend besides the depth of that point and the acceleration due to gravity at that place?
Answer: Density of liquid

Question 2. ‘The pressure at all points on the same horizontal plane inside a liquid are equal’—is the statement true or false?
Answer: True

Question 3. If two immiscible liquids in a U-tube are in equilibrium, then how are the heights of the two liquids from the surface of separation-related with the densities of the two liquids?
Answer: Inversely proportional

Question 4. Besides the volume of the immersed part of a body in a liquid and acceleration due to gravity, on what factor does the buoyancy acting on a body depend?
Answer: Density of the liquid

Question 5. For a body floating partly immersed in a liquid, state whether the density of the body will be greater or less than the density of the liquid. 
Answer: Less

Question 6. As a liquid can flow even under the influence of a minute tangential force, it is called a _________
Answer: Fluid

Question 7. ‘A liquid does not undergo change in density due to change in pressure.’ State whether the statement is true or false.
Answer: True

Question 8. Density of a substance = specific gravity of a substance x density of ________
Answer: Water at 4°C

Question 9. In all systems of units, the value of specific gravity is _____
Answer: Equal

Question 10. ‘The free surface of a liquid at rest is always horizontal’—Correct or incorrect?
Answer: Correct

Question 11. The density of the material of a body in its weightless
condition remains ______.
Answer: Unaltered

Question 12. ‘If the density of a body is more than the density of a liquid, then the body sinks in that liquid.’—State whether the statement is true or false.
Answer: True

Question 13. The weight of a body in air is 100 g and its weight in water is 40 g. Find the volume of the body.
Answer: 60 cm^-3

Question 14. Which quantity represents the normal force applied on unit area of a surface?
Answer: Pressure

Question 15. Is pressure a vector quantity?
Answer: No

Question 16. Write down the variation of the pressure at a point in a liquid to its density.
Answer: Directly proportional

Question 17. ‘During the construction of a dam, the base of the dam wall is usually made wider.’—Correct or incorrect?
Answer: Correct

Question 18. Due to the upthrust of a liquid, a body immersed in the liquid suffers an apparent ________ in weight.
Answer: Loss

Question 19. Standard atmospheric pressure is equal to pressure exerted by __________ cm of mercury column.
Answer: 76

Question 20. What is the height of homogeneous column of air?
Answer: 8 km (approx.)

Question 21. Does a hydraulic press work if the liquid in it is replaced by a gas?
Answer: Yes

Question 22. What is the name of the upward force exerted by displaced liquid or gas on a body when it is partly or totally immersed in it?
Answer: Buoyant force

Question 23. Does buoyancy depend on the depth up to which a body is completely immersed in a liquid?
Answer: No

Question 24. What fraction of floating ice remains below the surface of water?
Answer: 10/11 part

Question 25. In which direction does the buoyant force act?
Answer: In the direction opposite to the weight of the body

Question 26. What is the dimension of buoyancy?
Answer: [MLT^-2]

Question 27. Name the principle which is effective in a hydraulic press.
Answer: Multiplication of thrust

Question 28. Thrust = pressure x __________
Answer: Area

Question 29. Pascal’s law obeys the principle of conservation of _________
Answer: Energy

Question 30. In the tilted position of a floating body, the point at which the vertical line drawn through the centre of buoyancy cuts the central line, name the point of the body.
Answer: Metacentre

Question 31. Due to the upthrust of a liquid, the weight of a body immersed in that liquid decreases or increase?
Answer: Decreases

Question 32. 1 torr = ____________ dyn • cm^-2
Answer: 1332.8

Question 33. Is Archimedes’ principle applicable in the case of a freely falling body?
Answer: No

Question 34. Is Archimedes’ principle applicable inside an artificial satellite revolving around the Earth?
Answer: No

Question 35. A boat floating in a pond is carrying a number of stones. If the stones are dropped into the pond, will the water level rise or fall?
Answer: Fail

Question 36. Does Archimedes’ principle hold good in the case of a gas?
Answer: Yes

Question 37. State whether a lump of iron floats or sinks in mercury.
Answer: Floats

Question 38. Where does the centre of buoyancy lie in a displaced liquid.
Answer: At the centre of gravity

Question 39. If the weight of a body is greater than the weight of the liquid displaced by it, then the body _______ in that liq-uid.
Answer: Sinks

Question 40. What is the apparent weight of a floating body?
Answer: Zero

Question 41. For a floating body in unstable equilibrium, the meta-centre lies __________ the centre of gravity of the body.
Answer: Below

Question 42. When a ship enters a river from the sea. What will you find?
Answer: It sinks more

Question 43. A wooden block is floating on water in a closed vessel. If the air in the vessel is compressed, then the block will ________ more. [Fill in the blank]
Answer: Float up

Question 44. State whether the weight of a body measured in air will be more or less than its weight in vacuum.
Answer: Less

Question 45. For the stable equilibrium of a body, state whether its metacentre should lie above or below its centre of gravity.
Answer: Above

Question 46. A body of mass m is falling freely under gravity. What will be its weight in this condition?
Answer: Zero

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Match Column 1 With Column 2

Question 1. There are two points A and B inside a liquid as shown in Fig. Now the vessel starts moving upwards with an acceleration a.

Answer: 1. A, 2. A, 3. A

Question 2. A cube is floating in a liquid as shown in Fig.

Answer: 1. B, 2. C. 3. C

Question 3. A tube is inverted in a mercury vessel as shown in Fig. If the pressure p is increased.

Answer: 1. 2, 2. C, 3. C

Question 4.

Answer: 1. C, 2. B, 3. A

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Comprehension Type Question And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A small spherical ball of radius r is released from its completely submerged position in a liquid whose density varies with height h as \(\rho_L=\rho_0\left[4-\frac{3 h}{h_0}\right]\). The density of the ball is \(\frac{5}{2} \rho_0\). The height of the vessel is \(h_0=\frac{12}{\pi^2}\). Consider r <<h₀ and g = 10 m • s^-2

1. Where will the ball be in the equilibrium condition?

1. at a depth \(\frac{h_0}{2}\) from top
2. at the bottom of the vessel
3. at a depth \(\frac{3h_0}{4}\) from top
4. The ball will never be at equilibrium

Answer: 1. at a depth \(\frac{h_0}{2}\) from top

2. The motion of the ball in the vessel is

1. Oscillatory but not SHM
2. SHM with time period 2s
3. SHM with time period 1 s
4. The motion is not oscillatory

Answer: 2. SHM with time period 2s

3. Time taken by the ball to reach the bottom of the vessel is

1. 1s
2. 2s
3. 0.5s
4. The ball will not reach the bottom of the vessel.

Answer: 1. 1s

Question 2. A block of mass 1 kg and density 0.8 g · cm^-3 is held stationary with the help of a string as shown in Fig. The tank is accelerating vertically upwards with an acceleration a = 1.0 m · s^-2. Take g = 10 m • s^-2 and density of water = 10³ kg • m^-3.

1. What is the tension in the string?

1. 2.60 N
2. 2.85 N
3. 2.75 N
4. 3.10 N

Answer: 3. 2.75 N

2. If the string is now cut, find the acceleration of the block

1. 3.95 m • s^-2
2. 3.75 m • s^-2
3. 4.25 m • s^-2
4. 3.29 m • s^-2

Answer: 2. 3.75 m • s^-2

Hydrostatics Q&A for Students

Question 3. An engineering firm is assigned the job to design the cylindrical pressured water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 m • s^-2. The pressure at the surface of the wafer will be 130 kPa and depth of the water will be 14.2 m- The pressure of the air in the building outside the tank will be 93 kPa.

1. Find the net downward force on the tank’s flat bottom of area 2 m²

1. 179.4 kN
2. 365.4 kN
3. 105.36 kN
4. None

Answer: 1. 179.4 kN

2. What is the buoyant force on a wooden block of mass 2 kg and relative density 0.8, height 10 cm?

1. 5.94 N
2. 7.42 N
3. 1.48 N
4. None

Answer: 2. 7.42 N

Question 4. The vessel shown in Fig. has two sections of areas of cross-section A₁ and A₂. A liquid of density ρ fills both the sections, up to a height h in each. Neglect the atmospheric pressure.

1. What is the pressure at the base of the vessel?

1. 1/2 hρg
2. 2hρg
3. hρgA₂
4. None

Answer: 2. 2hρg

2. What is the force exerted by the liquid on the base of the vessel?

1. 2hρg A₁A₂
2. 1/2 hρg A₂
3. 2hρg A₂
4. hρg x (A₁+A₂)

Answer: 3. 2hpg A₂

3. Find the downward force exerted by the walls of the vessel at the level B.

1. 2hρg x (A₂-A₁)
2. 1/2hρg x (A₁ + A₂)
3. hρgA₂
4. hρg x (A₂-A₁)

Answer: 4. hρg x (A₂-A₁)

Question 5. A tank has a cylindrical hole H of diameter 2r at its bottom as shown in Fig. A cylindrical block B of diameter 4r and height h is placed on the hole H to prevent the flow of liquid through the hole. The liquid in the tank stands at a height h₁ above the top face of the block. The density of liquid is ρ and that of the block is \(\frac{\rho}{3} .\).

1. If the liquid is gradually taken out of the tank, the height h₁ of the liquid surface above the top face of the block for which the block just begins to rise is

1. 2h/4
2. 3h/4
3. 5h/3
4. 2h

Answer: 3. 5h/3

2. If the liquid level is further lowered so that it stands at a depth h₂ above the bottom face of the block as shown in the figure, then the maximum value of  h₂ so that the block does not move is

1. 4h/9
2. h/3
3. 2h/3
4. 5h/9

Answer: 1. 4h/9

3. If the liquid level is lowered below h₂, then

1. The block will never rise
2. The block will start rising if h₂ = h/3
3. The block will start rising if h₂ = h/4
4. The block will start rising if h₂ = h/5

Answer: 1. The block will never rise

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Integer Type Question And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. A candle of diameter d is floating on a liquid in a cylindrical container of diameter D(D >>d) as shown in Fig It is burning at the rate of 2 cm • h^-1. At what rate will the top of the candle fall?

Answer: 1

Question 2. A cylinder has radius 8 cm. Up to what height (in cm) should it be filled with water so that the thrust on its walls is equal to that on its bottom?
Answer: 8

Question 3. Mass of a balloon with its contents is 1.5 kg. It is descending with an acceleration equal to half of the acceleration due to gravity. If it is to go up with the same acceleration keeping the volume same, what amount of mass (in kg) should be decreased?
Answer: 1

Question 4. A ball whose density is 0.4 x 10³ kg · m^-3 falls into water from a height of 9 cm. To what depth does the ball sink?
Answer: 6

Question 5. A water tank is 20 m deep. If a water barometer reads 10 m at that place, then what is the pressure (in SI unit) at the bottom of the tank in the atmosphere?
Answer: 3

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Short Answer Type Questions

Question 1. The density of a body is d and that of air is ρ. If the body weighs w in air, what will be its actual weight?
Answer:

Given

The density of a body is d and that of air is ρ. If the body weighs w in air,

Let the actual weight of the body = w₀

∴ buoyant force of air, B = \(\frac{w_0}{d} \rho\)

So, the weight of the body in air,

⇒ \(w=w_0-B=w_0-\frac{w_0}{d} \rho=w_0\left(\frac{d-\rho}{d}\right)\)

∴ \(w_0=\frac{d}{d-\rho} w\)

Question 2. A drop of oil rises through water with an acceleration ag. If a is a constant quantity and g is the acceleration due to gravity, find the specific gravity of the oil. Neglect the friction of water.
Answer:

Given

A drop of oil rises through water with an acceleration ag. If a is a constant quantity and g is the acceleration due to gravity,

Let d and D be the density of water and oil respectively, and the mass of the oil drop is m.

∴ Volume of the oil drop = m/D

= volume of the displaced water by the oil drop or, the mass of the displaced water

= density of water x volume of the displaced water

= d x m/D

Now, buoyancy = weight of the displaced water = \(\frac{d m g}{D}\)

∴ Net upward force acting on the oil drop

= buoyant force – weight of the oil drop

= \(\frac{d m g}{D}-m g=\left(\frac{d}{D}-1\right) m g\)

∴ Acceleration of the oil drop inside water, \(\alpha g=\frac{\left(\frac{d}{D}-1\right) m g}{m}=\left(\frac{d}{D}-1\right) g\)

∴ Specific gravity of oil = \(/frac{D}{d}=\frac{1}{1+\alpha}\)

Question 3. A uniform rod is suspended horizontally from its mid-point. A piece of metal whose weight is W is suspended at a distance l from the mid-point. Another weight W₁ is suspended on the other side at a distance l₁ from the mid-point to bring the rod to a horizontal position. When W is completely immersed in water, W₁ needs to be kept at a distance l₂ from the mid-point to get the rod back into horizontal position. The specific gravity of the metal piece is

1. \(\frac{W}{W_1}\)
2. \(\frac{W l_1}{W l-W_1 l_2}\)
3. \(\frac{l_1}{l_1-l_2}\)
4. \(\frac{l_1}{l_2}\)

Answer:

Given

A uniform rod is suspended horizontally from its mid-point. A piece of metal whose weight is W is suspended at a distance l from the mid-point. Another weight W₁ is suspended on the other side at a distance l₁ from the mid-point to bring the rod to a horizontal position. When W is completely immersed in water, W₁ needs to be kept at a distance l₂ from the mid-point to get the rod back into horizontal position.

In equilibrium condition of rod in air, Wl = W₁l₁

In equilibrium condition of rod immersed in water, \(\left(W-F_B\right) l=W_1 l_2\)

Where, F_B = buoyant force on the body of weight W = \(\frac{W}{\rho}\)

So, \(\left(W-\frac{W}{\rho}\right) l=W_1 l_2=W_{\frac{l}{l}}^{l_1} l_2 \quad \text { or, } 1-\frac{1}{\rho}=\frac{l_2}{l_1}\)

∴ \(\rho=\frac{l_1}{l_1-l_2}\)

The option 3 is correct.

Question 4. A wooden block is floating on water kept in a beaker. 40% of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to g/2. The block will then

1. Sink
2. Float with 10% above the water surface
3. Float with 40% above the water surface
4. Float with 70% above the water surface

Answer:

Given

A wooden block is floating on water kept in a beaker. 40% of the block is above the water surface. Now the beaker is kept inside a lift that starts going upward with acceleration equal to g/2.

Weight of the body and the buoyancy of the liquid both depend on the effective acceleration. Hence there will be no change in the floating portion of the block.

The option 3 is correct.

Question 5. To determine the composition of a bimetallic alloy, a sample is first weighed in air and then in water. These weights are found to be w₁ and w₂ respectively. If the densities of the two constituent metals are p₁ and p₂ respectively, then the weight of the first metal in the sample is (where p_w is the density of water)

1. \(\frac{\rho_1}{\rho_{w^{(}}\left(\rho_2-\rho_1\right)}\left[w_1\left(\rho_2-\rho_w\right)-w_2 \rho_2\right]\)
2. \(\frac{\rho_1}{\rho_w\left(\rho_2+\rho_1\right)}\left[w_1\left(\rho_2-\rho_w\right)+w_2 \rho_2\right]\)
3. \(\frac{\rho_1}{\rho_w^{(}\left(\rho_2-\rho_1\right)}\left[w_1\left(\rho_2+\rho_w\right)-w_2 \rho_1\right]\)
4. \(\frac{\rho_1}{\rho_w\left(\rho_2-\rho_1\right)}\left[w_1\left(\rho_2-\rho_w\right)-w_2 \rho_1\right]\)

Answer:

⇒ \(w_1-w_2=V \rho_w g=\left(V_1+V_2\right) \rho_w g=\left[\frac{x}{\rho_1}+\frac{w_1-x}{\rho_2}\right] \rho_w\)

Hence, required weight,

x = \(\frac{\rho_1}{\rho_w\left(\rho_2-\rho_1\right)}\left[w_1\left(\rho_2-\rho_w\right)-w_2 \rho_2\right]\)

The option 1 is correct.

Question 6. A cylinder of height h is filled with water and is kept on a block of height h/2. The level of water in the cylinder is kept constant. Four holes numbered 1, 2, 3, and 4 are at the side of the cylinder and at heights 0, h/4 , h/2, and 3h/4 respectively. When all four holes are opened together, the hole from which water will reach farthest distance on the plane PQ is the hole number

1. 1
2. 2
3. 3
4. 4

Answer:

Net height of the block and the cylinder = \(\frac{h}{2}+h=\frac{3 h}{2}\)

If we consider a hole at a depth x from the top, height of that hole from the plane PQ, H = 3h/2 – x

Velocity of water flowing out of that hole horizontally, v = √2gx

Now, vertical velocity of water flowing out of that hole = 0

Hence, if the water takes t time to reach PQ plane,

H = \(\frac{1}{2} g t^2 \quad \text { or, } t=\sqrt{\frac{2 H}{g}}\)

For this time t, water will flow horizontally with the same velocity.

Hence, distance travelled on the PQ plane, \(D=v t=\sqrt{2 g x} \cdot \sqrt{\frac{2 H}{g}}=\sqrt{2 x \cdot 2 H}=\sqrt{2 x(3 h-2 x)}\)

For hole no. \(1, x=h ; D_1=\sqrt{2 h(3 h-3 h)}=h \sqrt{2}\)

For hole no. \(2, x=\frac{3 h}{4} ; D_2=\sqrt{\frac{3 h}{2}\left(3 h-\frac{3 h}{2}\right)}=h \cdot \frac{3}{2}\)

For hole no. \(3, x=\frac{h}{2}, D_3=\sqrt{h(3 h-h)}=h \sqrt{2}\)

For hole no. \(4, x=\frac{h}{4} ; D_4=\sqrt{\frac{5 h}{2}\left(3 h-\frac{5 h}{2}\right)}=h \sqrt{\frac{5}{4}}\)

∴ \(D_2>D_1=D_3>D_4\)

The option 2 is correct.

Question 7. There is a circular tube in a vertical plane. Two liquids that do not mix and of densities d₁ and d₂ are Filled in the tube. Each liquid subtends 90° angle at centre. Radius joining their interface makes an angle a vertical. Ratio d₁/d₂ is

1. \(\frac{1+\sin \alpha}{1-\sin \alpha}\)
2. \(\frac{1+\cos \alpha}{1-\cos \alpha}\)
3. \(\frac{1+\tan \alpha}{1-\tan \alpha}\)
4. \(\frac{1+\sin \alpha}{1-\cos \alpha}\)

Answer:

Equating pressure at A, (Rcosa + Rsina)d₂g (Rcosa – Rsina)d₁g

⇒ \(\frac{d_1}{d_2}=\frac{\cos \alpha+\sin \alpha}{\cos \alpha-\sin \alpha}=\frac{1+\tan \alpha}{1-\tan \alpha}\)

The option 3 is correct.

Question 8. An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

1. 16cm
2. 22 cm
3. 38 cm
4. 6 cm

Solution:

⇒ \(p+x=p_0\)

or, \(p=p_0-x\)

or, \(8 \times A \times 76=(76-x) \times A \times(54-x)\)

∴ x=38

So, length of air column =54-38 = 16 cm.

The option 1 is correct.

Question 9. Two non-mixing liquids of densities p and np(n > 1) are put in a container. The height of each liquid is h. A solid cylinder of length L and density d is put in this container. The cylinder floats with its axis vertical and length pL(p < 1) in the denser liquid. The density d is equal to

1. {2 + (n+ 1)p}ρ
2. {2 + (n- 1)p}ρ
3. {1 +(n-1)p}ρ
4. {1 + (n+1)p}ρ

Answer:

Let, the area of the cross-secton of the cylinder be A.

Weight of the cylinder = LAgd

The buoyant force on the cylinder due to the liquid of higher density = pLAnρg

and buoyant force on the cylinder due to the liquid of lower density = (1- p)LAρg

In equilibrium,

LAgd = (1 – p)LAρg+ pLAnρg

or, d = (1 -ρ)ρ + pnρ = ρ-pρ + pnρ

∴ d = [1 +(n- l)p]ρ

The option 3 is correct.

Hydrostatics Density

Bodies of equal volume may have different weights. Among pieces of iron, lead and wood of equal volume, the piece of lead is heavier than the piece of iron, and the piece of wood is the lightest.

It leads to the concept of density. A body with greater density weighs more, and a body with lower density weighs less — if both the bodies are of the same volume.

Density Definition: Density is the mass per unit volume of a substance.

If a volume V of a substance has a mass M, then the density of the substance, D = \(\frac{M}{V}\). Often, the greek letter ρ(rho) is used as the symbol of density.

If a body is not homogeneous, the average density of the body can be found out by dividing the total mass of the body by its total volume.

Units of density:

• g • cm^-3
• kg • m^-3

Relation: 1 kg • m^-3 = \(\frac{1 \mathrm{~kg}}{1 \mathrm{~m}^3}=\frac{10^3 \mathrm{~g}}{10^6 \mathrm{~cm}^3}=10^{-3} \mathrm{~g} \cdot \mathrm{cm}^{-3}\)

∴ 1g • cm^-3 = 1000kg • m^-3

Density and Specific Gravity in Everyday Life

Effect of pressure and temperature on density: The density of a substance depends on pressure and temperature. With change in pressure, the volume of a solid or a liquid does not change appreciably.

• So, to express the density of a solid or a liquid, the pressure need not be mentioned. But with change in pressure, the volume of a gaseous substance changes appreciably, and the density also changes accordingly.
• With change in temperature, the volume, and hence the density, of a substance changes. So, to express the density of a gaseous substance, both the temperature and the pressure of the substance have to be mentioned. Ib express the density of a solid or a liquid, the temperature needs to be mentioned to get sufficient accuracy.
• For example, at normal atmospheric pressure, air at 0°C has a density of 1.29 kg · m^-3  but at 10°C, the density of air is 1.25 kg · m^-3 which is slightly less.

Dimension of density: [D] = \(\frac{M}{L^3}=M L^{-3}\)

Density vs Specific Gravity Explained

Hydrostatics Specific Gravity

In many cases, the density of a substance is expressed with respect to the density of another substance. The density of a substance expressed in this way explains the concept of specific gravity. For the specific gravity of a liquid or a solid, the density of water at 4°C is taken as the standard. But, in the case of a gaseous substance, the standard is the density of hydrogen gas at STP.

Specific Gravity Definition: The ratio of the density of a solid or a liquid to that of water at 4°C is called the specific gravity of that substance.

Alternatively, the specific gravity of a solid or a liquid is defined as the ratio between the mass of a certain volume of the substance to the mass of an equal volume of water at 4°C.

If S denotes the specific gravity of a solid or a liquid, then according to the second definition,

S = \(\frac{\text { mass of volume } V \text { of the substance }}{\text { mass of volume } V \text { of water at } 4^{\circ} \mathrm{C}}\)

= \(\frac{\text { mass of unit volume of the substance }}{\text { mass of unit volume of water at } 4^{\circ} \mathrm{C}}\)

= \(\frac{\text { density of the substance }}{\text { density of water at } 4^{\circ} \mathrm{C}}\)

This is the earlier definition of specific gravity.

• Since the specific gravity of a substance is the ratio of two densities, it is also called relative density. Specific gravity  (relative density) tells us how dense a substance is in comparison to water at 4°C.
• For example, when we say that the relative density of brass is 8.4, it means a piece of brass of any volume has mass 8.4 times that of an equal volume of water at 4°C.
• When specific gravity of a substance is greater than one, it is heavier than water. Hence it will sink in water. When specific gravity is less than one, it is lighter than water. Hence it will float in water.
• Specific gravity is only a number and has no unit. For this reason, the value of the specific gravity of a substance is equal in all systems of units. It is a dimensionless physical quantity.

Comparison of density and specific gravity in terms of magnitude: From the definition of specific gravity, we can write,

Density of a substance = specific gravity of the substance x density of water at 4°C

In the CGS system, the density of water at 4°C is 1 g • cm^-3 and hence in this system, density of a substance = specific gravity of the substance x 1 g • cm^-3. So, in the CGS system, the density of a substance and its specific gravity are numerically equal.

In SI, the density of water at 4°C is 1000 kg • m^-3. So, in this system, density of a substance = specific gravity of the substance x 1000 kg • m^-3. Therefore, in SI, the density of a substance is numerically 1000 times that of the specific gravity of the substance.

Applications of Density and Specific Gravity

Temperature correction of specific gravity: in determining the specific gravity of a solid or a liquid, we usually use water at room temperature. But, actually water at 4°C should be taken as standard because the density of water at 4°C only is 1 g • cm^-3.

Since the density of water changes with temperature, a temperature correction of specific gravity should be considered for a better result.

Let the temperature of the water used be r°C. If the specific gravity of the substance is S, then,

S = \(\frac{\text { mass of volume } V \text { of the substance }}{\text { mass of volume } V \text { of water at } 4^{\circ} \mathrm{C}}\)

= \(\frac{\text { mass of volume } V \text { of the substance }}{\text { mass of volume } V \text { of water at } t^{\circ} \mathrm{C}}\)

x \(\frac{\text { mass of volume } V \text { of water at } t^{\circ} \mathrm{C}}{\text { mass of volume } V \text { of water at } 4^{\circ} \mathrm{C}}\)

This correction may be avoided where a fine measurement is not strictly necessary. This is because the change in the density of water with temperature is quite small.

Hydrostatics Differences Between Density And Specific Density

Differences Between Density And Specific Density

Differences Between Density And Specific Density Numerical Examples

Example 1. By mixing 210 g of salt in 1 L of water, 1.05 L of solution is produced. Determine the density of that solution.
Solution:

Given

By mixing 210 g of salt in 1 L of water, 1.05 L of solution is produced.

Mass of the solution = mass of water + mass of salt

= (1000 + 210) g [Mass of 1 L of water = 1000 g] = 1210 g

Volume of the solution = 1.05 L = 1050 cm³.

∴ Density of the solution = \(\frac{\text { mass of the solution }}{\text { volume of the solution }}\)

= \(\frac{1210}{1050}=1.152 \mathrm{~g} \cdot \mathrm{cm}^{-3}\)

Dimensionless Nature of Specific Gravity

Example 2. How much of a liquid of density 1.3 g • cm^-3 can be stored in a container of 5 kg of kerosene oil? The density of kerosene = 0.8 g • cm^-3. Give your answer in kg.
Solution:

Volume of kerosene oil = \(\frac{5000}{0.8}\) = 6250 cm³

∴ The container can hold 6250 cm³ of any liquid.

∴ Mass of the given liquid that can be stored in the container =6250 x 1.3 = 8125 g = 8.125 kg.

Example 3. A liquid of mass my and density ρ₁ is mixed with another liquid of mass m₂ and density ρ₂. If the volume of the mixture does not change, then what will be the density of the mixture?
Solution:

Given

A liquid of mass my and density ρ₁ is mixed with another liquid of mass m₂ and density ρ₂. If the volume of the mixture does not change

Mass of the mixture, m = m₁ + m₂

Volume of the first liquid, \(v_1=\frac{m_1}{\rho_1}\) and volume of the second liquid, \(v_2=\frac{m_2}{\rho_2}\)

∴ Volume of the mixture,

ν = \(v_1+v_2=\frac{m_1}{\rho_1}+\frac{m_2}{\rho_2}=\frac{m_1 \rho_2+m_2 \rho_1}{\rho_1 \rho_2}\)

∴ Density of the mixture, \(\frac{m}{v}=\frac{\left(m_1+m_2\right) \rho_1 \rho_2}{m_1 \rho_2+m_2 \rho_1}\).

Examples of Density and Specific Gravity Calculations

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Example 4. The specific gravity of a mixture of two substances of equal volumes is 4. The specific gravity of the mixture becomes 3 when these two substances are mixed in equal masses. Calculate the specific gravity of the substances.
Solution:

Given

The specific gravity of a mixture of two substances of equal volumes is 4. The specific gravity of the mixture becomes 3 when these two substances are mixed in equal masses.

Let the specific gravities of the two substances be S₁ and S₂ and the density of water at 4°C be ρ.

∴ Densities of the substances are S₁ρ and S₂ρ.

Let V = volume of each of the substances in the mixture.

∴ Total mass of the mixture = \(\left(V S_1 \rho+V S_2 \rho\right)\)

Total volume of the mixture = V+ V = 2 V

∴ Density of the mixture = \(\frac{V S_1 \rho+V S_2 \rho}{2 V}\)=\(\frac{\rho\left(S_1+S_2\right)}{2}\)

Specific gravity = \(\frac{S_1+S_2}{2}\) (according to the problem)

or, S₁ + S₂ = 8 ……….(1)

Again, let m = mass of each of the two substances in the mixture.

∴ Total mass of the mixture = m + m = 2 m

Total volume of the mixture = \(\left(\frac{m}{S_1 \rho}+\frac{m}{S_2 \rho}\right)\)

∴ Density of the mixture = \(\frac{2 m}{\frac{m}{S_1 \rho}+\frac{m}{S_2 \rho}}=\frac{2 S_1 S_2 \rho}{S_1+S_2}\)

∴ Specific gravity of the mixture = \(\frac{2 S_1 S_2}{S_1+S_2}=3\) (according to the problem)

or, \(S_1 S_2=\frac{3}{2}\left(S_1+S_2\right)=\frac{3}{2} \times 8=12\) ………(1)

∴ \(S_1-S_2=\sqrt{\left(S_1+S_2\right)^2-4 S_1 S_2}\)

= \(\sqrt{(8)^2-4 \times 12}=4\) ……..(2)

From equations (1) and (2), S₁ = 6 and S₂ = 2

∴ The specific gravities of the two given substances are 6 and 2.

Example 5. The ratio of the densities of three liquids is 1:2:3. If they are mixed in

1. Equal volume,
2. Equal mass, then what will be the densities of their mixtures?

Solution:

Given

The ratio of the densities of three liquids is 1:2:3.

1. Let the densities of the liquids be d, 2d and 3d.

Let volume V of each of the three liquids be mixed.

∴ Total volume of the mixture = V+ V+ V = 3 V and total mass of the mixture = V • d + V • 2d + V • 3d = 6 Vd.

∴ Density of the mixture = \(\frac{6 V d}{3 V}\) = 2d

So, the density of the mixture will be twice the density of the first liquid.

2. if the mixture is prepared by mixing m mass of each of the three liquids, then the total mass of the mixture m+ m + m = 3m.

Total volume of the mixture = \(\frac{m}{d}+\frac{m}{2 d}+\frac{m}{3 d}=\frac{11}{6} \cdot \frac{m}{d}\)

∴ Density of the mixture = \(\frac{3 m}{\frac{11}{6} \cdot \frac{m}{d}}=\frac{18}{11} d\)

So, the density of the mixture will be 18/11 times the density of the first liquid.

Example 6. What amount of concentrated sulphuric acid (specific gravity 1.8) should be mixed with 1 L of water so that the specific gravity of the mixture becomes 1.24?
Solution:

Let the required volume be ν cm³.

Mass of that volume of acid =1.8 ν g and mass of 1 L of water = 1000 g

∴ Total mass =(1.8 ν+ 1000) g and total volume = (ν+ 1000) cm³

∴ Specific gravity of the mixture = \(\frac{1.8 v+1000}{v+1000}\)

∴ \(\frac{1.8 v+1000}{v+1000}\) = 1.24

or, (1.8-1.24)ν = 1240-1000 or, 0.56 ν = 240

or, v= 428.6 cm³.

Example 7. The mass of 1 L of milk is 1032 g. The milk contains 4% butter by volume. If the specific gravity of butter is 0.865, then find the density of butter-free milk.
Solution:

Given

The mass of 1 L of milk is 1032 g. The milk contains 4% butter by volume. If the specific gravity of butter is 0.865

Volume of butter present in 1 L of milk = 40 cm³.

Its mass =40 x 0.865 = 34.6 g

∴ Mass of butter-free milk = 1032-34.6 = 997.4 g and its volume = 1000 – 40 = 960 cm³

∴ Density of butter free milk = \(\frac{997.4}{960}=1.039 \mathrm{~g} \cdot \mathrm{cm}^{-3}\)

Dimensionless Nature of Specific Gravity

Example 8. The mass of a coil of copper wire of diameter 1.2 mm is 150 g. If the density of copper is 8.9 g • cm^-3, then find the length of the copper wire.
Solution:

Given

The mass of a coil of copper wire of diameter 1.2 mm is 150 g. If the density of copper is 8.9 g • cm^-3

Let the length of the wire be l and its radius be r.

Its volume = πr²l= π(0.06)²l cm³, its mass = π(0.06)² l x 8.9 g = 150

∴ l = \(\frac{150}{\pi \times 0.0036 \times 8.9}=1490 \mathrm{~cm} \text { (approx.) }\)

Example 9. The volume and mass of a piece of iron-aluminium alloy are 100 cm³ and 588 g respectively. The specific gravities of iron and aluminium are 8 and 2.7 respectively. Calculate the ratio of the volumes of iron and aluminium in the alloy.
Solution:

Given

The volume and mass of a piece of iron-aluminium alloy are 100 cm³ and 588 g respectively. The specific gravities of iron and aluminium are 8 and 2.7 respectively.

Let the volume of iron =  ν₁ cm³, volume of aluminium = ν₂ cm³; density of iron = 8 g • cm^-3 and density of aluminium =2.7 g • cm^-3.

According to the problem,

⇒ \(8 v_1+2.7 v_2=588\) ….(1)

and \(v_1+v_2=100\)

Solving (1) and (2) we get, ν₁ = 60, ν₂ = 40

∴ \(\frac{v_1}{v_2}=\frac{60}{40}=\frac{3}{2}\)

∴ The ratio of the volumes of iron and aluminium =3:2.

Hydrostatics Multiple Questions And Answers

Hydrostatics MCQs with Answers

Question 1. Spheres of iron and lead having same mass are com-pletely immersed in water. Density of lead is more than that of iron. Apparent loss of weight is w₂ for iron sphere and w₂ for lead sphere. Then w₁/w₂  is

1. 1
2. Between 0 and 1
3. 0
4. >1

Answer: 4. >1

Question 2. A U-tube of uniform cross-section contains some mercury. In one of its limbs, water is poured and in the other oil is poured such that the mercury in both the limbs remains at the same level. If the heights of the water and oil columns are 0.10 m and 0.125 m respectively, then the specific gravity of the oil will be

1. 0.8
2. 1.0
3. 1.25
4. 0.6

Answer: 1. 0.8

Question 3. The radius of a cylindrical container is r. When a liquid is poured into the container up to a height h, then the thrust on the bottom and that on the lateral surface of the container becomes equal. In this case

1. h = r/2
2. h= r
3. h = 2r
4. h = 4r

Answer: 2. h= r

Question 4. The density ρ of water of bulk modulus B at a depth y in the ocean is related to the density at surface ρ₀ by the relation

1. \(\rho=\rho_0\left(1-\frac{\rho_0 g y}{B}\right)\)
2. \(\rho=\rho_0\left(1+\frac{\rho_0 g y}{B}\right)\)
3. \(\rho=\rho_0\left(1+\frac{B}{\rho_0 g y}\right)\)
4. \(\rho=\rho_0\left(1-\frac{B}{\rho_0 g y}\right)\)

Answer: 2. \(\rho=\rho_0\left(1+\frac{\rho_0 g y}{B}\right)\)

Question 5. Two identical cylindrical containers contain a liquid of density d. The bases of the containers lie on the same horizontal plane. The depths of liquid in the two containers are h₁ and h₂ respectively and the base areas of both containers is A. If the two containers are connected by a pipe, then the work done by gravity to equalize the levels of liquid in the two containers is

1. (h₁ – h₂)gd
2. (h₁-h₂)gAd
3. \(\frac{1}{2}\left(h_1-h_2\right)^2 g A d\)
4. \(\frac{1}{4}\left(h_1-h_2\right)^2 g A d\)

Answer: 4. \(\frac{1}{4}\left(h_1-h_2\right)^2 g A d\)

]Read And Learn More WBCHSE Class 11 Physics MCQs

Question 6. The dimensions of a block of iron of density 5 g · cm^-3 are 5 cm x 5 cm x 5 cm. It is weighed by immersing it completely in water. Its apparent weight will be

1. (5x5x5x5)xg
2. (4x4x4x5)xg
3. (3x5x5x5)xg
4. (4x5x5x5)xg

Answer: 2. (5x5x5x5)xg

Hydrostatic Pressure Questions and Answers

Question 7. There is a layer of oil over mercury in a container. The density of mercury is 13.6 g • cm^-3 and that of the oil is 0. 8 g-cm^-3. A homogeneous sphere remains immersed with half of its volume in mercury and the other half in oil. The density of the material of the sphere in g • cm^-3 is

1. 3.3
2. 6.4
3. 7.2
4. 12.8

Answer: 3. 7.2

Hydrostatic Law Multiple Choice Questions

Question 8. When a block is suspended in air from a spring balance, it reads 60 N. When the block is immersed completely in water, the spring balance reads 40 N. The specific gravity of the block is

1. 3/2
2. 6
3. 2
4. 3

Answer: 4. 3

Question 9. A wooden cubic block with a 200 g mass floats in water just fully immersed. When the mass is removed, the block floats up by 2 cm. The length of each side of the cube is

1. 5 cm
2. 10 cm
3. 15 cm
4. 20 cm

Answer: 2. 10 cm

Question 10. A piece of ice of density 900 kg · m^-3 floats on water. The part of the piece of ice that remains outside the water surface is

1. 20%
2. 35%
3. 10%
4. 25%

Answer: 3. 10%

Question 11. Two stretched membranes of area 2m² and 3m² are kept at the same depth in a liquid. The ratio of the pressures on them is

1. 1:1
2. 2:3
3. √2:√3
4. 2²:3²

Answer: 1. 1: 1

Question 12. Two containers A and B of different shapes, but having the same base area are filled with water up to the same height h. The effective force on the base of the container A is F_A and that on the base of container B is F_B. If the weights of the two containers are W_A and W_B respectively, then which of the following is correct?

1. F_A>F_B’ W_A> W_B
2. F_A = F_B: W_A> W_B
3. F_A = F_B‘; W_A<W_B
4. F_A = F_B‘ ; W_A<W_B

Answer: 2.

Question 13. If a pressure is applied on any part of an enclosed fluid, then that pressure

1. Acts on all parts of the fluid with increased magnitude
2. Acts on the wall of the container with decreased magnitude
3. Increases according to the mass of the fluid and is then transmitted
4. Is transmitted throughout the fluid with undiminished magnitude and acts normally on tire wall of the container in contact with the fluid

Answer: 4. Is transmitted throughout the fluid with undiminished magnitude and acts normally on tire wall of the container in contact with the fluid

Question 14. The pressure exerted by a liquid at the bottom of the container containing the liquid does not depend on

1. Acceleration due to gravity
2. Height of the liquid column in the container
3. Area of the bottom of the container
4. Density of the liquid

Answer: 3. Area of the bottom of the container

Question 15. Bar is a unit of pressure. 1 bar = ?

1. 10⁴ Pa
2. 10⁵ Pa
3. 10⁶ Pa
4. 10⁷ Pa

Answer: 2. 10⁵ Pa

Hydrostatic Forces Practice Questions

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Question 16. The working principle of a hydraulic press depends on

1. Boyle’s law
2. Pascal’s law
3. Dalton’s law of partial pressure
4. Newton’s law of gravitation

Answer: 2. Pascal’s law

Question 17. A thin uniform cylindrical shell, closed at both ends is partially filled with water. It is floating vertically in water in a half-submerged state. If pc is the relative density of the material of the shell with respect to water, then the correct statement is that the shell is

1. More than half filled if ρ_c is less than 0.5
2. More than half filled if ρ_c is less titan 1.0
3. Half filled if ρ_c is less than 0.5
4. Less than half filled if ρ_c is less than 0.5

Answer: 4. Less than half filled if ρ_c is less than 0.5

Question 18. Which of the following works on Pascal’s law?

1. Barometer
2. Hydraulic lift
3. Venturi meter
4. Sprayer

Answer: 2. Hydraulic lift

Question 19. If the liquid used in a hydraulic press is replaced by a gas, then the press

1. Continues to work as before
2. Does not work
3. Works but cannot be used for heavy duty
4. None of the above

Answer: 3. Works but cannot be used for heavy-duty

Question 20. The radii of two thin circular discs are 0.03 m and 0.04 m respectively. They are placed inside a liquid at the same depth. The ratio of the thrusts acting on them is

1. 9: 16
2. 3:4
3. 4:3
4. √3:2

Answer: 1. 9: 16

Question 21. A given shaped glass tube having uniform cross section is filled with water and is mounted on a rotable shaft as shown in Fig. If the tube is rotated with a constant angular velocity ω, then

1. Water level in both sections A and B go up
2. Water level in section A goes up and that in B comes down
3. Water level in section A comes down and that in B goes up
4. Water level remain same in both sections

Answer: 1. Water level in both sections A and B go up

Question 22. The diameters of the smaller and larger pistons of a hydraulic press are d₁ and d₂ respectively. If the force applied on the smaller piston is F₁, then the force developed on the larger piston will be

1. \(F_2=\frac{d_2^2}{d_1^2} \cdot F_1\)
2. \(F_2=\frac{d_1^2}{d_2^2}, F_1\)
3. \(F_2=\frac{d_1^2}{d_2^2} \cdot \frac{1}{F_1}\)
4. \(F_2=\frac{d_2^2}{d_1^2} \cdot \frac{1}{F_1}\)

Answer: 1. \(F_2=\frac{d_2^2}{d_1^2} \cdot F_1\)

Question 23. A hole is made at a depth h on the wall of a container containing a liquid. The velocity of efflux of the liquid through the hole is

1. Zero
2. √gh
3. √2gh
4. Dependent on the density of the liquid

Answer: 3. √2gh

Question 24. A cylindrical vessel contains water up to a height of H. A hole is made in the wall of the vessel. If the jet of water coming out from the hole traverses the maximum distance from the base of the vessel, then the hole is situated

1. At a depth H from the upper surface of water
2. At a depth H/2 from the upper surface of water
3. At a depth H/4 from the upper surface of water
4. At a depth 3H/4 from the upper surface of water

Answer: 2. At a depth H/2 from the upper surface of water

Question 25. A tank contains water up to a height H. A hole is made at a depth h from the upper surface. The horizontal range of the jet of water coming out from the hole is

1. \(2 \sqrt{h(H-h)}\)
2. \(4 \sqrt{h(H+h)}\)
3. \(4 \sqrt{h(H-h)}\)
4. \(2 \sqrt{h(H+h)}\)

Answer: 1. \(2 \sqrt{h(H-h)}\)

Question 26. Of the following substances, which one can be lifted upwards with more ease with the help of a hydrogen-filled balloon?

1. 1 kg steel
2. 1 kg feather
3. 1 kg lead
4. 1 kg water

Answer: 2. 1 kg feather

Question 27. There is a coin placed on a wooden block floating on water as shown in Fig. In this figure, l and h are directed. After sometime, the coin falls into the water. Then it is observed that

1. l decreases but h increases
2. l increases but h decreases
3. Both l and h increase
4. Both l and h decrease

Answer: 4. Both l and h decrease

Question 28. A man is carrying a bucket of water in one hand and a piece of plastic in the other. When the piece of plastic is dropped into the water of the bucket, then the man

1. Carries more weight than before
2. Carries less weight than before
3. Carries the same wei|ht as before
4. Carries more or less weight than before depending on the density of the plastic

Answer: 3. Carries the same wei|ht as before

Question 29. A body remains floating in a liquid kept in a beaker. If the entire system falls under gravity, then the upthrust of the liquid acting on the body will be

1. Zero
2. Equal to the weight of the displaced liquid
3. Equal to the weight of the part of the body exposed to air
4. Equal to the weight of the immersed part of the body in that liquid

Answer: 1. Zero

Step-by-Step Solutions to Hydrostatics Problems

Question 30. Figure is here shows the vertical cross-section of a vessel filled with a liquid of density ρ. The normal thrust per unit area on the walls of the vessel at point P, as shown will be

1. hρg
2. Hρg
3. (H-h)pg
4. (H- h)ρgcosθ

Answer: 3. (H-h)pg

Question 31. A body floats in a liquid fully submerged. If the body is pressed slightly downwards and then released, it will

1. Oscillate up and down
2. Remain at rest in its new position
3. Return to the previous position quickly
4. Return to the previous position slowly

Answer: 2. Remain at rest in its new position

Question 32. A bird is sitting on the floor of a wire cage. This cage is carried by a boy. In this situation, if the bird starts flying inside the cage, then the boy will feel that

1. The cage has become heavier than before
2. The cage has become fighter than before
3. The weight of the cage remains the same as before
4. The cage appears fighter first and then heavier

Answer: 2. The cage has become fighter than before

Question 33. A boat of length 3 m and breadth 2 m is floating on water in a lake. When a man rides on the boat, it sinks 1 cm more. The mass of the man is

1. 60 kg
2. 72 kg
3. 12 kg
4. 128 kg

Answer: 1. 60 kg

Question 34. A wooden block is floating on water in a closed vessel. If the entire space above the water is evacuated, then

1. The block floats up more
2. The block sinks more
3. The block will remain in the same position
4. None of the above

Answer: 2. The block sinks more

Question 35. A body is floating on a liquid. The upthrust by the liquid acting on the body is

1. Equal to the weight of the displaced liquid
2. Zero
3. Less than the weight of the displaced liquid
4. Weight of the body – weight of the displaced liquid

Answer: 1. Equal to the weight of the displaced liquid

Question 36. A block of wood and a solid lead sphere are kept in a container with water up to its brim. Now, if the lead sphere is placed on the floating block of wood, then the level of water

1. Will remain the same
2. Will fall
3. Will rise and some water will overflow from the container
4. Will change depending on the shape of the lead sphere

Answer: 3. Will rise and some water will overflow from the container

Question 37. Two solid pieces one of steel and other of aluminium when immersed completely in water have equal weights. When the solid pieces are weighed in air

1. The weight of aluminium is half the weight of steel
2. Steel piece will weigh more
3. They have the same weight
4. Aluminium piece will weigh more

Answer: 4. Aluminium piece will weigh more

Question 38. A man is sitting on a boat in a pond. If the man drinks some water from the pond, the level of water in the pond

1. Will remain the same
2. Will fall
3. Will rise
4. Will rise or fall depending on the amount of water drunk by the man

Answer: 1. Will remain the same

Question 39. A piece of ice is floating on a liquid of density 1.5 g- cm^-3 kept in a beaker. If the piece of ice melts completely, then the liquid level in the beaker

1. Will rise
2. Will fall
3. Will remain the same
4. Will fall down first and then return to the original level

Answer: 1. Will rise

Question 40. A wooden block of mass 120 kg floats on water. The density of the wooden block is 600 kg • m^-3. The weight that should be placed on the wooden block so that it just sinks will be

1. 80 kg
2. 50 kg
3. 60 kg
4. 30 kg

Answer: 1. 80 kg

Question 41. A body with a uniform cross-section floats on a liquid. The density of the liquid is thrice the density of the body. The part of the body that remains above the liquid is

1. 2/3
2. 5/6
3. 1/6
4. 1/3

Answer: 1. 2/3

Question 42. A body remains floating in a liquid with a part of it immersed in the liquid. If the body along with the liquid is taken to the moon, then the body

1. Will remain floating as before
2. Will remain floating with a greater part of it inside the liquid than before
3. Will remain floating with a lesser part of it inside the liquid than before
4. Will sink in the liquid

Answer: 1. Will remain floating as before

Question 43. With increase in temperature, the upthrust acting on a body immersed in a liquid

1. Increases
2. Decreases
3. Remains unchanged
4. None of these

Answer: 2. Decreases

Question 44. A block of wood of mass m and density ρ is tied with a thread and rigidly fixed at the bottom of a vessel. The vessel is filled with a liquid of density σ(σ > ρ). The tension acting on the string is

1. \(\left(\frac{\sigma-\rho}{\sigma}\right) m g\)
2. \(\left(\frac{\sigma-\rho}{\rho}\right) m g\)
3. \(\frac{\rho}{\sigma} m g\)
4. \(\frac{\sigma}{\rho} m g\)

Answer: 2. \(\left(\frac{\sigma-\rho}{\rho}\right) m g\)

In this type of questions more than one options are correct.

Question 45. A spring balance A reads 2 kg with a block suspended from it. Another balance B reads 5 kg when a beaker with liquid is put on the pan of the balance. When the block is immersed in water

1. The balance a will read more than 2 kg
2. The balance b will read more than 5 kg
3. The balance a will read less than 2 kg
4. The balance a will read 2 kg and the balance b will read 5 kg

Answer:

2. The balance b will read more than 5 kg

3. The balance a will read less than 2 kg

Question 46. A closed vessel is half filled with water. There is a hole near the top of the vessel and air is pumped out from this hole. As a result

1. The water level will rise up in the vessel
2. The pressure at the surface of the water will decrease
3. The force by the water on the bottom of the vessel will decrease
4. The density of the liquid will decrease

Answer:

2. The pressure at the surface of the water will decrease

3. The force by the water on the bottom of the vessel will decrease

Question 47. When a body of density ρ and volume V is floating in a liquid of density σ

1. Its true weight is vρg
2. Loss in its weight is vσg
3. Its apparent weight is zero
4. Its density ρ is less than that of liquid σ

Answer:

1. Its true weight is vρg

3. Its apparent weight is zero

4. Its density p is less than that of liquid σ

Question 48. Two liquids that do not react chemically are placed in a bent tube as shown in Fig. Which of the following statements are correct?

1. The pressure at the interface must be same, calculated via both tubes
2. The heights of the liquids above their surfaces of separation are inversely proportional to their densities
3. Both 1 and 2
4. Neither 1 nor 2

Answer:

1. The pressure at the interface must be same, calculated via both tubes
2. The heights of the liquids above their surfaces of separation are inversely proportional to their densities
3. Both 1 and 2

Question 49. A wire is found to have a length L when it is loaded with a block of mass M and relative density n. When the block is immersed in water, the length of the wire reduces by x, then

1. Weight of water displaced when the block is immersed in water is mg/ n
2. The apparent loss of weight due to immersion is \(m g\left(1-\frac{1}{n}\right)\)
3. The original length of the wire before it was loaded is L₁ = L -nx
4. The original length of the wire before it was loaded is L₁ =L- x/n

Answer:

1. Weight of water displaced when the block is immersed in water is mg/ n
2. The apparent loss of weight due to immersion is \(m g\left(1-\frac{1}{n}\right)\)
3. The original length of the wire before it was loaded is L₁ = L -nx

Question 50. A uniform cylinder of density ρ and cross-sectional area A floats in equilibrium in two non-mixing liquids of densities ρ₁ and ρ₂ as shown in Fig. The length of the part of the cylinder in air is h and the lengths of the part of the cylinder immersed in the liquid are h₁ and h₂ as shown in the figure. Which among the following are correct?

1. The net resultant force acting on the cylinder by the liquid of density ρ₁ is zero
2. h = \(h_1\left(\frac{\rho_1}{\rho}-1\right)+h_2\left(\frac{\rho_2}{\rho}-1\right)\)
3. The cylinder is depressed in such a way that its top surface is just covered by the liquid of density ρ₁ and then released. The restoring force acting on the cylinder is f = hAρ₂g
4. In choice 3 above, the acceleration of the cylinder is a = \(a=\frac{h+\rho_2 g}{\left(h+h_1+h_2\right) \rho}\)

Answer: All are correct

Question 51. Two solid spheres A and B of equal volumes but of different. densities d_A and d_B are connected by string. They are fully immersed in a fluid of density d_F. They get arranged into an equilibrium state as shown in Fig. with a tension in the string. The arrangement is possible only if

1. d_A< d_F
2. d_B > d_F
3. d_A >d_F
4. d_A+d_B = 2d_F

Answer:

1. dA< dp

2. dB> dp

4. dA+dB = 2dF

Hydrostatics Buoyancy And Buoyant Force

Hydrostatic Forces Explained

A body appears lighter when it is immersed totally or partly in a liquid or a gas. For example, a pitcher full of water appears lighter when immersed in water.

On drawing a bucket full of water from a well, it appears lighter as long as it is inside water. But once it rises above water, it appears heavier. In no case does the actual weight of a body decrease, but the body experiences an apparent loss of weight.

Buoyancy And Buoyant Force Definition: The ability of a liquid or a gas at rest to exert an upward force on a body immersed in that fluid is called buoyancy.

• When a body is immersed in a liquid or a gas, the liquid or the gas exerts pressure that is different on different parts of the body. The pressure at any point on the body depends on the depth of that point from the free surface or the upper surface of that liquid or gas.
• Since the immersed body experiences different forces at different points, we can say that the body experiences a resultant force or thrust which acts in the upward direction. This resultant upward thrust balances a part of the downward force due to the weight of the body and hence the body suffers an apparent loss of weight.

Read and Learn More: Class 11 Physics Notes

Definition: The upward thrust exerted by a liquid or a gas at rest on a body partly or totally immersed in it is called buoyant force.

The terms buoyancy and buoyant force usually have the same meaning.

The Magnitude Of Buoyancy On A Body Immersed In A Liquid: Let a cube ABCDEFGH be immersed in a liquid of density ρ. The depths of the surfaces ABCD and EFGH of the cube from the upper surface of the liquid are h₁ and h₂ respectively. Each side of the cube is length of l.

The surfaces ABCD and EFGH of the cube are horizontal but the other surfaces are vertical. The liquid exerts a normal thrust on each surface of the cube.

The lateral thrusts acting horizontally on the mutually opposite vertical faces ABHE and CDFG, being equal and opposite, balance each other. Similarly, the lateral thrusts acting on the side faces AEFD and BCGH balance each other.

So, no lateral resultant thrust acts on the cube in the horizontal direction. But, due to the difference in the depths of the surfaces ABCD and EFGH from the upper surface of the liquid, unequal thrusts act on these two surfaces.

The downward pressure of the liquid at any point on the surface ABCD = h₁ρg

The downward thrust on the surface ABCD = pressure x area = h₁ρg · l²

The upward pressure of the liquid at any point on the surface EFGH = h₂ρg

The upward thrust on the surface EFGH = h₂ρg· l²

Since h₂> h₁, the upward thrust acting on the cube becomes greater than the downward thrust.

∴The net upward thrust on the cube

= \(h_2 \rho g l^2-h_1 \rho g l^2=l^2 \rho g\left(h_2-h_1\right)=l^3 \rho g\)

[because \(h_2-h_1=l\)]

But, l³ = V = volume of the cube

∴ l³ρg = Vρg = weight of the liquid of a volume equal to that of the cube.

Buoyancy Principles in Fluid Mechanics

• So, the buoyancy on the cube due to the liquid = weight of the liquid displaced by the cube.
• This inference is applicable to any object of any shape whatsoever.
• Hence, we can say that when a body is immersed partly or totally in a liquid, then the body feels an upward thrust and this upthrust or buoyant force is equal to the weight of the liquid displaced by the body.
• From the presence of the factor g in the expression of the buoyant force, it is evident that buoyancy is a direct consequence of the effect of gravity on fluid pressure.

Characteristics Of Buoyant Force:

1. The buoyant force depends on

1. The volume of the immersed portion of the body,
2. The density of the displaced liquid and
3. The acceleration due to gravity. Gd the buoyant force acts in the direction opposite to the weight of the body.

2. For a totally immersed body, the buoyant force does not depend on the depth up to which the body is immersed inside the liquid.

3. If the volume of a body is kept unaltered, then

1. The buoyant force on a totally immersed body does not depend on the size, physical state (i.e., Solid, liquid, or gas), and mass of the body but
2. For a partly immersed body, the buoyant force on it does not depend upon its size or physical state but depends on its mass.
3. It is clear that there will be no buoyant force acting on a body if both the body and the liquid are in a weightless condition. This situation arises in an artificial satellite or for a freely falling system.

Units And Dimension Of Buoyant Force:

• dyne CGS System
• newton SI

The dimension of the buoyant force is MLT^-2.

Centre Of Buoyancy : The point where the centre of gravity of the liquid or gas is located before it is displaced by the immersed body is the centre of buoyancy or the centre of floatation of the immersed body.

The buoyant force on an immersed body acts through the centre of buoyancy. The centre of gravity and the centre of buoyancy coincide for a totally immersed body in a fluid. But for a partly immersed body, these two points are different.

Reaction Of Buoyant Force : when a body is immersed in a liquid, the displaced liquid exerts an upthrust or buoyant force on the body, and according to Newton’s third law of motion, the body also exerts an equal but opposite reaction on the liquid. The existence of this reaction can be realised from the following experiment.

Reaction Of Buoyant Force Experiment: A beaker containing water is placed on one pan of a common balance and by placing the required counterpoising weights on the other pan, the balance beam is made horizontal.

• Now, a piece of iron is tied by a thread and is immersed in water by holding the free end of the thread so that the piece of iron does not touch the wall of the beaker.
• It is seen that the pan with the beaker immediately goes down. As the liquid exerts an upthrust on the piece of iron, the iron piece at the same time applies a downward reaction on the liquid. This reaction ultimately acts on the pan and hence it goes down.

Apparent Weight Of A Body Immersed In A Liquid : Let the real weight of a body be W₁ this weight acts vertically downwards through the centre of gravity of the body. When the body is immersed in a liquid, then the buoyancy W₂ acts in the upward direction through the centre of buoyancy.

So, apparent weight of the body immersed in a liquid = real weight of the body – buoyant force acting on the body (i.e., weight of liquid displaced by the body)

= W₁-W₂

Hence, the apparent weight of the body < its real weight. So, a body suffers an apparent loss of weight due to the upthrust of the liquid when it is immersed in that liquid.

Hydrostatics Floatation And Immersion Of A Body In A Liquid At Rest

Hydrostatic Force on Submerged Surfaces

When a body is immersed in a liquid at rest, then the following two forces act on the body simultaneously.

1. The weight W₁ of the body which acts vertically downwards through the centre of gravity of the body.
2. The buoyancy of the liquid W₂ (weight of the liquid displaced by the body) which acts vertically upwards through the centre of buoyancy.

The value of W₁ may be greater than, equal to, or less than W_2, and hence any one of the three cases may arise.

Case 1: If W₁>W₂, i.e., the weight of the body is greater than the weight of the liquid displaced by the body, then the body will sink in the liquid.

In this case, the net downward force = W₁-W₂.

If the body is homogeneous and if its density and volume are ρ₁ and V₁ respectively, then W₁= V₁ρ₁g. If the density of the liquid is ρ₂, then the weight of the liquid displaced by the body, W₂ = V₂ρ₂g.

∴ If W₁ > W₂ , then, ρ₁ > ρ₂

So, if the density of the material of the body is greater than the density of the liquid, then the body sinks in the liquid. For example, a piece of stone or iron sinks in water.

Case 2: If W₁ = W₂, i.e., the weight of the body is equal to the weight of the liquid displaced by the body, then the body will float at rest in any position in the liquid and be totally immersed in that liquid.

In this case, the weight of the body and the buoyancy i balance each other and hence the net force acting on the body becomes zero. So, the body becomes appar- . ently weightless.

Since, \(W_1=W_2, V_1 \rho_1 g=V_1 \rho_2 g \text { or, } \rho_1=\rho_2\)

Hence, if the density of the material of the body is equal to the density of the liquid, then the body can float remaining totally immersed within the liquid.

Case 3: If W₁<W₂, i.e., the weight of the body is less than that of the liquid displaced by it, then the body floats remaining partially immersed in the liquid.

If the body is released after immersing it completely inside the liquid, then under the influence of the force (W₂-W₁), it will move up through the liquid and, after some time, a part of the body will gradually come out of the liquid.

As a result, the amount of liquid displaced by the body is less. When the weight of the body is equal to that of the liquid displaced by it, the body floats remaining partially immersed in the liquid.

Since, \(W_1<W_2, \rho_1<\rho_2\)

So, when the density of the material of the body is less than the density of the liquid, the body floats on the liquid. For example, the density of wood is 0.7 – 0.9 g · cm^-3 and that of pure water is 1 g · cm^-3.

So, a piece of wood floats on water. Again, as the density of iron (7-7.9 g · cm^-3) is less than that of mercury (13.68 g · cm^-3), a piece of iron floats on mercury.

A floating body has no apparent weight: when a body floats in a liquid, the weight of the body becomes equal to the weight* of the liquid displaced by the body. According to Archimedes’ principle, the apparent loss of weight of the body inside the liquid is equal to the weight of the liquid displaced by that body, and hence the apparent weight of the body in the liquid = weight of the body – weight of liquid displaced by the body = 0.

So, a floating body in a liquid has no apparent weight. The weight of the floating body is balanced by the buoyant force exerted by the liquid.

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Hydrostatics Properties Of A Floating Body

Conditions of Equilibrium of a Floating Body: Two equal but opposite forces act on a floating body—one is the weight of the body which acts downwards and another is the buoyancy of the liquid which acts in the upward direction.

To keep the floating body in equilibrium, these two forces must act along the same straight line. Otherwise, these two forces will constitute a couple and will produce a rotational motion in the body. Hence, for the equilibrium of a floating body, the following two conditions must be satisfied.

1. Condition of floatation: The weight of the floating body must be equal to the weight of the liquid displaced by the body.

2. Condition of equilibrium: The centre of gravity of the body and the centre of buoyancy must lie on the same vertical line.

In Fig, a body is floating in equilibrium. The centre of gravity of the body is G, centre of buoyancy is B, and weight of the body (W₁) = weight of the displaced liquid (W₂). In equilibrium, the vertical line ab passes through the centre of gravity G and the center of buoyancy B. The line ab is called the centre line.

Stability of a Floating Body: The equilibrium of a floating body may be of three types

1. Neutral equilibrium,
2. Stable equilibrium and
3. unstable equilibrium.

If the floating body is tilted slightly, then the state of its equilibrium can be studied.

• If a slight tilt of the floating body from its equilibrium position does not produce any shift of the centre of buoyancy, i.e., if the new centre of buoyancy coincides with its previous position, then the body is said to be floating in neutral equilibrium.
• In this case, the centre of gravity of the body and the centre of buoyancy lie on the same vertical line. A sphere floating on a liquid is always in neutral equilibrium,

State equilibrium and unstable equilibrium: Usually, when a floating body is tilted, the centre of buoyancy shifts towards the leaning side as more liquid is displaced on that side of the body.

• Let be the new position of the centre of buoyancy. In this situation, the weight of the body (W₁) and the buoyant force (W₂) do not act along the same vertical line. These two equal but parallel forces constitute a couple.
• If this couple can return the body to its original position, then the equilibrium is said to be stable. But if the couple does not return the body to its original position, but rather it tends to tilt the body further, then the equilibrium is said to be unstable.

Metacentre: In the tilted position of a floating body, the point at which the vertical line drawn through the centre of buoyancy (B₁) cuts the centre line ab is called the metacentre (M) of the body.

• From Fig, it is understood that if the metacenter (M) lies above the centre of gravity (G) of the body, then the equilibrium of the body is stable. But if the metacenter (M) lies below the centre of gravity (G) of the body, then the equilibrium is unstable.
• Hence, for stable equilibrium of a floating body, the metacentre of the body should be situated above its centre of gravity. In the case of neutral equilibrium, the points G and M coincide with each other.
• The distance GM is called metacentric height. The more the distance of the centre of gravity of the body below the metacentre, the stronger the moment of the restoring couple acting on the body and the equilibrium will be more stable.

Stability of ships and boats: Boats, ships, etc., are so constructed that they can float in water at stable equilibrium. For this, two conditions should be followed.

1. The external shapes of these vehicles are so designed that the metacentre lies at a suitable height. For this, the width of the vehicles are made greater at the top them at the bottom.
2. The centre of gravity of these vehicles is kept as low as possible by filling the bottom with heavy cargo. For this, some additional weights, called ballast, are kept loaded in the hold of an empty ship.

Two Important Relations in Connection with a Floating Body

1. Let a body of volume V and density D be allowed to float on a liquid of density d. If the volume of the immersed part of the body in that liquid is ν, then the volume of liquid displaced by the body will also be ν. So, according to the condition of floatation,

weight of the body = weight of the displaced liquid

or vDg or, \(\frac{v}{V}=\frac{D}{d}\)

∴ \(\frac{\text { volume of the immersded of part of a body }}{\text { total volume of the body }}\)

= \(\frac{\text { density of the body }}{\text { density of the liquid }}\)

2. if n is the fraction of volume of the body that remains immersed in the liquid, then

⇒ \(\frac{\nu}{V}=n=\frac{D}{d} \text { or, } D=n \times d\)

Application and Illustration of the Principle of Floatation

The upward motion of a balloon: The upward motion of a balloon depends on the upthrust of air. The net weight of an inflated balloon containing hydrogen gas is much less than the weight of air displaced by it (due to lower density of hydrogen).

• As a result, the buoyant force exerted by air on the balloon becomes greater than the weight of the balloon. Due to this reason, the balloon experiences a net upward force and it moves up. We know that with the increase in altitude from the surface of the earth, the density of air decreases and hence the upthrust exerted on the balloon decreases.
• After attaining a certain height, the weight of the balloon becomes equal to the buoyant force acting on it and then the balloon cannot rise any further; it remains floating at a particular altitude.
• Again, another situation may arise. Due to the lower atmospheric pressure at a higher altitude, the volume of the balloon gradually increases.
• As a result, the weight of air displaced by the balloon, i.e., the upthrust also increases and, in consequence, the balloon rises more and more and finally may burst due to excessive expansion in its volume.
• Besides hydrogen, helium is also used as a filling agent in a balloon. Hydrogen is lighter than helium, but is inflammable whereas helium is not. For this reason, helium is used in balloons.

Why does ice float on water: The mass of 1 cm³ of ice is 0.917 g. The mass of 1 cm³ of water is 1 g. So, the same volume of ice is lighter than water. Since the density of ice is less than that of water, it floats partially immersed in water. If nth part of a lump of ice remains submerged, then n = density of ice = 0.917 = 11/12 (approx.)

So, in its floating condition, 11/12 th part of a lump of ice remains immersed in water and 1/12 th part remains above the water surface. The density of sea water is 1.03 g· cm^-3. So, 1/9 th part (approx.) of an iceberg remains exposed to air while floating on sea water.

To float on water, we have to learn swimming, but to animals swimming comes naturally-explain why: The average density of the human body 1.01 g · cm^-3 (approx.).

• The weight of the body of a man is less than the weight of an equal volume of water. But the head is heavier than the weight of an equal volume of water.
• Hence, a man can float on water, but his head sinks in it. So, we have to learn swimming to float on water. A man strives to keep his head afloat by moving his limbs in water. This is the act of swimming.
• For animals, both their heads and bodies are lighter than an equal volume of water and hence they can float on water effortlessly.
• It is easier to float in sea water than in fresh river water because the density of sea water is more than that of river water. Hence, the upthrust of sea water is more than that of river water.

A piece of iron sinks into water but a ship made of iron floats on it—explain how: This is due to the special shape of a ship. The bottom of a ship is made wider and hollow. It is given a concave shape so that the weight of displaced water by the ship becomes equal to the weight of the ship.

According to the condition of floatation, if the weight of a body becomes equal to the upthrust exerted on it by the displaced liquid then the body floats on that liquid. That is why a ship floats on water. The weight of a piece of iron is more than the weight of water displaced by it and so it sinks in water.

How does a submarine submerge and float on water: A submarine can float on water, but can also be submerged. A submarine contains a large number of tanks which can be filled up with air or water.

• These tanks are called ballast tanks. When the ballast tanks are filled with air, the submarine floats on water as the weight of the submarine becomes less than the weight of water displaced by it.
• When water is admitted into the ballast tanks using a pump, tire submarine becomes slightly heavier than the water displaced by it and hence it submerges. To raise the submarine again to the surface of water, water is driven out of the ballast tanks by compressed air.

Why are life-belts used: Steamers and ships are provided with life-belts. If a steamer or a ship sinks, then the passengers can float by wearing these life-belts. A life-belt is nothing but an air-filled bag. The weight of this bag along with the weight of the person is less than the weight of the displaced water so the person is able to keep afloat.

1 kilogram of cotton is heavier than 1 kilogram of iron: For a body in air,

apparent weight = actual weight (i.e., weight in vacuum) – buoyancy of air

So, actual weight =  apparent weight + buoyancy of air

= apparent weight + weight of air displaced by the body

When we weigh 1 kg of cotton and 1kg of iron in air, those are their apparent weights. So the actual weight depends on the weight of air displaced by them.

Clearly, 1kg of cotton has a much larger volume, and it displaces a much larger amount of air. so the actual weight of 1kg of cotton is higher than that of 1kg of iron. Therefore, it is said that, 1kg of cotton is heavier than 1kg of iron.

Hydrostatics Pressure Difference And Buoyant Force In A Liquid Of A Uniformly Accelerated Container Filled With The Liquid

The mathematical expression used so far for the buoyant force acting on a body immersed in a liquid holds true for any liquid at rest. However, those equations are to be modified for a uniformly accelerated container filled with the liquid.

In the Case of a Moving Lift

Case 1: When a lift with a liquid-filled container moves upwards with an acceleration a (or moves downwards with a retardation a):

Difference: In pressure Let us imagine a liquid column of height h and cross-sectional area A inside the lift. If the density of the liquid is ρ, then the mass of the liquid column, m = hAρ.

Different forces exerted on the liquid column:

1. If p₁ is the pressure on the upper surface of the liquid column, then the thrust on that surface F₁ = p₁A; this force acts vertically downwards.
2. If p₁ is the pressure on the lower surface of the liquid column, then the thrust on that surface F₂ = ρ₂A; this force acts vertically upwards.
3. The weight of the liquid column mg = hAρg; this force acts vertically downwards.

Therefore, the equation of motion of the liquid column is

Factors Affecting Buoyancy in Fluids

⇒ \(F_2-F_1-m g=m a\)

or, \(p_2 A-p_1 A=m a+m g\)

or, \(\left(p_2-p_1\right) A=m(a+g)\)…..(1)

= \(h A \rho(a+g)\)

or, \(p_2-p_1=h \rho(g+a)\)…….(2)

So the difference in pressure increases by hρa.

Buoyant force: The buoyant force or upthrust on the liquid column, W’ = (p₂– p₁)A

From equation (1) we get, W’ = m(g+ a)…. (3)

So, the buoyant force increases by ma.

Case 2: When the lift with a liquid-filled container moves downwards with an acceleration a (or moves upwards with a retardation a):

Following the previous calculation, it can be seen that \(p_2-p_1=h \rho(g-a)\)

i.e., the difference in pressure, in this case, decreases by ρa.

Similarly, the buoyant force, W’ = m(g- a)

i.e., the buoyant force, in this case, decreases by ma.

Case 3: When the lift with a liquid-filled container moves downwards freely (i.e., a = g):

In this case, the difference in pressure, p₂– p₁ = 0 and the buoyant force, W’ = 0

i.e., both the pressure at any point inside a liquid and the upthrust on any immersed object become nil.

Case 4: When the lift with a liquid-filled container is at rest or moves up or down with uniform speed:

Here, a = 0.

∴ Difference in pressure, p₂-p₁ = hρg and buoyant force, W’ = mg.

Free Surface in a Liquid-filled Container Moving horizontally with Uniform Acceleration a: A horizontal cylinder of length l and cross-sectional area a is imagined inside a liquid.

If the density of the liquid is ρ, then the mass of the said liquid cylinder, m = lap.

Let the liquid exert pressures p_A and p_B respectively at the two ends A and B of the cylinder.

∴ The equation of motion of the liquid cylinder is \(p_A \alpha-p_B \alpha=m a\)

or, \(\alpha\left(p_A-p_B\right)=l \alpha \rho a \text { or, } p_A-p_B=l \rho a\)…..(1)

Pressures at two points on the same horizontal line inside a liquid-filled container moving horizontally with uniform acceleration are not the same.

Due to the absence of any vertical acceleration of the container, the pressure at any point inside the liquid is similar to that in the case of a liquid at rest. But in this case, the free surface of the liquid does not remain horizontal. Suppose the free surface makes angle 6 with the horizontal.

Now, \(p_A=h_1 \rho g\) and \(p_B=h_2 \rho g\)

∴ According to equation (1),

⇒ \(h_1 \rho g-h_2 \rho g=l \rho a\)

or, \(\frac{h_1-h_2}{l}=\frac{a}{g}\)

or, \(\tan \theta=\frac{a}{g}\)

or, \(\theta=\tan ^{-1}\left(\frac{a}{g}\right)\)

Hydrostatics Pressure Difference And Buoyant Force In A Liquid Of A Uniformly Accelerated Container Filled With The Liquid Numerical Examples

Hydrostatic Pressure Calculation Examples

Example 1. A body weighs 100gxg and 40gxg In air and water respectively. Calculate the weight of the body when Immersed in a liquid of specific gravity 0.8.
Solution:

Given

A body weighs 100gxg and 40gxg In air and water respectively.

Weight of the body in air = 100 g x g, weight of the body in water = 40 g x g.

∴ Weight of displaced water = (100 – 40) x g = 60 g x g.

∴ Volume of displaced water = 60 cm³

So, the volume of the body = 60 cm³

The specific gravity of the liquid =0.8

∴ Density of the liquid =0.8 g · cm^-3.

∴ Weight of the displaced liquid = (60 x 0.8) x g = 48 g x g.

Weight of the body in the liquid = weight of the bodyin air – weight of the displaced liquid

= (100 – 48) x g = 52 g x g = 50960 dyn .

Weight of the body in the liquid = 50960 dyn

Example 2. The mass of a steamer is 10 tonnes. When it enters a fresh water lake from the sea, it displaces 50 L more water than before. Calculate the density of seawater. (Mass of 1 L of fresh water = 1 kg)
Solution:

Given

The mass of a steamer is 10 tonnes. When it enters a fresh water lake from the sea, it displaces 50 L more water than before.

Mass of displaced seawater = mass of the streamer = 10 x 10³ kg = 10⁷ g.

∴ The volume of displaced seawater = \(\frac{10^7}{\rho} \mathrm{cm}^3\) [p = density of sea water)

Mass of displaced take water = 10⁷ g

As the density of fresh water is 1 g · cm^-3, the volume of dis¬placed lake water = 10⁷ cm³.

According to the problem,

⇒ \(10^7-\frac{10^7}{\rho}=50 \times 10^3 \text { or, } 10^7\left(1-\frac{1}{\rho}\right)-5 \times 10^4\)

or, \(1-\frac{1}{\rho}=\frac{5}{10^3} \text { or, } \frac{1}{\rho}=1-\frac{5}{1000}=\frac{995}{1000}\)

∴ \(\rho=\frac{1000}{995}=1.005 \mathrm{~g} \cdot \mathrm{cm}^{-3}\)

Example 3. The internal and external diameters of a sphere are 8cm and 10cm respectively. The sphere neither floats nor sinks in a liquid of specific gravity 1.5 g ⋅ cm^-3. Calculate the density of the material of the sphere.
Solution:

Given

The internal and external diameters of a sphere are 8cm and 10cm respectively. The sphere neither floats nor sinks in a liquid of specific gravity 1.5 g ⋅ cm^-3.

Let the density of the material of the given sphere be ρ.

According to the condition of floatation, volume of the sphere x density of the material of the sphere = volume of displaced liquid x density of that liquid.

or, \(\left\{\frac{4}{3} \pi(5)^3-\frac{4}{3} \pi(4)^3\right\} \times \rho=\frac{4}{3} \pi(5)^3 \times 1.5\)

or, \(\left\{(5)^3-(4)^3\right\} \times \rho=(5)^3 \times 1.5\)

or, \((125-64) \rho=125 \times 1.5\)

or, \(\rho=\frac{125 \times 1.5}{61}=3.07 \mathrm{~g} \cdot \mathrm{cm}^{-3}\)

Example 4. A body of density ρ is placed slowly on the surface of a liquid of density δ. If the depth of the liquid is d, then prove that the time taken by the body to reach the bottom of the liquid is \(\left[\frac{2 d \rho}{g(\rho-\delta)}\right]^{1 / 2}\) second.
Solution:

Given

A body of density ρ is placed slowly on the surface of a liquid of density δ.

Let the mass of the body be m.

∴ Its volume = \(\frac{m}{\rho}\).

Volume of liquid displaced = \(\frac{m}{\rho}\)

∴ Mass of liquid displaced = \(\frac{m}{\rho}\) x 8

If the downward acceleration of the body in the liquid is a, then by the problem

mg – \(\frac{m}{\rho}\).δg= ma

or, \(a=\left(1-\frac{\delta}{\rho}\right) g=\left(\frac{\rho-\delta}{\rho}\right) g\)

If the time taken by the body to reach the bottom of the liquid is t, then

d = \(u t+\frac{1}{2} a t^2=\frac{1}{2} a t^2 \quad(because u=0)\)

= \(\frac{1}{2} \frac{(\rho-\delta)}{\rho} g t^2 \text { or, } t^2=\frac{2 d \rho}{(\rho-\delta) g}\)

∴ t = \(\left[\frac{2 d \rho}{(\rho-\delta) g}\right]^{\frac{1}{2}}\)

Example 5. The weight of a body in air is 0.4 g x g. Its weight along with an sinker in water = 3.37 gx g.The weight of the sinker in air = 4 g x g. Find the specific gravity of the body. [Specific gravity of the sinker = 8]
Solution:

Given

The weight of a body in air is 0.4 g x g. Its weight along with an sinker in water = 3.37 gx g.The weight of the sinker in air = 4 g x g.

Let the density of the body be d g • cm-3.

Volume of the body = \(\frac{0.4}{d} \mathrm{~cm}^3\); volume of the sinker = 4/8 = 0.5 cm3

Volume of water displaced by the body along with the sinker = \(\left(0.5+\frac{0.4}{d}\right)\) cm

.-. Weight of displaced water = (\(\left(0.5+\frac{0.4}{d}\right)\)) g x g

Again, weight of displaced water = weight of the body along with sinker in air – weight of the body along with sinker in water

= (0.4 + 4) x g- 3.37 x g =(4.4-3.37) x g = 1.03g x g

.-. \(0.5+\frac{0.4}{d}=1.03 \text { or, } \frac{0.4}{d}=0.53\)

or, d = \(\frac{0.4}{0.53}\)= 0.75

∴ The specific gravity of the body = 0.75.

Example 6. A piece of wood of volume 20.5 cm³ is tied to a piece of lead of volume 1 cm³. State whether the combination will sink or float in water. [Specific gravity of wood and lead are respectively 0.5 and 11.4]
Solution:

Given

A piece of wood of volume 20.5 cm³ is tied to a piece of lead of volume 1 cm³.

Weight of the piece of wood = 20.5 x 0.5 x g = 10.25 g x g

Weight of lead = 1 x 11.4 x g = 11.4 g x g

∴ Total weight of the combination = (10.25 +11.4) x g = 21.65 g xg

Total volume of wood and lead = (20.5+1) = 21.5 cm³

∴ Weight of displaced water by wood and lead = 21.5 g x g

∴ Weight of combination > weight of displaced water

So, when the piece of wood and lead are tied together, the combination will sink in water.

Example 7. A piece of an alloy of gold and silver weighs 25 g xg in air and 23 g x g in water. Find out the amount of gold and silver in the piece of alloy. [The specific gravity of silver is 10.8 and that of gold is 19.3]
Solution:

Given

A piece of an alloy of gold and silver weighs 25 g xg in air and 23 g x g in water.

Let the amount of gold in the alloy be x g.

∴ Amount of silver =(25-x)g

Volume of gold = \(\frac{x}{19.3} \mathrm{~cm}^3 \text {; }\)

volume of silver = \(\frac{25-x}{10.8} \mathrm{~cm}^3\)

volume of alloy = \(\left(\frac{x}{19.3}+\frac{25-x}{10.8}\right) \mathrm{cm}^3\)

Again, weight of water displaced by the alloy = (25 – 23) x g = 2 g x g

i.e., mass of this displaced water = 2 g

∴ Volume of displaced water = 2 cm³

∴ \(\frac{x}{19.3}+\frac{25-x}{10.8}=2\)

or, 10.8x + 19.3 x 25 – 19.3x = 2 x 19.3 x 10.8

or, x = 7.72

∴ Amount of gold =7.72 g and amount of silver = 25-7.72 = 17.28 g.

Example 8. There is a cavity inside a piece of metal of mass 237.3 g. The apparent weight of the piece of metal immersed completely in water is 192.1 g x g. If the density of the metal is 7.91 g · cm^-3, then find the volume of the cavity.
Solution:

Given

There is a cavity inside a piece of metal of mass 237.3 g. The apparent weight of the piece of metal immersed completely in water is 192.1 g x g. If the density of the metal is 7.91 g · cm^-3,

Volume of the material of the metal piece = \(\frac{237.3}{7.91}=30 \mathrm{~cm}^3 \text {; }\)

mass of water displaced by the metal piece = (237.3-192.1)  x g = 45.2g x g

∴ Volume of displaced water = 45.2 cm³

∴ Volume of the metal piece including the cavity = 45.2 cm³

∴ Volume of the cavity =45.2-30 = 15.2 cm³

Example 9. A piece of wood weighs 40 g x g In air. A piece of brass of mass 12 g is tied to the wooden piece and the combination floats just totally immersed in water. If the specific gravity of brass is 8.5, calculate the specific gravity of wood. [Density of water = 1g · cm^-3]
Solution:

Given

A piece of wood weighs 40 g x g In air. A piece of brass of mass 12 g is tied to the wooden piece and the combination floats just totally immersed in water. If the specific gravity of brass is 8.5,

Let the density of the piece of wood be ρ g • cm^-3.

Total volume of the pieces of wood and brass = \(\left(\frac{40}{\rho}+\frac{12}{8.5}\right) \mathrm{cm}^3\)

Volume of displaced water = \(\left(\frac{40}{\rho}+\frac{12}{8.5}\right) \mathrm{cm}^3\)

By the condition, the weight of the piece of wood + weight of the piece of brass = weight of water displaced by both of them

or, 40 + 12 = \(\left(\frac{40}{\rho}+\frac{12}{8.5}\right) \times 1 \text { or, } 52=\frac{40}{\rho}+\frac{12}{8.5}\)

or, \(\frac{40}{\rho}=52-1.41 \text { or, } \rho=\frac{40}{50.59}=0.79\)

∴ Specific gravity of wood is 0.79.

Example 10. The apparent weights of two bodies suspended from the two ends of a balance beam and immersed in water are the same. The mass and density of one body are 32 g and 8 g · cm^-3 respectively. If the density of the other body is 5 g · cm^-3, then find its mass.
Solution:

Given

The apparent weights of two bodies suspended from the two ends of a balance beam and immersed in water are the same. The mass and density of one body are 32 g and 8 g · cm^-3 respectively.

Let the mass of the second body be mg.

Volume of the first body = 32/8 = 4 cm³

∴ Volume of water displaced by the first body = 4 cm³

Volume of the second body = m/5 cm³

∴ Volume of water displaced by the second body = m/5 cm3

By the problem, apparent weight of the first body = apparent weight of the second body

∴ Real weight of the first body – weight of water displaced by the first body

= real weight of the second body – weight of water displaced by the second body

or, 32 – 4 = m – m/5 or, 28 = 4m/5 or, m = 35

∴ Mass of the second body is 35 g.

Example 11. body of mass 25 g is underwater at a depth of 50 cm. If the specific gravity of the material of the body is 5 and g = 980 cm • s^-2, find the amount of work required to lift it very slowly to the surface.
Solution:

Given

Body of mass 25 g is underwater at a depth of 50 cm. If the specific gravity of the material of the body is 5 and g = 980 cm • s^-2,

The volume of water displaced by the body = volume of the body = 25/5 = 5 cm³

∴ Weight of displaced water = 5 g x g

∴ Downward resultant force = (25 x 980 – 5 x 980)

= 20 x 980 = 19600 dyn

∴ Required work done = 19600 x 50 = 9.8 x 10⁵ erg.

Example 12. A solid spherical ball having density d and volume v floats on the interface of two immiscible liquids. The density of the liquid in the upper portion is d₁ and that of the liquid in the lower portion is d₂. What parts of the ball will remain in the liquids in the upper and lower portions respectively, if d₁ < d < d₂?
Solution:

Given

A solid spherical ball having density d and volume v floats on the interface of two immiscible liquids. The density of the liquid in the upper portion is d₁ and that of the liquid in the lower portion is d₂.

Suppose a volume x of the ball remains in the liquid in the upper portion. So, a volume (ν- x) remains in the liquid in the lower portion.

According to the principle of floatation, weight of the body = weight of displaced liquid

or, \(v d=x d_1+(\nu-x) d_2 \text { or, } x\left(d_1-d_2\right)=v\left(d-d_2\right)\)

or, \(\frac{x}{v}=\frac{d-d_2}{d_1-d_2}=\frac{d_2-d}{d_2-d_1}\)

∴ The part of the ball that remains in the liquid in the upper portion = \(\frac{d_2-d}{d_2-d_1}\)

and the part of the ball that remains in the liquid in the lower = \(\left(1-\frac{d_2-d}{d_2-d_1}\right)=\frac{d-d_1}{d_2-d_1} .\)

Example 13. An object of density 12 g · cm^-3 is weighed with some counterpoising weights made of brass with the help of a common balance. Density of brass = 8 g · cm^-3. If the buoyancy of air is neglected in this measurement, then calculate the percentage error in the measurement of mass. Density of air = 1.2 x 10^-3 g · cm^-3.
Solution:

Given

An object of density 12 g · cm^-3 is weighed with some counterpoising weights made of brass with the help of a common balance. Density of brass = 8 g · cm^-3. If the buoyancy of air is neglected in this measurement

Let the real weight of the body be mgxg; real weight of the brass weights = m’ g x g.

Here, apparent weight of the body = apparent weight of the brass weights

or, real weight of the body – weight of air displaced by the body

= real weight of the brass weights – weight of air displaced by the brass weights

or, \(m-\frac{m}{12} \times 1.2 \times 10^{-3}=m^{\prime}-\frac{m^{\prime}}{8} \times 1.2 \times 10^{-3}\)

or, \(m-m \times 10^{-4}=m^{\prime}-m^{\prime} \times 1.5 \times 10^{-4}\)

or, \(\frac{m^{\prime}}{m}=\frac{1-10^{-4}}{1-1.5 \times 10^{-4}}\)

or, \(\frac{m^{\prime}-m}{m}=\frac{(1.5-1) \times 10^{-4}}{1-1.5 \times 10^{-4}}=0.5 \times 10^{-4}\)

∴ Error = \(\frac{m^{\prime}-m}{m} \times 100=0.5 \times 10^{-4} \times 100=0.005 \%\).

Example 14. A robber ball of mass 10 g and radius 2 cm is submerged in water up to a depth of 5 cm and released. Find the height up to which the ball pops up above the surface of water, neglecting the resistance of water and air. The density of water = 1 g · cm^-3.
Solution:

Given

A robber ball of mass 10 g and radius 2 cm is submerged in water up to a depth of 5 cm and released.

Volume of the rubber ball,

V = \(\frac{4}{3} \times \frac{22}{7} \times(2)^3 \mathrm{~cm}^3=\frac{704}{21} \mathrm{~cm}^3\)

Resultant upward force = buoyant force – weight of the rubber

or, 10a = V x 1 x g – 10 x g [a = upward accleration]

or, \(10 a=\frac{704 \mathrm{~g}}{21}-10 \mathrm{~g} \text { or, } a=\frac{494}{210} \mathrm{~g} \mathrm{~cm} \cdot \mathrm{s}^{-2}\)

Let the velocity of the ball when it reaches the surface of water be v, then

⇒ \(\nu =\sqrt{2 a s}=\sqrt{\frac{2 \times 494}{210} \times 980 \times 5}\)

= \(\sqrt{\frac{2 \times 494 \times 14 \times 5}{3}}=151.83 \mathrm{~cm} \cdot \mathrm{s}^{-1}\)

If the ball pops up to a height x above the surface of water, then

⇒ \(\nu^2=2 g x \text { of, } x=\frac{v^2}{2 g}=\frac{(151.83)^2}{2 \times 980}=11.76 \mathrm{~cm} .\)

Example 15. The density of the material of a hollow sphere of radius 11 Is ρ. Prove that the sphere can float on water If the thickness (t) of Its wall satisfies the relation: \(t<\frac{R}{3 \rho}\).
Solution:

Given

The density of the material of a hollow sphere of radius 11 Is ρ.

According to the condition of floatation, weight of the sphere ≤ weight of the same volume of water

or, \(\left[\frac{4}{3} \pi R^3-\frac{4}{3} \pi(R-t)^3\right] \rho\) ≤ \(\frac{4}{3} \pi R^3 \times 1\)

or, \(\left[\frac{4}{3} \pi R^3-\frac{4}{3} \pi R^3\left(1-\frac{t}{R}\right)^3\right] \rho\) ≤ \(\frac{4}{3} \pi R^3\)

or, \(\frac{4}{3} \pi R^3\left[1-\left(1-\frac{t}{R}\right)^3\right] \rho\) ≤ \(\frac{4}{3} \pi R^3\)

or, \(\left[1-\left(1-\frac{3 t}{R}\right)\right] \rho\) ≤1

[t ≤ R terms containing \(\frac{t^3}{R^3}, \frac{t^2}{R^2}\) are neglected]

or, \(\frac{3 t}{R} \rho\) ≤ 1 or, \(t ≤ \frac{R}{3 \rho}\).

Example 16. When a body is Immersed separately into three liquids of specific gravities S₁, S₂, and S₃, its apparent weights become W₁, W_2, and W₃ respectively. Show that \(S_1\left(W_2-W_3\right)+S_2\left(W_3-W_1\right)+S_3\left(W_1-W_2\right)=0\).
Solution:

Given

When a body is Immersed separately into three liquids of specific gravities S₁, S₂, and S₃, its apparent weights become W₁, W_2, and W₃ respectively.

Let the real weight of the body be W and the volume of the body be V.

∴ \(W_1=W-V \cdot S_1\)……..(1)

∴\(W_2=W-V \cdot S_2\)…..(2)

∴\(W_3 =W-V \cdot S_3\)…..(3)

Subtracting (2) from (1), we get

∴ \(W_1-W_2=V\left(S_2-S_1\right)\)

Subtracting (3) from (2), we get \(W_2-W_3=V\left(S_3-S_2\right)\)

Subtracting (1) from (3), we get \(W_3-W_1=V\left(S_1-S_3\right)\)

∴ \(S_1\left(W_2-W_3\right)+S_2\left(W_3-W_1\right)+S_3\left(W_1-W_2\right)\)

= \(S_1 V\left(S_3-S_2\right)+S_2 V\left(S_1-S_3\right)+S_3 \cdot V\left(S_2-S_1\right)\)

= \(V\left[S_1\left(S_3-S_2\right)+S_2\left(S_1-S_3\right)+S_3\left(S_2-S_1\right)\right]\)

= 0

\(S_1\left(W_2-W_3\right)+S_2\left(W_3-W_1\right)+S_3\left(W_1-W_2\right)=0\)

Hence Proved.

Example 17. A body Is floating on mercury keeping its – th part inside mercury. Now, water is poured into the container such that the body remains just immersed totally in water. Now what part of the body will remain immersed in mercury? (Density of mercury = 13.6 g · cm^-3]
Solution:

Given

A body Is floating on mercury keeping its – th part inside mercury. Now, water is poured into the container such that the body remains just immersed totally in water.

Let the volume of the body be V; density of its material be d.

In the case of floatation in mercury, \(V \times d=\frac{V}{4} \times 13.6 \text { or, } d=\frac{13.6}{4}=3.4 \mathrm{~g} \cdot \mathrm{cm}^{-3}\)

When water is added, suppose x part of the volume of the body remains immersed in mercury and (1 – x) part of its volume in water.

∴  x 3.4 = V x x x 13.6 + V(1 – x) x 1

or, 3.4 = 13.6 + 1 -x

or, 12.6x = 2.4

or, \(x=\frac{2.4}{12.6}=\frac{4}{21}\)

So, 4/21 th part of the given body will remain immersed in

Example 18. A sugar crystal of mass 40 g is coated with wax of mass 5.76 g. The specific gravity of wax is 0.96. If the wax-coated sugar crystal weighs 14.76 g xg inside i water, calculate the specific gravity of sugar.
Solution:

Given

A sugar crystal of mass 40 g is coated with wax of mass 5.76 g. The specific gravity of wax is 0.96. If the wax-coated sugar crystal weighs 14.76 g xg

Let the density of sugar be \(d \mathrm{~g} \cdot \mathrm{cm}^{-3}\).

∴ Volume of sugar = \(\frac{40}{d} \mathrm{~cm}^3\);

volume of wax = \(\frac{5.76}{0.96} \mathrm{~cm}^3 \text {. }\)

Mass of water displaced by the coated sugar

=(40+5.76) x g -14.76 x g

= 31 g x g.

∴ Volume of displaced water = 31 \(\mathrm{~cm}^3\)

∴ \(\frac{40}{d}+\frac{5.76}{0.96}=31\)

or, \(\frac{40}{d}+6=31\)

or, \(\frac{40}{d}=25\)

or, \(d=\frac{40}{25}=1.6\)

∴ Specific gravity of sugar =1.6

Example 19. When a loaded cargo ship enters a river from the sea, it sinks by a length x. When the ship is totally unloaded, it rises by a length y. When the unloaded ship again enters the sea, it rises by z cm more. If the density of water of the river is ρ_w and the body of the ship is vertical, show that the density of sea water is \(\frac{y \rho_w}{(z-x+y)}\).
Solution:

Given

When a loaded cargo ship enters a river from the sea, it sinks by a length x. When the ship is totally unloaded, it rises by a length y. When the unloaded ship again enters the sea, it rises by z cm more. If the density of water of the river is ρ_w and the body of the ship is vertical,

Let the weight of the loaded ship be w; the weight of cargo be w’ the base-area of the ship be A and the density of seawater be p.

Let us also assume that the loaded ship sinks a length h in sea water.

∴ In the case of floatation in seawater, w = Ahρ ……(1)

In the case of floatation in river water, w = A(h + x)ρ_w …(2)

When the ship is unloaded, w- w’ = A(h + x-y)ρ_w ….(3)

When the unloaded ship enters the sea, w-w’ = A(h + x-y-z)ρ….(4)

From (1) and (2), we get

⇒ \(A h \rho=A(h+x) \rho_w \text { or, } h+x=\frac{h \rho}{\rho_w}\)

From (3) and (4), we get

or, \(h+x-y=(h+x-y-z) \frac{\rho}{\rho_w}\)

or, \(\frac{h \rho}{\rho_w}-y=\frac{h \rho}{\rho_w}+(x-y-z) \frac{\rho}{\rho_w}\)

or, \(y=(y+z-x) \frac{\rho}{\rho_w}\)

or, \(\rho=\frac{y \rho_w}{(z-x+y)}\).

Example 20. A cubical steel block floats on mercury erectly. If each side of the cube is 10 cm,

1. How much of the body is above the mercury surface?
2. If water is poured over mercury and it just covers the upper surface of the cube, then what will be the depth of the water column? Densities of steel and mercury are 7.8 g · cm^-3 and 13.6g · cm^-3 respectively.

Solution:

Given

A cubical steel block floats on mercury erectly. If each side of the cube is 10 cm,

1. Suppose x cm of the cube remajps immersed in mercury.

So, according to the condition of floatation, volume of the cube x density of the material of the cube = volume of mercury displaced x density of mercury

∴ 103³x 7.8 = 10² x xx 13.6 or, x = 7.8/13.6 = 5.74

∴ (10-5.74) = 4.26 cm height of the cube remains above mercury.

2. Let the length of the cube that remains in water be h cm and that in mercury be (10 – h) cm.

∴ According to the condition of floatation, volume of the cube x density of the material of the cube = volume of displaced water x density of water + volume of displaced mercury x density of mercury

∴ 10³ x 7.8 = 10² x h x 1 + 10² x (10-h) x 13.6

or, 78 = h + (10- h) x 13.6 or, h =

∴ Depth of the water column = 4.6 cm.

Example 21. A and B are two cylinders identical in size but made of different materials. A floats with half of itself immersed in a liquid. When B is placed on A, both j the cylinders remain floating in the liquid, but completely immersed in it. Compare the density of the liquid and the densities of the materials of A and B.
Solution:

Given

A and B are two cylinders identical in size but made of different materials. A floats with half of itself immersed in a liquid. When B is placed on A, both j the cylinders remain floating in the liquid, but completely immersed in it.

Let the volume of each of the cylinders A and B be V. Also, let the densities of the materials of A and B and that of the liquid be d₁, d_2, and d₃ respectively.

When cylinder A alone floats on the liquid, then volume of the cylinder x density of its material = volume of displaced liquid x density of the liquid

or, \(V \times d_1=\frac{V}{2} \times d_3 \text { or, } d_1=\frac{d_3}{2}\)

When cylinders A and B together float in the liquid, then volume of the cylinder A x its density + volume of the cylinder B x its density = volume of displaced liquid x its density

or, \(V d_1+V d_2=2 V \times d_3\)

or, \(d_1+d_2=2 d_3\)

or, \(\frac{d_3}{2}+d_2=2 d_3\)

or, \(d_2=2 d_3-\frac{d_3}{2}=\frac{3 d_3}{2}\)

∴ \(d_1: d_2: d_3=\frac{d_3}{2}: \frac{3 d_3}{2}: d_3=1: 3: 2\)

Example 22. The apparent weight of a body In a liquid of density d₁ is m₁ and, in another liquid of density d₂, its apparent weight is m₂. What will be its apparent weight in a liquid of density d₃?
Solution:

Given

The apparent weight of a body In a liquid of density d₁ is m₁ and, in another liquid of density d₂, its apparent weight is m₂.

Let the real weight of the body be M and its volume be V.

∴ M- Vd₁ = m₁ ………(1)

and M- Vd₂ = m₂ ……..(2)

From equations (1) and (2), we get

V = \(\frac{m_1-m_2}{d_2-d_1}\)

Let the apparent weight of the body in the liquid of density d₃ be m₃.

∴ \(m_3=M-V d_3=m_1+V d_1-V d_3\) [with the help of equation (1)]

= \(m_1+V\left(d_1-d_3\right)\)

= \(m_1+\frac{\left(m_1-m_2\right)\left(d_1-d_3\right)}{\left(d_2-d_1\right)}\)

= \(\frac{m_1\left(d_2-d_1\right)+\left(m_1-m_2\right)\left(d_1-d_3\right)}{\left(d_2-d_1\right)}\)

= \(\frac{m_1\left(d_2-d_3\right)-m_2\left(d_1-d_3\right)}{\left(d_2-d_1\right)}\)

Example 23. A glass ball of density 2.6 g · cm^-3 is coated with a thick layer of wax of density 0.8 g · cm^-3. If the combination floats in water remaining completely submerged, find the ratio of the volume of wax to that of the glass ball.
Solution:

Given

A glass ball of density 2.6 g · cm^-3 is coated with a thick layer of wax of density 0.8 g · cm^-3.

Let volume of glass be V₁ cm³ and volume of wax be V₂ cm³.

According to the problem, weight of glass ball + weight of wax = weight of displaced water

or, \(V_1 \times 2.6 \times g+V_2 \times 0.8 \times g=\left(V_1+V_2\right) \times 1 \times g\)

or, \(V_1(2.6-1)=V_2(1-0.8)\)

or, \(\frac{V_2}{V_1}=\frac{1.6}{0.2}=\frac{8}{1}\)

∴ Volume of wax: volume of glass =8:1.

Example 24. A body can float on a liquid keeping J th of It vol-ume outside the liquids The body is completely immersed inside the liquid and then released. What will be its upward acceleration at that time?
Solution:

Given

A body can float on a liquid keeping J th of It vol-ume outside the liquids The body is completely immersed inside the liquid and then released.

Let volume of the body be V and density be ρ; density of the liquid be ρ’.

According to the condition of floatation,

⇒ \(V \times \rho \times g=\frac{3}{4} V \times \rho^{\prime} \times g \text { or, } \rho^{\prime}=\frac{4 \rho}{3}\)

When the body is completely immersed in the liquid, the net upward force = buoyant force – weight

or, \(V \rho a=V \rho^{\prime} g-V \rho g\)

[a= upward acceleration]

or, \(V \rho a=V \cdot \frac{4}{3} \rho \cdot g-V \rho g=\frac{1}{3} V \rho g\)

or, \(a=\frac{1}{3} g\).

Example 25. A drop of oil rises within water with an upward acceleration of ag. If or is a constant and g is the acceleration due to gravity, find the specific gravity of the oil Neglect the friction of water.
Solution:

Given

A drop of oil rises within water with an upward acceleration of ag. If or is a constant and g is the acceleration due to gravity

Let the densities of oil and water be D and d respectively and the mass of the oil drop be m.

∴ Volume of the oil drop = \(\frac{m}{D}\) = volume of water displaced by the oil drop

∴ Mass of displaced water = density of water x volume of displaced water = \(\frac{d m}{D}\)

∴ Buoyancy = weight of displaced water = \(\frac{d m g}{D}\)

∴ Net upward thrust on the oil drop = buoyant force – weight of the oil drop

= \(\frac{d m g}{D}-m g=\left(\frac{d}{D}-1\right) m g\)

∴ Acceleration of the oil drop inside water,

⇒ αg= \(\frac{\left(\frac{d}{D}-1\right) m g}{m}\) or, α =\(\left(\frac{d}{D}-1\right)\)

or, \(\frac{d}{D}=1+\alpha\)

Example 26. A stone of density 2.5 g · cm^-3 Is completely  Immersed In sen water and Is allowed to sink from rest. Calculate the depth uttalned by the stone In 2 s. Neglect the effect of friction. The specific gravity of seawater Is 1.025, acceleration due to gravity = 980 cm · s^-2.
Solution:

Given

A stone of density 2.5 g · cm^-3 Is completely  Immersed In sen water and Is allowed to sink from rest

Let the mass of the stone be m g, So, the volume of the stone = 3/2.5 cm³.

When the stone is allowed to sink in seawater, the resultant downward force weight of the stone upthrust

or, \(m a=m g-\frac{m}{2.5} \times 1.025 \times g\)

[a = downward acceleration of the stone]

or, \(a=980-\frac{1.025}{2.5} \times 980=578.2 \mathrm{~cm} \cdot \mathrm{s}^{-2} .\)

If the stone attains a depth h in 2 8, then

h = \(\frac{1}{2} a t^2=\frac{1}{2} \times 578.2 \times(2)^2=1156.4 \mathrm{~cm}=11.564 \mathrm{~m} \text {. }\)

Example 27. A uniform cube of side 10 cm weighs 880 g x g. It is floating in saline water (specific gravity of saline water = 1.1). What will be the thrust on each face of the cube?
Solution:

Given

A uniform cube of side 10 cm weighs 880 g x g. It is floating in saline water (specific gravity of saline water = 1.1).

Density of the material of the cube

= \(\frac{880}{10^3}=0.88 \mathrm{~g} \cdot \mathrm{cm}^{-3} \text {. }\)

Since the density is less than the density of saline water, some part of the cube remains above water. Let the lower surface of the cube be at a depth of x cm from the upper surface of saline water.

So, according to the condition of floatation, weight of the cube = weight of the saline water displaced by the cube

or, 880 x g = x x 10 x 10 x l .l x g or, x = 8 cm

So, the area of the part of a side of the cube that remains immersed in the saline water, A = 8×10 = 80 cm²

and the average depth that of the surface = \(\frac{0+x}{2}=\frac{x}{2}\)

Hence, the thrust on each lateral surface of the cube

= \(\frac{x}{2} \times 1.1 \times g \times A=\frac{8}{2} \times 1.1 \times 981 \times 80\)

Example 28. A glass test tube with a plane base has a diameter of 4 cm and mass 30 g. Centre of gravity of the empty test tube is at a height of 10 cm from the base. Find the amount of water to be taken in the test tube so that when it floats vertically, its centre of gravity shifts to the midpoint of the immersed portion of the tube.
Solution:

Given

A glass test tube with a plane base has a diameter of 4 cm and mass 30 g. Centre of gravity of the empty test tube is at a height of 10 cm from the base

The centre of gravity P of the empty tube is 10 cm above the point O. Let the height of water level taken in the tube be h cm. Midpoint of the water column is Q.

Let the centre of gravity of the test tube-water system be at R. Hence, part of the tube immersed in water, AC = 2x OR.

Volume of water in the tube = \(\pi(2)^2 h=4 \pi h \mathrm{~cm}^3\)

Mass of the tube + water in it = \((30+4 \pi h) \mathrm{g}\)

From the condition of floatation,

⇒ \(\pi\left(2^2\right) \times A C=30+4 \pi h\)

∴ AC = \(2 \times O R=\frac{30}{4 \pi}+h=d \text { (say). }\)

Taking a moment about the point R, \(4 \pi h \times R Q \stackrel{\circ}{=} 30 \times P R\)

or, \(4 \pi h[O R-O Q]=30 \times[O P-O R]\)

or, h = 8.806 cm

Hence the required mass of water -n x 8.806 = 110.66 g

or, \(4 \pi h\left[\frac{d}{2}-\frac{h}{2}\right]=30\left[10-\frac{d}{2}\right]\)

or, \(4 \pi h\left[\frac{1}{2}\left(\frac{30}{4 \pi}+h\right)-\frac{h}{2}\right]=30\left[10-\frac{1}{2}\left(\frac{30}{4 \pi}+h\right)\right]\)

or, h=8.806 cm

Hence the required mass of water = \(4 \pi \times 8.806=110.66 \mathrm{~g}\).

Example 29. A body of uniform cross-section remains floating in a liquid. The density of the liquid is 3 times that of the body. What part of that body remains outside the liquid?
Solution:

Given

A body of uniform cross-section remains floating in a liquid. The density of the liquid is 3 times that of the body.

Let the volume of the body be V, density be ρ and density of the liquid be 3ρ.

Let us assume that x part of the body remains immersed in the liquid.

So, according to the condition of floatation, weight of the body = weight of displaced liquid

or, \(V \rho g=V x \cdot 3 \rho \cdot g \text { or, } x=\frac{1}{3}\)

∴ \(\left(1-\frac{1}{3}\right) \text { or } \frac{2}{3}\) part of the body remains outside the liquid.

Example 30. When a 300 g mass is placed over a wooden cube, the cube just floats in water. When the mass is removed, the cube comes out by 4 cm from the water. Determine the length of each side of the cube.
Solution:

Given

When a 300 g mass is placed over a wooden cube, the cube just floats in water. When the mass is removed, the cube comes out by 4 cm from the water.

Let the mass of the cube be mg; the length of each side of the cube be l cm.

∴ According to the problem for the first case, (m + 300)g = l³ x 1 x g ….(1)

For the second case, mg = l²(l-4) x 1 x g….(2)

∴ From (1) and (2), we get

300 = l³ – l³ + 4l²

or, l² = 75

or, l = 5√3

∴ Each side of the cube is 5√3 cm in length.

Example 31. From the two arms of a beam balance, a metal body weighing 20 g and a piece of glass are suspended. The apparent weights of these two bodies when measured in water are found to be the same. If immersed in alcohol instead of water, the mass of the metal needs to be increased by 0.84 g for balancing the beam. Calculate die mass of the glass piece. [Given, density of water =1 g·cm^-3; density of alcohol = 0.96 g·cm^-3]
Solution:

Given

From the two arms of a beam balance, a metal body weighing 20 g and a piece of glass are suspended. The apparent weights of these two bodies when measured in water are found to be the same. If immersed in alcohol instead of water, the mass of the metal needs to be increased by 0.84 g for balancing the beam.

Let us assume that the weight of the glass piece is mg

Let the density of the metal body = d₁ g · cm^-3, and the density of glass = d₁ g · cm^-3

According to the question, when the bodies are immersed in water, apparent weight of the metal body = apparent weight of the glass piece

or, real weight of the metal body – weight of the water displaced by the metal body

= real weight of the glass piece – weight of the water displaced by the glass piece

or, \(\left(20-\frac{20 \times 1}{d_1}\right) \times g=\left(m-\frac{m \times 1}{d_2}\right) \times g\)

or, \(\frac{m}{d_2}-\frac{20}{d_1}=m-20\)

When immersed in alcohol,

⇒ \(\left(20-\frac{20}{d_1} \times 0.96\right) \times g+0.84 \times g=\left(m-\frac{m}{d_2} \times 0.96\right) \times g\)

or, \(m\left(1-\frac{0.96}{d_2}\right)-20\left(1-\frac{0.96}{d_1}\right)=0.84\)

or, \(m-\frac{m \times 0.96}{d_2}-20+\frac{20 \times 0.96}{d_1}=0.84\)

or, \(m-0.96\left(\frac{m}{d_2}-\frac{20}{d_1}\right)=20.84\)

or, \(m-0.96(m-20)=20.84\) [From equation (1)]

or, \(0.04 m=20.84-19.2\)

or, \(m=\frac{1.64}{0.04}=41\)

∴ The mass of the glass piece is 41g.

Example 32. A piece of silver weighing 105 gf and another piece of glass weighing 130 gf are placed on the right and left pan of a balance respectively. Which pan will move down when the die balance is immersed in water? [Given, the density of silver, ρ₁ = 10.5 g • cm^-3 and density of glass, ρ₂ = 2.6 g · cm^-3]
Solution:

Given

A piece of silver weighing 105 gf and another piece of glass weighing 130 gf are placed on the right and left pan of a balance respectively.

Here weight of the piece of silver, W₁ = 105 gf, and weight of the piece of glass, W₂ = 130 gf.

Now, volume of the piece of silver,

⇒ \(V_1=\frac{105}{\rho_1}=\frac{105}{10.5}=10 \mathrm{~cm}^3\)

and volume of the piece of glass,

⇒ \(V_2=\frac{130}{\rho_2}=\frac{130}{2.6}=50 \mathrm{~cm}^3\)

Since density of water is 1 g · cm^-3, weight of the water displaced by the pieces of silver and glass will be 10 gf and 50 gf respectively.

Therefore on immersing the balance in water, the apparent weight of the piece of silver,

W’₁ = W₁ – 10 = 105-10 = 95 gf

and the apparent weight of the piece of glass,

W’₂ = W₂ -50 = 130-50 = 80 gf

Since W’₁ < W’₂, the right pan i.e. the pan having piece of silver on it, will move down.

Example 33. A body Is fully immersed in water. Calculate the change in potential energy of the body, raised to a height h inside the water.
Solution:

Given

A body Is fully immersed in water.

Let ρ and ρ₀ be the density of the material of the body and the water respectively. If V is the volume of the body, then apparent weight of the body

= Vρg- Vρ₀g = V(ρ-ρ₀)g

Therefore, increase in potential energy of the body,

ΔU = work done in raising the body to a height h = V(ρ-ρ₀)g x h

Example 34. A solid uniform ball of volume V and density ρ floats on the interface of two immiscible liquids as shown In the figure. The density of the upper liquid is ρ₁ and that of the lower liquid is ρ₂ (ρ₂>ρ>ρ₁).What fraction of the volume of the ball will be in the upper liquid and what fraction will be in the lower one?

Solution:

Given

A solid uniform ball of volume V and density ρ floats on the interface of two immiscible liquids as shown In the figure. The density of the upper liquid is ρ₁ and that of the lower liquid is ρ₂ (ρ₂>ρ>ρ₁).

Let V be the volume of the ball and a fraction x of its volume is in the upper liquid.

Then, volume of the ball inside the upper liquid =xV, and volume of the ball inside the lower liquid = V- xV = (1-x)V

Due to buoyancy, the net force on the ball in upward direction = [xρ₁+(1-x)ρ₂]V_g

At equilibrium, \(\left[x \rho_1+(1-x) \rho_2\right] V g\)=\(V \rho g\)

or, \(x \rho_1+(1-x) \rho_2=\rho\)

or, \(x\left(\rho_2-\rho_1\right)=\rho_2-\rho\)

Hence, the fraction of the volume of the ball inside the upper liquid,

x = \(\frac{\rho_2-\rho}{\rho_2-\rho_1}\)

Example 35. To what height should a cylindrical vessel be filled with a homogeneous liquid such that the force exerted by the liquid on the wall of the vessel be equal to the force exerted by the liquid on the bottom of the vessel?
Solution:

Let h be the height of the liquid of density p in the cylinder of radius r. The pressure of the liquid column at the bottom of the cylinder is hρg.

Therefore, the force exerted by the liquid column on the bottom of the cylinder is

F₁ = pressure x area of the bottom = hρg x πr²

The pressure of the liquid column on the wall of the vessel varies from zero at the top of the column to hρg at the bottom.

Thus, the average pressure on the wall is \(\frac{0+h \rho g}{2}=\frac{1}{2} h \rho g\)

Therefore, the force exerted by the liquid column on the wall of the cylinder is

F₂ = pressure x area of wall = 1/2 hρgx 2πrh

For F₂ = F₁ we should have 1/2 hρg x 2πrh = hρg x πr²

or, h = r

i.e., the height of the liquid column in the cylinder should be equal to the radius of the cylinder.

Hydrostatics Transmission Of Fluid Pressure Pascal’s Law

If pressure is applied to a small region of a solid, then it is not transmitted throughout the entire volume. It only produces a local deformation in the solid. Since a fluid is not rigid like a solid, any similarly applied pressure is transmitted throughout the fluid.

A change in pressure at any point in a confined fluid produces an equal change in pressure at all points in the fluid. Pascal, a French scientist, determined the law of transmission of fluid pressure, which is known as Pascal’s law.

Pascal’s law: The pressure applied at any point of a confined fluid is transmitted with undiminished magnitude in all directions throughout the fluid and acts normally on the surface in contact with the fluid.

Read and Learn More: Class 11 Physics Notes

Demonstration of Pascal’s law

1. Take a rubber ball and fill it with water. Tie its mouth tightly. Take a pin and make fine holes at many points on its surface. If the ball is squeezed, it is seen that jets of water come out in the direction normal to the surface of the ball.

Observe carefully that the jets are all of the same speed and that this speed depends on the squeezing force applied. This shows that the applied pressure is transmitted equally in all directions throughout the liquid and acts normally at every point on the surface.

Examples of Pascal’s Law in Everyday Life

In the above demonstration, the effect of gravity has been ignored. This is justified only if the height h of the rubber ball is sufficiently small, so that the pressure hρg at its bottom is negligible compared to the pressure due to the externally applied squeeze.

2. A spherical vessel with four circular openings is taken. The holes A, B, C, D have cross-sectional areas a, b, c, and d respectively. Each opening is fitted with a piston. When a force is applied on a piston, it can move in or out The vessel is then filled with water or any other liquid. When an inward pressure is applied on any one of the pistons, it is seen that the other pistons move outwards.

This proves that pressure applied to any part of a confined fluid is transmitted in all directions. If one of the pistons is moved inwards, then to keep the other pistons at rest in their respective positions, an external force needs to be applied on each of them from outside.

If a force F₁ is applied on the piston A, then the pressure applied on it = \(\frac{F_1}{a}\). To keep the pistons B, C, and D intact in their respective positions, forces F₂, F₃, and F₄ respectively have to be applied. It is seen that

Understanding Pascal’s Law with Diagrams

i.e., the pressure exerted on all the pistons is the same. It indicates that the pressure applied on the piston A is transmitted with undiminished magnitude in all directions throughout the fluid.

Principle of Multiplication of Thrust from Pascal’s Law: According to Pascal’s law, by applying a small force at any part of a confined liquid, a large force can be obtained at another part of it. This increase in the force is known as the principle of multiplication of thrust.

• A narrow and a wide cylinder are connected by a tube. Let the cross-sectional area of the narrow cylinder be a and that of the wider cylinder be b.
• The cylinders are filled with water and they are fitted with two water-tight frictionless pistons of negligible weights. The pistons have platforms for placing weights on their tops.

Now, a load w₁ is placed on the smaller piston. So, the force applied on the smaller piston = w₁

Therefore, the pressure exerted by the smaller piston on water = \(\frac{w_1}{a}\)

According to Pascal’s law, this pressure is transmitted invariably through water in all directions and it acts upwards on the larger piston.

So, the thrust or force exerted on the larger piston

= \(\text { pressure } \times \text { area }=\frac{w_1}{a} \times b\)

• So, a thrust that is n times the applied force will be effective on the larger piston. The larger the value of n, i.e., the larger the value of b compared to that of a, the heavier the weight w₂ compared to wl.
• For example, if a = 1 cm², b = 200 cm² and w₁ = 1 kg x g, then w₂ = 200kg x g. So, by applying a force of 1 kg x g only on the smaller piston, a 200 kg x g force will be obtained on the larger piston.
• So, a small force acting at any part of a confined liquid can be multiplied at some other part of it. This is called the principle of the multiplication of thrust.

Conservation of energy in the case of multiplication of thrust: It may appear at first sight that the principle of the multiplication of thrust violates the principle of conservation of energy.

But, though we get a greater force on the large piston by applying a lesser force on the small piston, we cannot gain energy. Suppose a force wx on the smaller piston moves it downwards through a distance x. Consequently, the larger piston moves upwards through a distance y.

Since water is incompressible, the volume of water going out of the narrow cylinder = the volume of water added to the wider cylinder.

∴ ax = by or,  \(\frac{x}{y}=\frac{b}{a}\)

Now, the work done on the smaller piston = \(w_1 \times x\) and the work done by the larger piston = \(w_2 \times y\).

But, \(w_2 y=w_1 \times \frac{b}{a} \times y=w_1 \times \frac{x}{y} \times y=w_1 x\)

∴ Work done by the larger piston = work done on the smaller piston,

i. e., the energy expended on the smaller piston can be recovered as the work done by the larger piston. So, the principle of conservation of energy is valid in this case. In the above example, the effect of friction is not considered.

In actual practice, however, some amount of energy is dissipated as heat in performing work against friction and therefore, work done by the larger piston < work done on the smaller piston.

It should be noted that as w₂ >>w₁, y << x and hence, the
displacement of the larger piston is much less than that of the smaller one.

• Hydraulic Press
• Hydraulic Brake
• Hydraulic Lift

Hydraulic press: The working of a hydraulic press is based on the principle of multiplication of thrust. The British engineer Bramah made some improvements on it and so these days it is also called Bramah’s press. In this machine, a large thrust can be developed.

Hydraulic press Description: A schematic diagram of a hydraulic press is shown in Fig It consists of two iron cylinders C₁ and C₂ connected by a tube T₁.

• Two water-tight pistons P₁ and P₂ are fitted inside the cylinders. The cross-sectional area β of the larger piston P₂ is much larger than the cross-sectional area α of the smaller piston P₁. With the help of a lever OL, the piston P₁ can be moved up and down the cylinder C₁.
• A platform P is fitted at the upper end of the piston P₂. The object B which is to be compressed is placed on this platform. The technical details of the apparatus has been omitted in our description.

Hydraulic Press Action: The object which is to be compressed is placed on platform P. According to Pascal’s law, the pressure applied to the piston P₁ with the help of the lever is transmitted to the larger piston P₂.

Since the area of the cross-section of the piston P₂ is much larger than that of the piston P₁, a large thrust is exerted on the piston P₂. As a result, the piston P₂ moves upwards and any object placed between the platform P and the plate is compressed.

Hydraulic Press Total thrust developed: In this machine, thrust is multiplied in two stages. Initially, thrust is multiplied according to the principle of a lever and then it is multiplied again according to the principle of multiplication of thrust by Pascal’s law.

Let a force W be applied at the end of the lever which exerts a force F₁ on the piston P₁. Let the length of the longer arm of the lever = y and the length of the shorter arm = x.

∴ According to the principle of action of the lever

⇒ \(F_1 \cdot x=W \cdot y \quad \text { or, } F_1=W \cdot \frac{y}{x}\) ……(1)

α and β are the cross-sections of the smaller piston P₁ and the larger piston P₂ respectively. Therefore, the pressure exerted on water by the piston P₁ = \(\frac{F_1}{a}\). According to Pascal’s principle, this pressure creates a thrust F₂ on the piston P₂.

∴ Thrust on the piston P₂, \(F_2=\frac{F_1}{\alpha} \times \beta=W \cdot \frac{y}{x} \cdot \frac{\beta}{\alpha}\)…(2)

∴ \(y>x \text { and } \beta \gg \alpha,\)

⇒ \(F_2 \gg W\)

So, the thrust thus developed increases manifold.

Hydraulic Systems Explained by Pascal’s Law

Hydraulic Press Mechanical advantage: Mechanical advantage

= \(\frac{\text { thrust developed }}{\text { applied force }}=\frac{F_2}{W}=\frac{y}{x} \cdot \frac{\beta}{\alpha}\)

As y > x and β >>α,

the mechanical advantage is much greater than 1. It should be mentioned here that a part of the energy is dissipated due to friction and, so, the mechanical advantage becomes slightly less.

Hydraulic Press Uses: A hydraulic press is used for compressing bales of cotton, clothes, paper, or jute, in extracting oil from seeds, and for testing the strength of iron and steel beams.

The use of a hydraulic press is less common nowadays—it is still in use mainly in jute and similar industries.

Hydraulic brake: The principle is essentially that used in the construction of a hydraulic press. It uses a confined brake fluid, usually ethylene glycol, to transfer thrust from the controlling unit to the brake mechanism. The controlling unit is kept near the driver of a vehicle, whereas the actual brake mechanism is near the wheels of the vehicle. The controlling unit is also aided with a first-class lever system that offers a mechanical advantage.

Hydraulic Brake Action: A hydraulic brake uses two cylinders for the multiplication of thrust. The area of the piston attached to the slave cylinder near the wheels is usually four to six times that of the piston in the master cylinder iff control of the driver. The mechanical advantage of the lever is about 3.

• So, to produce a 1cm displacement of the slave piston, the driver has to push down the lever by about 15 cm. The movement of the slave piston then applies a force on the brake pads, which in turn push against the rotating wheel. The friction between the pads and the wheels generates a braking torque that slows down the vehicle.
• A series of springs are set in the system so that, as soon as the driver releases the lever, the brake pads release the wheels.

Some important points to be noted:

1. Hydraulic brakes are smaller and cheaper than air or vacuum brakes.
2. The brake fluid must be incompressible so that a high multiplication of thrust is obtained from a single stroke of the master lever—in this sense, the operation is easy.
3. Water vaporizes easily due to the heat produced from friction inside the system, and also produces corrosion of the metal parts. So water is never used as the brake fluid—light organic oils are more common.

Hydraulic lift: At present times, the use of the device as a hydraulic lift is very common. For example, an automobile workshop uses it to lift motor cars and heavier vehicles for their maintenance and repair.

Hydraulic lift Working principle: Let the area of cross-sections of the smaller piston be A₁ and of the larger piston be A₂(A₂>>A₁) . If the applied force on the smaller piston is F, then the pressure on the liquid is p = \(\frac{F_1}{A_1}\)

According to Pascal’s law, this pressure is transferred to the larger piston with its value unchanged and F₂ thrust is generated.

So the thrust on the larger piston,

F₂ = p x A₂ = (F₁/A₁) X A₂

A₂>> A₁ and so F₂ >> F₁.

So, if some heavy object or car is placed on the larger piston, it can be lifted easily by the increased thrust F₂. Normally the plate on the larger piston is kept at the ground level.

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Hydrostatics Transmission Of Fluid Pressure Pascal’s Law Numerical Examples

Step-by-Step Guide to Understanding Pascal’s Law

Example 1. In a hydraulic press, a 10⁷ dyn force is applied on water by a piston of area 100 cm². If the other piston can raise a car of mass 2000 kg, then find Its base area.
Solution:

Given

In a hydraulic press, a 10⁷ dyn force is applied on water by a piston of area 100 cm². If the other piston can raise a car of mass 2000 kg

We know that, \(\frac{F_1}{\alpha}=\frac{F_2}{\beta}\)

Here, F₁ = 10⁷ dyn, a = 100 cm²,

F₂ = 2000 kg x g = 2000 x .1000 x 980 dyn.

∴ \(\beta=\frac{F_2 \alpha}{F_1}=\frac{2000 \times 1000 \times 980 \times 100}{10^7}=19600 \mathrm{~cm}^2 .\)

Example 2. The diameters of the two pistons of a hydraulic press are 0.1 m and 1 m. The lengths of the two arms of a first-class lever are 0.15 m and 0.9 m. A 50 N force is applied at the end of the lever. Determine the total thrust developed on the larger piston.
Solution:

Given

The diameters of the two pistons of a hydraulic press are 0.1 m and 1 m. The lengths of the two arms of a first-class lever are 0.15 m and 0.9 m. A 50 N force is applied at the end of the lever.

We know that the thrust developed on the larger piston,

⇒ \(F_2=W \cdot \frac{y}{x} \cdot \frac{\beta}{\alpha}\)

Here, force applied at the end of the lever arm, W = 50 N

Area of the larger piston, β = π(0.5)² = 0.25π m²

Area of the smaller piston, α = π(0.05)² = 0.0025π m²

Length of the longer arm of the lever, y = 0.9 m

Length of the smaller arm of the lever, x = 0.15 m

∴ \(F_2=50 \times \frac{0.9}{0.15} \times \frac{0.25 \pi}{0.0025 \pi}=3 \times 10^4 \mathrm{~N} .\)

Hydrostatics Transmission Of Fluid Pressure Problems

Example 3. A bottle completely filled with water is corked. The areas of the mouth and the bottom of the bottle are 10 cm² and 100 cm² respectively and the height of the bottle is 40 cm. If the cork is pressed with a 10 N force, then calculate the total thrust on its bottom.
Solution:

Given

A bottle completely filled with water is corked. The areas of the mouth and the bottom of the bottle are 10 cm² and 100 cm² respectively and the height of the bottle is 40 cm. If the cork is pressed with a 10 N force

Area of a cross-section of the cork = 10 cm²; area of the bottom of the bottle = 100 cm².

Pressure on the cork = 1 N · cm^-2.

According to Pascal’s law, the pressure exerted on the bottom of the bottle, p = 1N · cm^-2.

∴ Thrust on the bottom of the bottle due to the applied pressure, F₁ =1 x 100 = 100 N

Thrust on the bottom of the bottle due to the water in it, F₂ = hρgxA

Here, h = 40cm = 0.4m, ρ = 10³kg · m^-3, g = 9.8m · s^-2,

A = 100cm² = 10^-2m²

∴ F₂ = 0.4 x 10³ x 9.8 x 10^-2 = 39.2 N

∴ Total thrust on the bottom = F₁ + F₂ = 100 + 39.2 = 139.2N

Example 4. A hydraulic automobile lift is used to lift a car of mass 3000 kg. The cross-sectional area of the piston on which the car is supported is 425 cm². What pressure would the smaller piston have to bear if the bigger piston with the car is 3 m above the smaller piston? The density of the oil filling the hydraulic machine is 800 kg · m^-3. Let the pistons be of equal mass.
Solution:

Given

A hydraulic automobile lift is used to lift a car of mass 3000 kg. The cross-sectional area of the piston on which the car is supported is 425 cm².

The positions of the pistons (with a car) of a hydraulic automobile lift is shown in Fig.

Let the area of the larger piston = A₂ and the area of the smaller piston = A₁

If the pressure applied on the smaller piston is p₁, then according to Pascal’s law,

p₁ = pressure on the larger piston + pressure exerted by a 3m liquid column

= \(\frac{W}{A_2}\) + pressure exerted by a 3 m liquid column A2 [W = weight of the car]

= \(\frac{3000 \times 9.8}{425 \times 10^{-4}}+3 \times 800 \times 9.8=7.155 \times 10^5 \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Example 5. The diameters of the larger and the smaller pistons of a hydraulic press are 45 cm and 5 cm respectively. Find the magnitude of the force that has to be applied on the smaller piston to produce a 4050N thrust on the larger piston.
Solution:

Given

The diameters of the larger and the smaller pistons of a hydraulic press are 45 cm and 5 cm respectively.

According to Pascal’s law, \(\frac{F_1}{\pi r_1^2}=\frac{F_2}{\pi r_2^2}\)

[F₁ = force on the smaller piston, r₁ = radius of the smaller piston, F₂ = force on the larger piston, r₂ = radius of the larger piston]

∴ \(\frac{F_1}{r_1^2}=\frac{F_2}{r_2^2} \text { or, } \frac{F_1}{\left(\frac{5}{2}\right)^2}=\frac{4050}{\left(\frac{45}{2}\right)^2}\)

or, \(F_1=\frac{4050 \times 25 \times 4}{4 \times 45 \times 45}=50 \mathrm{~N}\)

Hydrostatics Pressure

Hydrostatic Pressure Definition and Examples

Pressure Definition: The force acting normally on unit area of a surface is called pressure.

Let the force or thrust acting normally on an area A be F. Therefore, pressure,

p = \(\frac{\text { normal force }}{\text { area }}=\frac{F}{A}\)

Examples:

1. The area of the cross-section of the pointed edge of a nail is very small. When its head is hammered, due to a large force exerted on the small area of its point, the pressure is high and the nail enters the surface easily.
2. The area of the cross-section of the sharpened edge of a knife is small. So when a force is applied on the blade, it cuts through easily because of the large pressure developed.
3. When a person stands on his feet, the contact area with the ground becomes less than that when he lies down. That is why when a person stands on soft clay or sand, his feet sink into the surface.

Read and Learn More: Class 11 Physics Notes

Pressure and Thrust of a Liquid: When a liquid is kept in a container, it exerts a force on the bottom and on the walls of the container.

• This force acts normally at every point of contact in the container. From our experience, we know that if there is a hole in the wall of a container of water, the water comes out of the hole at a high speed.
• If we try to block the flow of water with a finger, we feel the pressure. So, we can infer that water exerts pressure on the walls of the container.

Pressure of a liquid: The pressure at a point inside a liquid is defined as the normal force exerted by the liquid on a unit area surrounding the point.

Sometimes, liquid pressure is also called hydrostatic pressure, if the liquid is at rest and no pressure is developed due to its motion.

Let F be the normal force exerted by a liquid on a surface area A surrounding a point inside a liquid. The pressure p at that point is

p = F/A or, F = pA

Thrust of a liquid: The normal force exerted by a liquid on any surface in contact with it is called the thrust of the liquid.

Thrust (F) = pressure (p) x area (A)

Thrust is a force and hence it is a vector quantity. But pressure is a scalar quantity. The direction of thrust \(\vec{F}\) is defined as the direction of the area vector \(\vec{A}\), where A is a plane surface i.e., \(\vec{F}\) = p\(\vec{A}\).

Dimension of pressure: \([p]=\frac{[F]}{[A]}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}=\mathrm{ML}^{-1} \mathrm{~T}^{-2} .\)

Dimension of thrust: As thrust and force are the same physical quantity, the dimension of thrust is MLT^-2 .

Units of pressure and thurst

Absolute unit

Pressure:

• dyn • cm^-2 CGS System
• N · m^-2 or pascal(pa) SI

Thurst:

• dyn
• N

Relation: 1N = 10⁵ dyn

⇒ \(1 \mathrm{~N} \cdot \mathrm{m}^{-2}\)

or, \(1 \mathrm{~Pa}=\frac{1 \mathrm{~N}}{1 \mathrm{~m}^2}\)

= \(\frac{10^5 \mathrm{dyn}}{10^4 \mathrm{~cm}^2}\)

= \(10 \mathrm{dyn} \cdot \mathrm{cm}^{-2}\)

The magnitude of pressure at a point within a Liquid: Let us consider a liquid of density ρ kept in a vessel. We have to determine the pressure exerted by the liquid at a point C at a depth h below the surface of the liquid. Let us imagine a vertical cylinder of small horizontal area A surrounding the point C inside the liquid. The weight of the liquid column inside the cylinder is the thrust or normal force applied on the area A.

Now, weight of the liquid column

= mass of the liquid column x g

= volume of the liquid column x density of the liquid column x g = Ah x ρ x g

∴ Thrust exerted on the area A, F = Ahρg

∴ Pressure exerted by the liquid at the point C,

p = \(\frac{F}{A}=\frac{A h \rho g}{A}=h \rho g\) …….(1)

∴ The pressure exerted by the liquid = depth x density x acceleration due to gravity

So, at a particular place (where g is a constant), the pressure at a point in a liquid is

1. directly proportional to the depth of the point when the density of the liquid is a constant (p ∝ h) and
2. directly proportional to the density of the liquid when the depth is kept constant (p ∝ ρ).

If the surface of the liquid is exposed to air, then the atmospheric pressure also acts on the surface of the liquid If the atmospheric pressure is R, then the net pressure at a point inside the liquid is given by

p = B+ hρg ………(2)

Sometimes, pressure is expressed in terms of the height of a liquid column. The pressure exerted by a liquid of height h is hρg.

So, ‘atmospheric pressure is 76 cm of mercury’ means that in the CGS system, it is equal to a pressure of 76 x 13.6 x 980 = 1.013 x 10⁶ dyn · cm^-2.

Pressure at a Point in a Liquid at Rest

Experiment: A glass funnel (F) is taken whose mouth has been closed with a stretched rubber sheet. It is connected by means of a rubber tube (R) to a glass tube of Fine bore (T). The glass tube is kept horizontal with a scale (S) attached to it.

• A drop of a coloured liquid (l) is j introduced in the tube and it serves as an index. VY’ hence the rubber sheet is pressed, the air inside the tube gets compressed and pushes the index towards right. From the movement of the drop, we can study the variation of pressure on the rubber sheet.
• If the funnel is introduced in a liquid kept in a vessel and is gradually immersed, it will be seen that the index (l) gradually moves towards right. As the funnel is submerged deeper into the liquid, the index moves more towards right. This proves that with the increase in depth within the liquid, the pressure of the liquid increases.

Visualizing Hydrostatic Pressure with Diagrams

Factors Affecting Hydrostatic Pressure

Now, keeping the funnel fixed at a particular depth below the liquid surface, if its mouth is turned in diluent directions, the index remains stationary. It indicates that at a point inside a liquid, the pressure exerted by the liquid is equal in all directions.

If the funnel is now moved to different points at the same horizontal level, the position of

the index will still remain stationary. From this, it may be concluded that the pressure exerted by a liquid at different points at the same depth is equal.

From this experiment, we can infer that the characteristics of pressure exerted by a liquid at rest are:

1. A liquid exerts pressure at a point inside it equally in all directions.
2. The pressure exerted at different points on the same horizontal plane inside it is the same.
3. The pressure exerted by a liquid at a point inside it increases with the increase in its depth.

Action of a small hole on the wall of a container: Let a small hole be made on the wall at a depth h of a vessel containing a liquid. We know that for a liquid of density ρ, pressure at a depth h of the liquid is hρg.

So, the lateral pressure exerted by the liquid on the mouth of the hole = hρg.

Due to this lateral pressure, the liquid comes out through the hole. Naturally, with the ejection of liquid through the hole, the level of water goes down in the container and hence the lateral pressure exerted by the liquid gradually decreases. For this reason, the rate of flow of the liquid through the hole also decreases gradually.

Nature Of The Free Surface Of A Liquid At Rest:

1. The free surface of a liquid at rest always remains horizontal.

Take a container with some liquid at rest. Let us assume that the free surface of the liquid is not horizontal but way. Two points A and B are taken on the same horizontal plane inside the liquid, which are at depths h₁ and h₂ respectively from the free surface of the liquid.

If the density of the liquid is p then, pressure of the liquid at the point A = h₁ρg and pressure of the liquid at the point B = h₂ρg.

We know that, at all points on the same horizontal plane inside a liquid at rest, the pressure exerted by the liquid is the same.

So, liquid pressure at the point A = liquid pressure at the point B

or, h₁ρg = h₂ρg or, h₁ = h₂, i.e., h₁ and h₂ cannot be unequal. Since A and B lie on the same horizontal plane and h₁ is equal to h₂, the free surface of a liquid at rest always remains horizontal.

2. The free surfaces of a liquid at rest in connected vessels always remain on the same horizontal plane: A system of connected vessels of different shapes is taken and a liquid is poured into it. Though the shapes and volumes of the vessels are different, it is seen that the liquid rises up to the same horizontal level in all the vessels.

Let us consider a horizontal plane passing through the common interconnecting tube of the system. We know that at different points (for example, A, B, C, D, E) on that horizontal plane, the pressure exerted by the liquid is the same.

Since the liquid pressure at a point depends on the depth of that point from the free surface of the liquid, we can say that the free surfaces of the liquid in the vessels must remain on the same horizontal plane. Hence, a liquid seeks its own level— which is regarded as a general property of a liquid.

If the diameters of the different vessels become very small (2 mm or less that), then due to the surface tension of the liquid, it rises to different heights in different vessels. In that case, the free surfaces of the liquid in different vessels will not remain on the same horizontal plane.

Equilibrium Of Two Different Liquids In A U-Tube: When two immiscible liquids are poured into the two limbs of a U-tube, it is found that the free surfaces of the liquids in the two limbs remain horizontal, but they attain different heights.

Let ADBEC be a U-tube made of glass. Into one of the limbs (say, the right one), mercury is poured and it is seen that mercury enters the other limb and rises up to the same horizontal level in the two limbs.

Now, another liquid lighter than mercury is poured into the left limb of the U-tube and it is observed that the mercury level falls in this limb and rises in the other limb through the same height.

In equilibrium, the interface of the liquids in the left arm is at D, the free surface of the liquid is at A and that of mercury is at C. In equilibrium, at the same horizontal level DE, liquid pressure at the point D = mercury pressure at the point E.

Now, the pressure at the point D = h₂ρ₂g

[AD = h₂ , density of the liquid = ρ₂, acceleration due to gravity = g]

and pressure at the point E = h₁ρ₁g[CE = h₁, density of mercury = ρ₁]

∴ \(h_2 \rho_2 g=h_1 \rho_1 g \text { or, } \frac{h_1}{h_2}=\frac{\rho_2}{\rho_1}\) ……(1)

Therefore, in equilibrium, the height of the liquid column from the level of the interface is inversely proportional to the respective density.

If the density of one liquid is known, then with the help of equation (1), the density of the other liquid can also be found out.

Lateral Thrust On A Surface Immersed Vertically In A Liquid: Consider a horizontal surface in a liquid at rest. At every point on that surface, the pressure exerted by the liquid will be the same and the pressure multiplied by the area of the immersed surface gives the total thrust.

• But if the surface is placed in a vertical or an inclined position inside the liquid, then different points of the surface will lie at different depths and hence the pressure will vary from point to point.
• In this case, the average pressure exerted on the surface is calculated. The total thrust acting on the surface can be obtained by multiplying the average pressure with the area of the surface.

Let the length and breadth of a rectangular surface ABCD, immersed vertically inside a liquid of density p, be a and b respectively.

Let us assume that the depth of the side AB from the free surface of the liquid is h.

∴ Depth of the side CD = (h+ b).

∴ Pressure at any point on the side AB = hρg and that at any point on the side CD = (h+ b)ρg.

∴ Average pressure on the surface ABCD

= \(\frac{h \rho g+(h+b) \rho g}{2}=\frac{2 h \rho g+b \rho g}{2}\)

= \(h \rho g+\frac{1}{2} b \rho g=\left(h+\frac{b}{2}\right) \rho\)

So, the lateral thrust on the surface ABCD

= average pressure x area = \(\left(h+\frac{b}{2}\right) \rho g \times a b\)

Unit 7 Properties Of Bulk Matter Chapter 2 Hydrostatics Pressure Numerical Examples

Hydrostatic Pressure Calculation Problems

Example 1. Calculate the pressure at the bottom of a freshwater lake of depth 10 m. The atmospheric pressure = 76 cm of mercury and the density of mercury = 13.6 g · cm^-3.
Solution:

Given

The atmospheric pressure = 76 cm of mercury and the density of mercury = 13.6 g · cm^-3.

Pressure at the bottom of the lake = atmospheric pressure + pressure exerted by a 10 m water column

= 76 x 13.6 x 980 + 10 x 100 x 1 x 980

= 980(76 x 13.6 + 1000)

= 980 x 2033.6 = 1.993 x 10  dyn ·cm^-2.

The pressure at the bottom of a freshwater lake of depth 10 m = 1.993 x 10  dyn ·cm^-2.

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Example 2. The length of a right circular cylinder filled with water is 4 m and its diameter is 1 m. At first, it is held upright on its base, and then is positioned laterally. Find the ratio of the thrusts exerted on its circular base in the above two cases.
Solution:

Given

The length of a right circular cylinder filled with water is 4 m and its diameter is 1 m. At first, it is held upright on its base, and then is positioned laterally.

When the cylinder is held upright, the thrust exerted on its circular base,

⇒ \(F_1=\text { pressure } \times \text { area }=4 \times 1000 \times 9.8 \times \pi\left(\frac{1}{2}\right)^2 \mathrm{~N}\)

When the cylinder is kept laterally, the thrust exerted on its circular base,

F₂ = average pressure x area

= \(\frac{1}{2} \times 1000 \times 9.8 \times \pi\left(\frac{1}{2}\right)^2 \mathrm{~N}\)

∴ \(\frac{F_1}{F_2}=\frac{4 \times 1000 \times 9.8 \times \pi\left(\frac{1}{2}\right)^2}{\frac{1}{2} \times 1000 \times 9.8 \times \pi\left(\frac{1}{2}\right)^2}\)

∴ \(F_1: F_2=8: 1\)

Example 3. A hollow right circular cone of height h and of semi-vertical angle θ is placed with its base on a horizontal table. If the cone is filled with a liquid of density ρ and the weight of the empty cone is equal to the weight of the liquid that it contains, find the thrust of the liquid on the base of the cone and the pressure exerted on the table.
Solution:

Given

A hollow right circular cone of height h and of semi-vertical angle θ is placed with its base on a horizontal table. If the cone is filled with a liquid of density ρ and the weight of the empty cone is equal to the weight of the liquid that it contains

According to Fig, r = h tan θ = radius of the circular base of the cone.

Volume of the cone, V = \(\frac{1}{3} \pi r^2 h\)

Mass of the liquid inside the cone, M = \(\frac{1}{3} \pi \rho r^2 h\)

Liquid pressure on the base of the cone = \(h \rho g\)

∴ Thrust exerted by the liquid on the base of the cone

= pressure x area = \(h \rho g \times \pi r^2=\pi \rho g h^3 \tan ^2 \theta\)

Again, mass of the empty cone = M = \(\frac{1}{3} \pi \rho r^2 h\)

∴ Total force exerted on the table = weight of the cone + weight of the liquid

= \(\frac{1}{3} \pi \rho r^2 h \cdot g+\frac{1}{3} \pi \rho r^2 h \cdot g=\frac{2}{3} \pi \rho g r^2 h\)

∴ Pressure on the table = \(\frac{\text { force }}{\text { area }}=\frac{\frac{2}{3} \pi \rho g r^2 h}{\pi r^2}=\frac{2}{3} h \rho g\).

Step-by-Step Guide to Hydrostatic Pressure Calculations

Example 4. The lock gate of a canal is 4.8 m broad. The depth of water on one side of the lockgate is 4.5 m and on the other side is 3 m. Calculate the total thrust on the lock gate. The density of water = 1000 kg · m³.
Solution:

Given

The lock gate of a canal is 4.8 m broad. The depth of water on one side of the lockgate is 4.5 m and on the other side is 3 m.

Thrust exerted by the water column of depth 4.5 m on one side of the lock gate

= 1/2 x 4.5 x 1000 x (4.5 x 4.8) xg = 48600 kg x g

Thrust exerted by the water column of depth 3 m on the other side of the lock gate

= 1/2 x 3 x 1000 x (3 x 4.8) xg = 21600 kg x g

Since, these two thrusts are mutually opposite in direction, the net thrust on the lockgate

= (48600-21600) xg = 2.65 x 10⁵ N.

The total thrust on the lock gate = 2.65 x 10⁵ N.

Example 5. An air bubble of diameter 1 mm is formed at the bottom of a lake and when it rises to the surface of the water, its diameter becomes 2 mm. If the atmospheric pressure is 76 cm of mercury, then find the depth of the lake. (Density of mercury = 13.6 g · cm^-3)
Solution:

Given

An air bubble of diameter 1 mm is formed at the bottom of a lake and when it rises to the surface of the water, its diameter becomes 2 mm. If the atmospheric pressure is 76 cm of mercury

Pressure at the bottom of the lake = (76 x 13.6 x 980 + h x 1 x 980) dyn · cm^-2

[h = depth of the lake; p = atmospheric pressure =76 x 13.6 x 980 dyn · cm^-2]

Volume of the bubble at the bottom of the lake = 4/3π(0.05)³ cm³ and volume of the bubble at the surface of the lake = 4/3π(0.1)³ cm³.

If the temperature of the lake water is uniform, then pV = constant.

∴ \((76 \times 13.6+h) \times 980 \times \frac{4}{3} \pi(0.05)^3\)

= \(76 \times 13.6 \times 980 \times \frac{4}{3} \pi(0.1)^3\)

∴ h = 7235 cm = 72.35 m.

The depth of the lake = 72.35 m.

Example 6. A large container with a cylindrical mouth is filled with water and closed with a piston. A vertical tube is now inserted through the piston. The radius of the 1 tube is 5 cm, the radius of the piston is 10 cm and the mass of the piston is 20 kg. To what height will the water rise inside the tube?
Solution:

Given

A large container with a cylindrical mouth is filled with water and closed with a piston. A vertical tube is now inserted through the piston. The radius of the 1 tube is 5 cm, the radius of the piston is 10 cm and the mass of the piston is 20 kg.

Area of the piston

= \(\pi\left(r_2^2-r_1^2\right)=\pi\left(10^2-5^2\right)=75 \pi \mathrm{cm}^2\).

Weight of the piston = 20 x 1000 x g dyn

Let the height of the water column in the tube be hem.

Now, pressure of the piston = pressure of the water column in the tube

or, \(\frac{20 \times 1000 \times g}{75 \pi}=h \times 1 \times g\)

∴h = \(\frac{20000}{75 \pi}=84.88 \mathrm{~cm}\)

Example 7. At what depth below the surface of a lake will the total pressure be twice the atmospheric pressure? (Atmospheric pressure = 76 cm Hg and the density of mercury = 13.6 g · cm^-3)
Solution:

Given

If the required depth is h and the atmospheric pressure is p, then 2p = p + hρg

or, p = hρg or, 76 x 13.6 x g = h x 1 x g

∴ h = 76 x 13.6 = 1033.6 cm.

Example 8. A U-tube is partially filled with mercury. Kerosene oil is poured into one of its limbs and glycerine into the other. It is observed that, when the height of the kerosene oil becomes 20 cm and that of glycerine becomes 12.68 cm, the levels of the mercury column in the two limbs are at the same horizontal level If the density of kerosene oil is 0.8 g · cm^-3, then find that of glycerine.
Solution:

Given

A U-tube is partially filled with mercury. Kerosene oil is poured into one of its limbs and glycerine into the other. It is observed that, when the height of the kerosene oil becomes 20 cm and that of glycerine becomes 12.68 cm, the levels of the mercury column in the two limbs are at the same horizontal level If the density of kerosene oil is 0.8 g · cm^-3

In the U-tube, mercury levels in the two limbs are at the same horizontal level.

So, pressure exerted by 20 cm of kerosene oil = pressure exerted by 12.68 cm of glycerine. Let the density of glycerine be ρ.

∴ 20 x 0.8 x g = 12.68 x ρ x g

or, \(\rho=\frac{20 \times 0.8}{12.68}=1.26 \mathrm{~g} \cdot \mathrm{cm}^{-3}\)

Example 9. A vertical U-tube of uniform cross-section contains mercury. Through one of its limbs, some water is poured such that the mercury level in that limb goes down by 2 cm. What will be the height of the water column?
Solution:

Given

A vertical U-tube of uniform cross-section contains mercury. Through one of its limbs, some water is poured such that the mercury level in that limb goes down by 2 cm.

Let the height of the water column be h. If the mercury level falls down by 2 cm in one limb, then it will rise by 2 cm in the other limb.

From fig, AC = 2cm, BE = 2cm

∴ DE = 2 + 2 = 4 cm

Now, pressure at the point C = pressure at the point D

∴ h x l x g = 4 x 13.6 x g or, h = 54.4 cm

Example 10. The cross-sections of the two limbs of a U-tube are 3 cm² and 1 cm² respectively. Keeping the tube vertical, some mercury is poured into it. Now 60 cm³ of water is poured into the wider limb. To what height will the mercury rise in the narrow limb? The density of mercury = 13.6 g · cm^-3.
Solution:

Given

The cross-sections of the two limbs of a U-tube are 3 cm² and 1 cm² respectively. Keeping the tube vertical, some mercury is poured into it. Now 60 cm³ of water is poured into the wider limb.

In Fig, we can see the final state after water is poured. The mercury column in the wider limb will fall and that in the narrow limb will rise.

The height of the water column in the wider limb

= \(\frac{\text { volume of water }}{\text { area of cross-section of the wider limb }}=\frac{60}{3}=20 \mathrm{~cm}\)

∴ FC = 20 cm

After water is poured into the wider limb, suppose the mercury column Tails by x. So, mercury will rise through 3x in the narrower limb (v the area of the cross-section of the wider limb is 3 times that of tire narrower limb).

∴ AC = x and BE = 3x

∴ DE = x+3x = 4x

Now, pressure at the point C = pressure at tire point D

or, 20 x 1 x g = 4x x 13.6 x g

or, x = \(\frac{20}{4 \times 13.6}=0.368 \mathrm{~cm}\)

∴ The mercury column will rise through 0.368 x 3 = 1.1 cm in the narrower limb.

Example 11. One of the two limbs of a U-tube of uniform bore is closed by means of a cork. Some water and paraffin oil are in the tube. Water level in the open tube is 2 cm higher than that in the closed tube and a 10 cm column of paraffin is above it. If the cork is removed, then by what height will the level of water in the closed limb rise or fall? (Density of paraffin = 0.8 g · cm^-3)
Solution:

Given

One of the two limbs of a U-tube of uniform bore is closed by means of a cork. Some water and paraffin oil are in the tube. Water level in the open tube is 2 cm higher than that in the closed tube and a 10 cm column of paraffin is above it.

Let us assume that, when the cork is removed, the water level below the oil column will fall by h with respect to the level of water in the other limb.

Now,pressure at D = pressure at A’

or, h x 1 x g = 10 x 0.8 x g or, h = 8 cm .

Here, the cross-section of the U-tube is uniform. So, if the height of the water column falls through x cm in one limb, the difference in the level of water in the two limbs will increase by 2x cm.

So, the initial difference in the levels of water = -2 cm, and the final difference = + 8 cm.

∴ 2x = 8-(-2) = 10

∴ x = 5 cm

So, when the cork is removed, the height of the water column will rise by 5 cm in that limb.

Example 12. A vertical U-tube of uniform cross-section contains mercury in both Its arms. A glycerine (density 1.3 g · cm^-3) column of height 10 cm is introduced into one of the arms. Oil of density 0.8 g · cm^-3 is poured into the other arm until the upper surfaces of oil and glycerine are at the same level. Find the height of the oil column. The density of mercury is 13.6 g · cm^-3.
Solution:

Given

A vertical U-tube of uniform cross-section contains mercury in both Its arms. A glycerine (density 1.3 g · cm^-3) column of height 10 cm is introduced into one of the arms. Oil of density 0.8 g · cm^-3 is poured into the other arm until the upper surfaces of oil and glycerine are at the same level.

Let the height of the oil column be h

According to the figure, AC = ED

or, h + BC = 10

or, BC = (10 – h) cm

Now,pressure at the point C = pressure at the point D

or, h x 0.8 x g + (10 – h) x 13.6 x g =10 x 1.3 x g

or, h x 0.8 + 136 – 13.6h = 13 or, 12.8 h = 123

or, h =\(\frac{123}{12.8}=9.61 \mathrm{~cm}\)

Example 13. The cross-sections of the two sides of a U-tube are 1 cm² and 0.1 cm² respectively. Some water is poured inside the tube when it rises to the same height in both the limbs. What volume of a liquid of density 0.85 g · cm^-3 should be poured into the wider limb so that the water level rises by 15 cm in the narrow limb?
Solution:

Given

The cross-sections of the two sides of a U-tube are 1 cm² and 0.1 cm² respectively. Some water is poured inside the tube when it rises to the same height in both the limbs.

At first, water rises up to the same level in both the arms. When the liquid is poured into the wider limb, sup¬pose the level of waterfalls by x cm and rises by 15 cm in the narrow limb.

AC = BF = x

volume of the part AC = volume of the part BD

or, x x 1 = 15×0.1 or, x = 1.5 cm

Let EC = h.

Pressure at the point C = pressure at the point F

or, h x 0.85 x g = (15 + 1.5) x 1 x g

or, h = \(\frac{16.5}{0.85}\) = 19.41 cm

So, the volume of liquid poured into the wider limb = 19.41 x 1 = 19.41 cm³.

Example 14. The cross-sectional area of the left limb of a U-tube is one-third of that of its right limb. It contains some mercury. The empty space in the left limb measures to 40 cm. If water is poured to fill it up, then find the rise of the mercury column in the right limb. (Specific gravity of mercury = 13.6)
Solution:

Given

The cross-sectional area of the left limb of a U-tube is one-third of that of its right limb. It contains some mercury. The empty space in the left limb measures to 40 cm.

At first, the mercury is at the same level in both the limbs. When water is poured, suppose the fall in the mercury level in the narrow limb is x and the corresponding rise in the mercury level in the wider limb is y.

If the cross-sectional area of the narrow limb is A, then, according to the problem, the cross-sectional area of the wider limb will be 3A.

From Fig. 2.14, we see,

volume of the part A’ C – volume of the part FB

or, x x A = y x 3A or, x = 3y

Now, pressure at point C = pressure at point D

or, EC x 1 x g = FD x 13.6 x g or, (40 + x) = (x + y) x 13.6

or, 40 + 3y= (3y+y)x 13.6

or, (13.6 x 4 – 3)y = 40

or, y = \(\frac{40}{54.4-3}=0.78 \mathrm{~cm} .\)

Example 15. A U-tube of uniform cross-section contains some mercury. If water is poured into one of the limbs up to a height of 13.4 cm, then find the rise in the mercury level in the other limb. The density of mercury = 13.4 g · cm^-3.
Solution:

Given

A U-tube of uniform cross-section contains some mercury. If water is poured into one of the limbs up to a height of 13.4 cm

At first, the mercury will attain the same height in both the limbs. Now a column of water 13.4 cm high is poured into the left limb and, due to this, the mercury level falls through a depth of x cm from its initial position.

Since the cross-sectional areas of both the limbs are the same, the mercury level will rise through a height of x cm from its original position in the other limb.

Now, pressure at point C = pressure at point E

or, 13.4 x 1 x g = 2JC X 13.4 xg or, x = 1/2 = 0.5 cm

The mercury level in the other limb will rise by 0.5 cm from its original position.

Example 16. Two cylindrical vessels of the same type contain a liquid of density ρ. The bases of both the vessels lie on the same horizontal plane. The depth of the liquid in the left vessel is h₁ and that in the right vessel is h₂. The area of the cross-section of the base of each vessel is A. If the vessels are connected by a tube, then how much is the work done by gravity to equalize the levels of the liquid in the vessels (suppose h₁ >h₂)?
Solution:

Given

Two cylindrical vessels of the same type contain a liquid of density ρ. The bases of both the vessels lie on the same horizontal plane. The depth of the liquid in the left vessel is h₁ and that in the right vessel is h₂. The area of the cross-section of the base of each vessel is A.

When the vessels are connected, the level of the liquid in the left-hand vessel will fall, and that in the right-hand one will go up.

The initial difference in the liquid levels in the vessels = h₁-h₂. When the levels in both vessels become the same, the level of liquid in the left-hand vessel will fall by 1/2(h₁-h₂), and that in the right-hand vessel will rise by 1/2(h₁-h₂).

Mass of this liquid = 1/2(h₁-h₂)Aρ.

The centre of gravity of this liquid moves up against gravity by 1/2(h₁-h₂).

∴ Increase in potential energy of this liquid = work done against gravity

= \(\frac{1}{2}\left(h_1-h_2\right) A \rho \times g \times \frac{1}{2}\left(h_1-h_2\right)=\frac{1}{4} A \rho g\left(h_1-h_2\right)^2\).

Example 17. A tank closed tightly with a lid, is completely filled with water and is rigidly fixed on a car. The car moves with a constant acceleration of 20 cm · s^-2. Determine the pressure at a point 10 cm below the lid and at a distance of 10 cm from the front wall of the tank.
Solution:

Given

A tank closed tightly with a lid, is completely filled with water and is rigidly fixed on a car. The car moves with a constant acceleration of 20 cm · s^-2.

Let us consider a vertical disc of unit area kept at a depth h from the upper surface of the tank. The disc is at a distance x from the front wall of the car and its width is dx. Let the pressures at distances x and (x + dx) be p and (p+dp) respectively.

∴ p + dp – p = p – dx- a

[ρ = density of the liquid, ρdx = mass of the liquid displaced by the disc, and a = acceleration of the car]

or, dp = aρdx

∴ ∫dp = ap ∫dx

or, p = aρx+ c [c = integration constant]

Now, if x = 0, then p = hρg.

∴ hρg = c

∴ p = aρx+ hρg = ρ(ax+ hg)

∴ The pressure at a point 10 cm below the lid and at a distance of 10 cm from the front wall of the tank,

p = 1(20 x 10 + 10×980) =200 + 9800

= 10000 dyn · cm^-2.

The pressure at a point 10 cm below the lid and at a distance of 10 cm from the front wall of the tank = 10000 dyn · cm^-2.

Elasticity Property Of Matter

Elastic Limit Formula

From our common experience, we know that when a rubber cord is pulled, it increases in length. But in order to increase the length of a steel wire of same diameter by the same amount, a greater force needs to be applied.

• When equal and opposite external forces act on a body, the different point masses of the body undergo relative displacements. As a result, the body undergoes a change in its shape or size, or both.
• In this condition, a reaction force develops inside the body which opposes the change. If this change lies within a definite limit then the body regains its original state once the forces are withdrawn.
• The property which opposes deformation is present in all materials. This general property of matter is known as elasticity (or the elastic property of matter).

Elastic Definition: The property by virtue of which a body resists the deformation in shape or volume or both due to external forces acting on it and regains its original shape or volume when these external forces are withdrawn is called elasticity. This property is present in every material, irrespective of whether it is solid, liquid, or gas.

Read and Learn More: Class 11 Physics Notes

Elasticity of rubber and Steel: From the viewpoint of physics, a body is said to be more elastic if it has a greater ability to resist deformation against the external force.

• The greater the external force necessary to produce a definite change in the size or shape of a body, the more elastic is the material of the body.
• As mentioned earlier, in order to produce an equal deformation in a steel wire and a rubber wire of the same dimensions, a greater force is necessary in the of the steel wire. For this reason, steel is more elastic than rubber.

Factors Affecting Elasticity:

1. The presence of impurities in a metal changes its elastic property.
2. If a metal is deformed frequently, then its elasticity decreases. For example, if a thick copper wire is often twisted, it becomes hard and brittle.
3. The elasticity of a metal changes with temperature. Usually, its elasticity decreases when temperature increases, and vice versa.
4. An exception to this rule occurs in the case of invar —whose elasticity does not change with any change in temperature. Again if a metal is first heated and then cooled, i.e., it is softened, its elasticity gradually decreases

Elasticity Some Useful Definitions

Perfectly elastic body: If after the withdrawal of external forces, a body completely regains its original shape and volume, then it is called a perfectly elastic body.

• In real life, however, a body cannot be perfectly elastic for all magnitudes of external forces. Up to a certain limiting value of external force, a body behaves as a perfectly elastic body.
• This limit is known as the elastic limit for the material of the body. Different materials have different elastic limits. For example, the elastic limit of steel is very high while that of rubber is very low.

Perfectly plastic or inelastic body: if a body, elongated (or compressed) by external forces remains in that deformed state even after the withdrawal of these deforming forces, it is called a perfectly plastic or inelastic body. Actually, no material is perfectly inelastic, though clay comes very close to it.

Partly elastic body: If after the withdrawal of external forces, a deformed body only partially regains its original shape and volume then it is called a partly elastic body. Practically all materials are partly elastic.

Strain: Under the influence of external forces, when a body gets deformed, the different parts of the body suffer relative displacements. As a result, the body undergoes a change in length, volume, or shape.

Strain of a body is defined as the change of its length, volume, or shape relative to its original length, volume, or shape before deformation. So, it is a ratio of two identical quantities and hence it has no unit, i.e., strain is a dimen¬sionless quantity.

Stress: When a body gets deformed under the influence of external forces, a reaction force develops in the body because of elasticity.

This force tries to resist the external forces and helps to bring the body back to its unstrained condition after the deforming forces are withdrawn. The reaction force acting per unit area of the cross-section of the body is called stress.

Since action and reaction are opposite but equal, stress is equal in magnitude to the force applied per unit area of the deformed body.

Stress = \(\frac{\text { applied force }}{\text { area of cross-section of the body }}\)

Units and dimension of stress:

• dyn. cm^-2 CGS system
• N. m^-2 SI

Dimensional of stress = \(\frac{\text { dimension of force }}{\text { dimension of area }}=\frac{M L T^{-2}}{L^2}\) = ML^-1T^-2

We shall see that the units and dimension of stress is the same as that of pressure.

Normal stress and shearing or tangential stress: An applied force can act normally or obliquely on the surface of a body. The component of the reaction force perpendicular to a unit area of the surface is called normal stress.

• When there is any change in the length of a wire or in the volume of a body, normal stress is developed. The stress associated with an increase in length is called tensile stress and that associated with a decrease is called compressive stress.
• On the other hand, the component of the reaction force parallel to a unit area of the surface is called shearing stress or tangential stress. Usually, shearing stress is developed during any change in the shape of a body.

Breaking load and breaking stress: if the external force exceeds the elastic limit, then a strained body cannot return to its original size or shape even after the deforming force is withdrawn.

• In such a case, the body gets perma¬nently deformed. If the amount of the external force is increased gradually, then, for a particular value of the applied force, the body breaks or snaps.
• The magnitude of force, or load, for which the body breaks or snaps is called the breaking load of that body. In that condition, the maximum reaction force developed per unit area of the surface of the body is called breaking stress. Every material has its characteristic breaking stress.

Elasticity Hooke’s Law

The fundamental law of elasticity was propounded by Robert Hooke in 1676. Later Thomas Young expressed this lawin the following way.

Hooke’s Law Experiment Procedure

Statement: Within the elastic limit of a substance, stress is directly proportional to strain.

Hence, stress ∝ strain or, \(\frac{\text { stress }}{\text { strain }}\) = constant

This constant is called the modulus of elasticity for the material of the body. Therefore, the stress developed for unit strain is defined as the modulus of elasticity.

This modulus depends on the material of the body. The modulus of elasticity changes with temperature, in general. Usually, its value decreases with an increase in temperature.

Units and dimension of the modulus of elasticity: Since strain has no unit, the modulus of elasticity has the unit of stress.

Relation: \(1 \mathrm{~N} \cdot \mathrm{m}^{-2}(1 \mathrm{~Pa})=\frac{1 \mathrm{~N}}{1 \mathrm{~m}^2}=\frac{10^5 \mathrm{dyn}}{10^4 \mathrm{~cm}^2}\)

= \(10 \mathrm{dyn} \cdot \mathrm{cm}^{-2}\)

Similarly, the dimensions of the modulus of elasticity and of
stress are identical, which is ML^-1T^-2.

ElasticityStress Strain Graph (Load Exten Sion Graph Of A String)

Applications of Hooke’s Law in Experiments

Suppose a wire of uniform cross-section is clamped at its upper end with rigid support and a load is applied at its lower end which is then gradually increased.

As a result, the length of the wire goes on increasing. In the stress- strain graph of a ductile metallic wire has been shown. The different parts of this graph are described below.

1. Straight line OA: In this section, the stress on the wire is proportional to the strain, which means that the metal follows Hooke’s Aaftr. The wire behaves like a completely elastic body up to the point A.

2. Point A: This point indicates the proportionality limit.

3. Line segment AB: In this section, the ratio of stress and strain is comparatively less, which means that the metal does not follow Hooke’s law. However, after reaching this section of strain, if the stress is removed, then the wire will regain its original length, which means that the strain will again be zero.

4. Point B: This point indicates the elastic limit. In the case of most metals, the two points A and 6 are found to be quite close to each other. In the case of glass, A, and B are identical points and in the case of rubber, the distance between A and B is quite high.

5. Line segment BC: In this section, stress divided by strain becomes even less and the metal gradually loses its elastic property and becomes plastic. After reaching this section of strain, if the stress is removed, then the wire is unable to regain its original length. So, the wire undergoes a permanent deformation.

6. Point C: This point is called the yield point or the upper yield point. At this point, the stress level is known as yield stress. The yield point for most substances can¬not be determined accurately.

Limitations of Hooke’s Law in Experiments

7. Line segment CD: In this section, stress divided by strain is negative. It implies that even if stress is decreased, strain will increase.

8. Point D: This point is called the lower yield point. If the stress is gradually decreased after the strain reaches this point, the return graph is not along DO, but along DO’. In this case OO’ indicates the permanent deformation.

For bodies with nearly perfect elasticity, the points A, B, C, and D are situated so close to one another that practically the four points can be assumed to be identical.

9. Line segment DE: In this section, stress divided by strain is the least and the metal becomes plastic. The cross-section of certain parts of a wire becomes comparatively lower than that of the remaining parts.

10. Point E: At point E the stress reaches its maximum value. At this point, the material of the wire flows like a viscous liquid and the wire becomes thin. Now, even if the load is decreased, the wire goes on thinning down.

11. Line segment EF: In this section, the area of the cross-section at different parts of the wire starts decreasing fast.

12. point F: At this point, the wire snaps from its weakest part. The stress corresponding to the point F is called breaking stress or the ultimate stress. This point F is called fracture or breaking point.

Ductile material: The materials which have large plastic range of extension are called ductile materials. As shown in the stress-strain curve in Fig, their fracture or breaking point is widely separated from the elastic limit. Such materials undergo an irreversible increase in length before snapping. So they can be drawn into thin wires. Copper, silver, iron, aluminum, etc., are examples of ductile materials.

Brittle material: The materials which have very small range of plastic limit of extension are called brittle materials.

Such materials break as soon as the stress is increased beyond the elastic limit. Their breaking point lies just close to their elastic limit as shown. Cast iron, glass, ceramic, etc., are examples of brittle materials.

Necessity of the stress-strain graph: For practical purposes, knowledge of the load-extension graph of a metal is absolutely essential. From this graph, the elastic limit of the material can be known.

For example, during the use of a machine, the stress developed on the axle or the other parts of the machine should be kept below the elastic limit of its material. For this, the stress-strain or the load-extension graphs of different materials are highly useful.

Elastic fatigue: If the force (or load) applied on an elastic body rises and falls rapidly and this periodic fluctuation continues for a long time, then the elastic property of the body gets degraded, even if the elastic limit for the material is not exceeded.

• It means that the body remains permanently deformed in some respect, i.e.„ some part of the body becomes thinner and weaker, even after the deforming force is withdrawn. The body may then break or snap at a load less than the normal breaking load.
• This kind of degradation of the elastic property of material due to rapid changes in stress is called elastic fatigue.

Elasticity Experimental Verification Of Hooke’s Law Determination Of Young’s Modulus

Calculating Spring Constant from Hooke’s Law

Searle’s Experiment: A uniform metal wire, about 2 to 3 m long, is hung from the roof of the laboratory. Initially, a small weight, usually called zero-load, is attached to the lower end of the wire. With the help of a screw gauge, the average diameter of the experimental wire is measured. Its length (L) is determined using a metre scale.

The load at the lower end of the wire is then gradually increased. Measuring the corresponding extensions with suitably fitted verniers, a graph can be plotted with the load along the X-axis and the elongation along the Y-axis. The graph passes through the origin and is a straight line.

Experimental Verification of Hooke’s Law

As the graph is a straight line, we can say that, load ∝ elongation, i.e., stress ∝ strain (within the elastic limit). This proves the validity of Hooke’s law.

Calculation: From any point P on the graph, two perpendiculars are drawn on the axes.

Here, OQ = load (mg); OR = elongation (l).

Therefore, the longitudinal stress \(\frac{m g}{\pi r^2}\) and the longitudinal.

∴ Y = \(\frac{\text { longitudinal stress }}{\text { longitudinal strain }}=\frac{\frac{m g}{\pi r^2}}{\frac{l}{L}}=\frac{m g L}{\pi r^2 l}\)

Since the quantities on the right-hand side of the above expression are known, the value of Young’s modulus (Y) can be determined.

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Elasticity Synopsis

The property by virtue of which a body resists the defor¬mation either in shape or in volume or both, due to application of external forces on it and recovers its original shape or volume when the deforming force is withdrawn, is called elasticity.

• A body cannot be perfectly elastic for all magnitudes of external force. Only up to a certain limit of external force a body can be perfectly elastic. This limit is known as the elastic limit for the material of that body.
• Actually there is no perfectly elastic or perfectly inelastic material, rather all materials can be treated as partially elastic.

Under the influence of external deforming force the reaction force developed per unit area of cross section of a body is called stress. Stress of a body can be expressed as

stress = \(\frac{\text { applied force }}{\text { area of cross-section of the body }}\)

• Relative displacements of different parts of an elastic body occur under the influence of external deforming force. As a result, the body undergoes change in length, volume, or shape. The fractional change in these quanti¬ties is known as strain.
• Strain is always represented by the ratio of two identical quantities. It has no unit its dimension = 1
• The applied force or load for which a body snaps or breaks is called the breaking load.

Hooke’s law: Within elastic limit, stress is directly proportional to strain. The amount of stress developed per unit strain is called elastic modulus.

Dimensions of elastic modulus and stress are identical and is ML^-1T^-2.

• The amount of force applied to produce unit elongation in a spring is called its force constant. Force constant of a spring represents its stiffness.
• The loss in elastic capability of a body due to repeated and rapid increase or decrease in the force applied on it is known as elastic fatigue.

Elasticity Useful Relations For Solving Numerical Problems

Hooke’s law: \(\frac{\text { stress }}{\text { strain }}=\text { elastic modulus }\)

Y = \(3 K(1-2 \sigma)=2 n(1+\sigma), \sigma=\frac{3 K-2 n}{6 K+2 n}\)

⇒ \(\frac{9}{Y}=\frac{1}{K}+\frac{3}{n}\)

Work done in stretching wire,

Hooke’s Law

W = \(\frac{1}{2} F l=\frac{1}{2} \frac{Y A l^2}{L}\)

In deforming a body, the work done per unit volume potential energy stored per unit volume of the body (energy density) = 1/2 x stress x strain .

For a force F applied on a spring, if its increase in length is x, then F∝x or, F = kx (where k = force constant of the spring]

 

 Elasticity Multiple Correct Answers Type Questions

“Elasticity Of Demand MCQ With Answers”

Question 1. The most elastic among the following substances is

1. Rubber
2. Glass
3. Steel
4. Copper

Answer: 3. Steel

Question 2. Which is most elastic?

1. Iron
2. Copper
3. Quartz
4. Wood

Answer: 3. Quartz

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 3. The ratio of the lengths of two wires made of the same metal and of equal radius is 1:2. If both the wires are stretched by the same force, then the ratio of the strains will be

1. 1:1
2. 1:2
3. 2:1
4. 1:4

Answer: 1. 1:1

Question 4. Shearing strain is expressed by

1. Shearing force
2. Angle of shear
3. Increase in area
4. Decrease in volume

Answer: 2. Angle of shear

Question 5. Breaking the stress of a wire depends on

1. The radius of the wire
2. The length of the wire
3. The shape of the cross-section
4. The nature of its material

Answer: 4. The nature of its material

Question 6. The elongation of an elastic material is very low. What should be the shape of a body for which the longitudinal strain will be appreciable?

1. A thin but long wire
2. A thick block having any cross-section
3. A thin block having a rectangular cross-section
4. A thin and short wire

Answer: 1. A thin but long wire

Question 7. The unit of elastic modulus is

1. N · m^-3
2. N · m^-2
3. N · m^-1
4. N · m

Answer: 2. N · m^-2

Question 8. Four wires made of the same metal are stretched by the same load. Their dimensions are given below. Which one will be elongated most?

1. Length 1 m, diameter 1 mm
2. Length 2 m, diameter 2 mm
3. Length 3 m, diameter 3 mm
4. Length 4 m, diameter 0.5 mm

Answer: 4. Length 4 m, diameter 0.5 mm

Question 9. Keeping the length of a wire unchanged, its diameter is doubled. Young’s modulus for the material of the wire will

1. Increase
2. Decrease
3. Remain the same
4. None of these

Answer: 3. Remain the same

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Question 10. Keeping one end of a wire of length L and radius r fixed, a force F is applied at the other end to elongate it by l. Another wire made of same metal but of length 2 L and radius 2r is stretched by a force 2 F. Its increase in length will be

1. l
2. 2l
3. l/2
4. 4l

Answer: 1. l

“Practice Questions On Elasticity Of Demand”

Question 11. The ratios of the lengths and diameters of two metallic wires A and B of the same material are 1: 2 and 2 :1 respectively. If both the wires are stretched by the same tension, then the ratio of the elongations of A and B will be

1. 1:2
2. 4:1
3. 1:8
4. 1:4

Answer: 3. 1:8

Question 12. A uniform rod weighs W, has a length L and has a cross-sectional area a. The rod is suspended from one of its ends. The Young’s modulus of its material is Y. An Increase in length of the rod will be

1. zero
2. \(\frac{W L}{2 a Y}\)
3. \(\frac{W L}{a Y}\)
4. \(\frac{2 W L}{a Y}\)

Answer: 2. \(\frac{W L}{a Y}\)

Question 13. If the ratio of diameters, lengths and Young’s modulus of steel and copper wires shown in the Fig are p, q and s respectively, then the corresponding ratio of increase in their lengths would be

1. \(\frac{5 q}{7 s p^2}\)
2. \(\frac{7 q}{5 s p^2}\)
3. \(\frac{2 q}{5 s p}\)
4. \(\frac{7 q}{5 s p}\)

Answer: 2. \(\frac{7 q}{5 s p^2}\)

Question 14. Two rods of equal length and cross-sectional area have their Young’s moduli Y₁ and Y₂ respectively. If the rods are joined end to end, then the equivalent Young’s modulus of the combined rod system is

1. \(\frac{2 Y_1 Y_2}{Y_1+Y_2}\)
2. \(\frac{Y_1 Y_2}{Y_1+Y_2}\)
3. \(\frac{1}{2\left(Y_1+Y_2\right)}\)
4. \(Y_1+Y_2\)

Answer: 1. \(\frac{2 Y_1 Y_2}{Y_1+Y_2}\)

Question 15. A rubber string of length 8 m is hanging vertically with one end fixed. If the density of rubber is 1.5 x 10³ kg · m^-3 and Young’s modulus is 5 x 10⁶ N · m^-2, then the increase of its length due to its own weight will be [g = 10 m · s^-2]

1. 9.6 x 10^-2 m
2. 19.2 x 10^-3 m
3. 9.6 x 10^-3 m
4. 9.6 m

Answer: 1. 9.6 x 10^-2 m

Question 16. The modulus of rigidity of steel is n and its Young’s modulus is Y. A steel wire of cross-sectional area A is so elongated that its area of cross-section becomes A/10. As a result

1. Y increases but n decreases
2. Y and n both remain the same
3. Y decreases but n increases
4. Both Y and n will increase

Answer: 2. Y and n both remain the same

“Calculating Elasticity Of Demand Questions”

Question 17. One end of a wire of length 1 m is rigidly fixed at the ceiling and a load W is applied at the other end of it. Load-elongation graph of it is shown in the diagram. If the area of the cross-section of the wire is 10^-6 m², then Young’s modulus for the material of the wire will be

1. 2 x 10¹¹ N · m^-2
2. 2 x 10^-11 N · m^-2
3. 3 x 10¹² N · m^-2
4. 2 x 10^-13 N · m^-2

Answer: 1. 2 x 10¹¹ N · m^-2

Question 18. In stretching a wire, the amount of work done per unit volume of the wire is

1. Stress x strain
2. 1/3 x strees x strain
3. \(\frac{stress}{strain}\)
4. 1/2 x strees x strain

Answer: 4. 1/2 x strees x strain

Question 19. A stretched rubber has

1. Increased kinetic energy
2. Increased potential energy
3. Decreased kinetic energy
4. Decreased potential energy

Answer: 2. Increased potential energy

Question 20. A wire of initial length L and area of cross-section A has Young’s modulus Y of its material. The wire is stretched by a stress S within its elastic limit. The stored energy density in the wire will be

1. \(\frac{S}{2 Y}\)
2. \(\frac{2 Y}{s^2}\)
3. \(\frac{s^2}{2 Y}\)
4. \(\frac{S^2}{Y}\)

Answer: 3. \(\frac{s^2}{2 Y}\)

“Elasticity Calculation Practice Problems”

Question 21. A wire of initial length L and area of cross-section A, having Young’s modulus Y of its material, is stretched to be elongated by an amount x. Work done in stretching the wire is

1. \(Y A \frac{x^2}{2 L}\)
2. \(Y A \frac{x^2}{L}\)
3. \(Y A \frac{x}{2 L}\)
4. \(Y A \frac{2 x^2}{L}\)

Answer: 1. \(Y A \frac{x^2}{2 L}\)

Question 22. One end of a wire is rigidly fixed and a force of 200 N is applied at its other end. The wire undergoes an elongation of 1 mm. The potential energy stored in the wire is

1. 0.1 J
2. 0.2 J
3. 10 J
4. 20 J

Answer: 1. 0.1 J

Question 23. Two wires A and B are made of the same metal. The diameter of A is double that of B and the length of B is thrice that of A. If both the wires are stretched by the same force to elongate them equally within elastic limit, then the ratio of energy stored in the wires A and B will be

1. 2:3
2. 3:4
3. 3:2
4. 6:1

Answer: 2. 3:4

Question 24. A spring of force constant k is cut into two equal parts. The force constant of each part of the spring will be

1. k/2
2. k
3. 2k
4. 4k

Answer: 3. 2k

Question 25. Force constants of two springs are k₁ and k₂. One end of a spring is connected with one end of the other. The equivalent force constant of the spring system will be

1. \(\frac{k_1+k_2}{2}\)
2. \(2\left(k_1+k_2\right)\)
3. \(\frac{k_1+k_2}{k_1 k_2}\)
4. \(\frac{k_1 k_2}{k_1+k_2}\)

Answer: 4. \(\frac{k_1 k_2}{k_1+k_2}\)

Question 26. A wire of length L and area of cross-section A, having Young’s modulus Y of its material, behaves like a spring of force constant k. The value of k will be

1. \(k=\frac{Y A}{L}\)
2. \(k=\frac{2 Y A}{L}\)
3. \(k=\frac{Y A}{2 L}\)
4. \(k=\frac{Y L}{A}\)

Answer: 1. \(k=\frac{Y A}{L}\)

“Interactive Quizzes On Elasticity Concepts”

Question 27. An elastic spring of length L and of force constant k is stretched to increase its length by an amount x. The spring is further stretched to elongate it by y. The amount of work done in stretching the spring in the second case (x and y are very small) is

1. \(\frac{1}{2} k y^2\)
2. \(\frac{1}{2} k\left(x^2+y^2\right)\)
3. \(\frac{1}{2} k(x+y)^2\)
4. \(\frac{1}{2} k y(2 x+y)\)

Answer: 4. \(\frac{1}{2} k y(2 x+y)\)

Question 28. An ideal spring with spring constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released when the spring is initially unstretched. Then the maximum extension in the spring is

1. \(\frac{4 M g}{k}\)
2. \(\frac{2 M g}{k}\)
3. \(\frac{M g}{k}\)
4. \(\frac{M g}{2 k}\)

Answer: 2. \(\frac{2 M g}{k}\)

Question 29. TWo springs P and Q of force constant k_P and k_Q \(\left(k_Q=\frac{k_P}{2}\right)\) are stretched by applying forces of equal magnitude. If the energy stored in Q is E, then the energy stored in P is

1. E
2. 2 E
3. E/4
4. E/2

Answer: 4. E/2

Question 30. Before snapping, a wire can bear a load of 100 kg. The wire is cut into two equal parts. Now the maximum load that can be withstood by each part of the wire is

1. 100 kg
2. 40 kg
3. 200 kg
4. 50 kg

Answer: 1. 100 kg

Question 31. A uniform wire of length I and weight W is rigidly fixed at one end and a load W₁ is applied at its other end. If the area of cross-section of the wire is S, then stress developed in the wire at a distance 3L/4 from its lower end will be (assume that increase in length of the wire is very small)

1. \(\frac{W_1}{S}\)
2. \(\frac{\left(\frac{3 W}{4}\right)}{S}\)
3. \(\frac{\left(W_1+\frac{3 W}{4}\right)}{S}\)
4. \(\left(\frac{W_1+\frac{W}{4}}{S}\right)\)

Answer: 3. \(\frac{\left(W_1+\frac{3 W}{4}\right)}{S}\)

Question 32. A steel ring of radius r and cross-section A is fitted onto a wooden disc of radius R(R> r). If Young’s modulus be E, then the force with which the steel ring is expanded is

1. \(A E \frac{R}{r}\)
2. \(\frac{E r}{A R}\)
3. \(\frac{E}{A}\left(\frac{R-r}{A}\right)\)
4. \(A E\left(\frac{R-r}{r}\right)\)

Answer: 4. \(A E\left(\frac{R-r}{r}\right)\)

In the type of questions more than one options are correct.

“Online Tests For Elasticity Of Demand”

Question 33. A wire of length L and cross-section A hung from a rigid support is loaded with a mass M. The elongation produced is

1. Inversely proportional to L
2. Directly proportional to M
3. Directly proportional to young’s modulus
4. Inversely proportional to A

Answer:

2. Directly proportional to M

4. Inversely proportional to A

Question 34. Which of the following statements are correct regarding elasticity?

1. Rubber does not obey Hooke’s law
2. Elasticity can be different for tensile and compressive stress
3. Elasticity is independent of temperature
4. Poisson’s ratio is a modulus of elasticity

Answer:

1. Rubber does not obey Hooke’s law

2. Elasticity can be different for tensile and compressive stress

Question 35. Potential energy per unit volume of a stretched wire is

1. 1/2 stress x strain
2. \(\frac{1}{2} \frac{\text { stress }}{\text { strain }}\)
3. 1/2 Young’s modulus x strain²
4. 1/2 x Young’s modulus x strain

Answer:

1. 1/2 stress x strain

2. \(\frac{1}{2} \frac{\text { stress }}{\text { strain }}\)

“Elasticity Of Demand For Specific Products Questions”

Question 36. Two wires A and B have equal lengths and are made of the same material, but the diameter of A is twice that of wire B. Then, for a given load

1. The extension of B will be four times that of A
2. The extension of A and B will be equal
3. The strain in B is four times that in A
4. The strain in A and B will be equal

Answer:

1. The extension of B will be four times that of A

3. The strain in B is four times that in A

Question 37. The figure shows the stress-strain graphs for materials A and B. From the graph it follows that

1. Material A has a higher Young’s modulus
2. Material B is more ductile
3. Material B can withstand greater stress
4. Material A can withstand greater stress

Answer:

1. Material A has a higher Young’s modulus

4. Material A can withstand greater stress

Question 38. Two wires A and B have the same cross-section and are made of the same material but the length of wire A is twice that of B. Then for a given load

1. The extension of A will be twice that of B
2. The extension of A and B will be equal
3. The strain in A will be half that in B
4. The strains in A and B will be equal

Answer:

1. The extension of A will be twice that of B

4. The strains in A and B will be equal

“Price Elasticity Scenarios For Economics Students”

Question 39. Choose the correct statement (s) from the following.

1. Steel is more elastic than rubber
2. The stretching of a coil spring is determined by the Young’s modulus of the wire of the spring
3. The frequency of a tuning fork is determined by the shear modulus of the material of the fork
4. When a material is subjected to a tensile (stretching) stress the restoring force is caused by interatomic attraction

Answer:

1. Steel is more elastic than rubber

4. When a material is subjected to a tensile (stretching) stress the restoring force is caused by interatomic attraction

Elasticity Long Answer Type Questions And Answers

Basic Elasticity Questions for Students

Question 1. Which is more elastic steel or diamond?
Answer:

Diamond is more elastic than steel

Magnitude of elastic limit is the degree of elasticity of a material. Higher the elastic limit of a material, greater is its degree of elasticity. The elastic limit of diamond is more than that of steel and hence diamond is more elastic than steel.

Question 2. Explain why the temperature of a wire under tension will change if it snaps suddenly.
Answer:

The temperature of a wire under tension will change if it snaps suddenly

During elongation of a wire the work done remains stored as elastic potential energy in the wire. When the wire snaps suddenly, that stored potential energy is transformed into heat energy resulting in increase in temperature.

Read And Learn More WBCHSE Solutions for Class 11 Physics

Question 3. Two bodies M and N of equal mass are hung separately from two lightweight springs. Force constants of the springs are k₁ and k₂. The bodies are set to vibrate so that their maximum velocities are equal. Find the ratio of the amplitudes of vibration of the two bodies.
Answer:

Given

Two bodies M and N of equal mass are hung separately from two lightweight springs. Force constants of the springs are k₁ and k₂. The bodies are set to vibrate so that their maximum velocities are equal.

A body suspended from a spring will acquire maximum velocity at the equilibrium position in its path of vibration. At that position potential energy of the body becomes zero and its energy becomes totally kinetic.

Let the amplitude of vibration of the body M be x₁ and that of N be x₂. Since they are of equal mass and their maximum velocities are also the same, their maximum kinetic energies will also be equal.

[Maximum kinetic energy = 1/2 x mass x (maximum velocity)²]. This kinetic energy transforms into the potential energy (1/2kx²) of the body at the end of its amplitude.

∴ \(\frac{1}{2} k_1 x_1^2=\frac{1}{2} k_2 x_2^2 \text { or, } \frac{x_1}{x_2}=\sqrt{\frac{k_2}{k_1}} \text {. }\)

Question 4. Springs are usually made of steel but not of copper. Why?
Answer:

Springs are usually made of steel but not of copper.

Elasticity of steel is more than that of copper. It means that the elastic limit of steel is greater. Let us consider two springs of the same size, one of steel and another of copper.

An equal tensile force is applied on both springs. On slowly increasing the magnitude of the applied force on the two springs it is seen that, at a certain stage the steel spring still behaves like an elastic body but the copper spring undergoes a permanent deformation.

For this reason, springs are usually made of steel but not of copper. Moreover, copper is costlier than steel. However, copper springs may be used where the applied force is not very high.

Question 5. On the basis of the moduli of elasticity distinguish between solid, liquid, and gaseous substances.
Answer:

We know that there are three moduli of elasticity

1. Young’s modulus,
2. Bulk modulus and
3. Modulus of rigidity.

Since solids have definite length, volume, and shape, it has all the three moduli of elasticity. Liquid and gaseous substances have no definite length or shape, but have volume only.

Hence a liquid or a gaseous substance only has bulk modulus. This modulus of a gas is much less than that of a liquid. But the bulk moduli of a solid and a liquid are nearly equal in magnitude.

Question 6. The Poisson’s ratio of a wire is σ. Show that if e is the longitudinal strain due to an applied force, the volume strain will be e( 1 – 2σ).
Answer:

The Poisson’s ratio of a wire is σ.

Suppose, the length of the wire is l, radius is r and its volume is V.

∴ V = πr²l

or, dV = \(\pi r^2 d l+2 \pi l r d r=\pi r^2 d l-2 \pi r^2 \sigma d l\)

[\(\sigma=\frac{-d r / r}{d l / l}\), dr = \(-r \sigma \frac{d l}{l}\)]

= \(\pi r^2 d l(1-2 \sigma)\)

∴ Volume strain = \(\frac{d V}{V}=\frac{\pi r^2 d l(1-2 \sigma)}{\pi r^2 l}=\frac{d l}{l}(1-2 \sigma)\)

= \(e(1-2 \sigma)\left[\text { given that } \frac{d l}{l}=e\right]\)

Practice Questions on Elasticity Concepts

Question 7. In the case of an elastic body which one is more fundamental—stress or strain?
Answer:

When an elastic body gets strained under the influence of external forces, a reaction force develops inside the body. This is the source of stress. This stress helps the deformed body to regain its original shape. It means that only when strain is produced in a body, a stress is developed within it. So, elastic strain is more fundamental than stress.

Question 8. Within elastic limit the Poisson’s ratio depends only on the nature of the material but not on the stress applied”—explain.
Answer:

With the increase in stress, longitudinal strain as well as lateral strain will increase proportionately. As a result, Poisson’s ratio remains fixed, because

Poisson’s ratio = \(\frac{\text { lateral strain }}{\text { longitudinal strain }}\); hence with in elastic limit Poisson’s ratio is independent of the stress applied.

Question 9. For a steel wire, if the diameter is larger, it can withstand a greater load. Why?
Answer:

We know that, breaking stress = \(\frac{\text { breaking load }}{\text { area of cross-section of the wire }}\)

or, breaking load = breaking stress x area of cross-section of the wire

= breaking stress x \(\frac{\pi d^2}{4}\)

[d = diameter of the wire]

Since for a particular material, the breaking stress is fixed, a wire with a larger diameter (d) is able to withstand a greater load.

Elasticity Problems and Solutions

Question 10. In the Fig, load-elongation graphs of two wires made of two different materials A and B are shown. The wires have the same length and the same area of cross-section. Which material has a greater value of F?
Answer:

Given

In the Fig, load-elongation graphs of two wires made of two different materials A and B are shown. The wires have the same length and the same area of cross-section.

For the same load F, elongations in the wires made of materials A and B are l₁ and l₂ respectively.

Since their initial lengths are the same, (longitudinal strain)_A < (longitudinal strain)B.

Now Y = \(\frac{\text { longitudinal stress }}{\text { longitudinal strain }}\)

Hence, YA> YB.

Question 11. Load-elongation graphs of two wires A and B, made of the same material and of equal initial length are shown in the Fig. Which wire is thicker?
Answer:

Given

Load-elongation graphs of two wires A and B, made of the same material and of equal initial length are shown in the Fig.

Y = \(\frac{F L}{\alpha l}\)

or, Y = \(\frac{F L}{\alpha l}\)(α = area of cross-section of the wires)

F, L, and Y are the same for the two wires

∴ \(\alpha \propto \frac{1}{l}\)

So the wire of smaller elongation has a larger cross-section.

∴ (α)_A>(α)_B

Hence, the wire A is thicker than the wire B.

Question 12. Young’s moduli of two rods of equal length and equal cross-section are Y₁ and Y₂. These rods are joined end to end forming a composite rod-system. Prove that the equivalent Young’s modulus of the composite system of rods = \(\frac{2 Y_1 Y_2}{Y_1+Y_2} .\)
Answer:

Given

Young’s moduli of two rods of equal length and equal cross-section are Y₁ and Y₂. These rods are joined end to end forming a composite rod-system.

Let for applied force F, elongations of the two rods be /j and /2 respectively.

∴ \(l_1=\frac{F L}{Y_1 A} ; l_2=\frac{F L}{Y_2 A}\)

 

A = area of cross-section of each rod;

L = original length of each rod.

Now, equivalent Young’s modulus of the composite rod-system,

Y = \(\frac{\frac{F}{A}}{\frac{l_1+L_2}{2 L}}=\frac{2 F L}{A\left(l_1+l_2\right)}\)

= \(\frac{2 F L}{A\left[\frac{F L}{Y_1 A}+\frac{F L}{Y_2 A}\right]}\)

= \(\frac{2}{\frac{1}{Y_1}+\frac{1}{Y_2}}=\frac{2 Y_1 Y_2}{Y_1+Y_2}\)

Question 13. State whether the values of Young’s moduli for thin and thick iron wires of equal length will be different.
Answer:

Young’s moduli of two iron wires of the same length but of different thicknesses cannot be different because it depends only on the material of the wire.

Question 14. Can a steel wire be elongated to twice its initial length by hanging a load from its end?
Answer:

By hanging a load from its free end, a steel wire can¬not be elongated to twice its initial length. This is because the wire snaps before attaining that elongation, as it crosses its breaking load.

Question 15. How does the value of modulus of elasticity change due to increase in temperature?
Answer:

In most cases, the value of elastic modulus decreases slightly due to increase in temperature

Question 16. A hanging wire of length L is elongated by an amount l with a load M attached to Its free end. Prove that the elastic potential energy stored In the wire Is 1/2 Mgl.
Answer:

Given

A hanging wire of length L is elongated by an amount l with a load M attached to Its free end.

Elastic potential energy stored in the wire

= 1/2 x force x increase in length

= 1/2 x Mgx l = 1/2 Mgl.

Question 17. A spring balance gives erroneous readings if It is used frequently over a long period of time. Explain.
Answer:

A spring balance gives erroneous readings if It is used frequently over a long period of time.

On using a spring balance frequently over a long period of time, its elastic property degrades and it deforms permanently. As a result, the elongation of the spring is more than the elongation it should suffer for a given load suspended at its free end. So, we get a wrong reading.

Question 18. An elastic wire is cut into two equal halves. Deter – mine whether there will be any change in the maxi¬mum load that each half can carry.
Answer:

An elastic wire is cut into two equal halves.

Young’s modulus, Y = \(\frac{\text { stress }}{\text { strain }}\). For maximum load, the stress is the breaking stress. For wires of the same material, both Y and the breaking stress have fixed characteristic values.

So, when a wire carries the maximum load, the strain acquires a particular value. This value is independent of the initial length of the wire. Therefore, two wires of the same material of lengths L₁ and L₂, can carry the same maximum load.

Question 19. The breaking force for a wire is F. What will be the breaking force for

1. two parallel wires of same size and
2. for a single wire of double the thickness?

Answer:

1. When two wires of the same size are suspended in parallel; a force equal to 2F has to be applied on the parallel combination, so that a force equal to the breaking force for the wire acts on each of the two wires.

2. Now, F = \(\frac{Y a l}{L}=\frac{Y\left(\pi d^2\right) l}{4 L}\)

or, F ∝ d²

If the wire is of double the thickness i.e., of double the diameter then breaking force will be 4F.

Elasticity Short Questions And Answers

Question 1. A spring is cut into two equal pieces. What is the spring the constant of each part if the spring constant of the original spring is k,
Solution:

Given

A spring is cut into two equal pieces.

Let us consider that the spring elongates by x when a force F is applied on it. So, the force constant of the spring, k = F/x.

Now, if the spring is cut into two equal parts, then on the application of the same force F, each part of the spring will elongate by x/2.

The force constant each part, \(k^{\prime}=\frac{F}{\frac{x}{2}}=\frac{2 F}{x}=2 k\)

Question 2. A spring having spring constant k is cut into two parts in the ratio 1:2. Find the spring constants of the two parts.
Solution:

Given

A spring having spring constant k is cut into two parts in the ratio 1:2.

Let the initial length of the spring be x.

The spring constant of a particular spring is inversely proportional to its length.

∴ kx = constant.

When the spring is cut into two parts in the ratio 1:2, the length of the two parts are x/3 and 2x/3 respectively.

⇒\(k_1 \frac{x}{3}=k x \text { or, } k_1=3 k\)

and \(k_2 \cdot \frac{2 x}{3}=k x \text { or, } k_2=\frac{3 k}{2}\)

Question 3. The length of a metal wire is L₁. when the tension is T₁ and L₂ when the tension is T₂ The unstretched length of the wire is

1. \(\frac{L_1+L_2}{2}\)
2. \(\sqrt{L_1 L_2}\)
3. \(\frac{T_2 L_1-T_1 L_2}{T_2-T_1}\)
4. \(\frac{T_2 L_1+T_1 L_2}{T_2+T_1}\)

Answer:

Given

The length of a metal wire is L₁. when the tension is T₁ and L₂ when the tension is T₂

Young’s modulus, Y = \(\frac{\text { stress }}{\text { strain }}\)

or, strain = \(\frac{\text { stress }}{\text { strain }}\)

If the length of the wire is L₀ when there is no tension in the string, then in the first case, stress = \(\frac{T_1}{\alpha}\) and strain = \(\frac{L_1-L_0}{L_0}\)

[a = area of cross-section = constant (approximately)]

So, \(\frac{L_1-L_0}{L_0}=\frac{T_1}{\alpha Y} \quad \text { or, } \frac{1}{\alpha Y}=\frac{1}{T_1}\left(\frac{L_1}{L_0}-1\right)\)

Similarly in the second case, \(\frac{1}{\alpha Y}=\frac{1}{T_2}\left(\frac{L_2}{L_0}-1\right)\)

So, \(\frac{1}{T_1}\left(\frac{L_1}{L_0}-1\right)=\frac{1}{T_2}\left(\frac{L_2}{L_0}-1\right)\)

or, \(\frac{1}{L_0}\left(\frac{L_1}{T_1}-\frac{L_2}{T_2}\right)=\frac{1}{T_1}-\frac{1}{T_2}\)

or, \(\frac{1}{L_0} \frac{T_2 L_1-T_1 L_2}{T_1 T_2}=\frac{T_2-T_1}{T_1 T_2}\)

∴ \(L_0=\frac{T_2 L_1-T_1 L_2}{T_2-T_1}\)

The option 3  is correct

Question 4. A liquid of bulk modulus k is compressed by applying an external pressure such that its density increased by 0.04% . The pressure applied on the liquid is

1. k/10000
2. k/10000
3. 1000k
4. 0.01k

Answer:

k = \(\frac{p}{\frac{\Delta V}{V}}\)

or, \(p=k \times \frac{\Delta V}{V}=k \times \frac{\Delta \rho}{\rho}=k \times 0.01 \%=\frac{k}{10000}\)

The option 1 is correct.

Real-Life Applications of Elasticity Questions

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 5. The stress along the length of a rod (with rectangular cross section) is 1% of the Young’s modulus of its material. What is the approximate percentage of change of its volume? (Poisson’s ratio of the material of the rod is 0.3)

1. 3%
2. 1%
3. 0.7%
4. 0.4%

Answer:

Let, volume of the rod, V = xyz and Young’s modulus of its material of the rod = Y

Now, \(\frac{F}{A}=Y \times 1 \%\)

or, \(Y \times \frac{\Delta x}{x}=\frac{Y}{100}\)

or, \(\frac{\Delta x}{x}=0.01\)

∴ \(\frac{\Delta V}{V}=\frac{\Delta x}{x}+\frac{\Delta y}{y}+\frac{\Delta z}{z}\)

= \(\frac{\Delta x}{x}-\sigma \frac{\Delta x}{x}-\sigma \frac{\Delta x}{x}\) ……..(1)

[Poisson’s ratio, \(\sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}=\frac{\frac{\Delta y}{y}}{\frac{\Delta x}{x}}=\frac{\frac{\Delta z}{z}}{\frac{\Delta x}{x}}\)]

Negative symbol in equation (1) implies that, as length increases due to stress, value of y and z decreases simultaneously.

∴ From equation (1),

∴ \(\frac{\Delta V}{V}=0.01-2 \times 0.3 \times 0.01=0.004=0.4 \%\)

The option 4 is correct.

Question 6. When a rubber band is stretched by a distance x, it exerts a restoring force of magnitude F = ax + bx², where a and b are constants. The work done in stretching the unstretched rubber band by L isothermal

1. \(a L^2+b L^3\)
2. \(\frac{1}{2}\left(a L^2+b L^3\right)\)
3. \(\frac{a L^2}{2}+\frac{b L^3}{3}\)
4. \(\frac{1}{2}\left(\frac{a L^2}{2}+\frac{b L^3}{3}\right)\)

Answer:

⇒ \(\int d W=\int F d l\)

W = \(\int_0^L a x d x+\int_0^L b x^2 d x=\frac{a L^2}{2}+\frac{b L^3}{3}\)

The option 3 is correct.

Question 7. A man grows into a giant such that his linear dimensions increase by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of

1. 9
2. 1/9
3. 81
4. 1/81

Answer:

According to the question, \(\frac{V_f}{V_i}=(9)^3\)

So, \(\frac{m_f}{m_i}=(9)^3\)

Also, \(\frac{A_f}{A_i}=(9)^2\)

Stress = \(\frac{\text { force }}{\text { area }}=\frac{m \times g}{A}\)

∴ \(\frac{S_f}{S_i}=\frac{m_f}{m_i} \times \frac{A_i}{A_f}=(9)^3 \times \frac{1}{(9)^2}=9\)

The option 1 is correct.

Question 8. An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by

1. \(\frac{P}{3 \alpha K}\)
2. \(\frac{P}{\alpha K}\)
3. \(\frac{3 \alpha}{P K}\)
4. \(3 P K \alpha\)

Answer:

Bulk modulus, K= \(\frac{P}{\left(\frac{\Delta V}{V}\right)}\)

∴ \(\frac{\Delta V}{V}=\frac{P}{K}[\Delta V= change in volume]\)

If we bring back the cube to its original size by increasing the temperature Δt,

⇒ \(\Delta V=V \cdot \gamma \Delta t\)

or, \(\Delta t=\frac{\Delta V}{V} \cdot \frac{1}{\gamma}=\frac{\Delta V}{V} \cdot \frac{1}{3 \alpha}=\frac{P}{3 k \alpha}\)

The option (1) is correct.

Key Terms in Elasticity with Questions

Question 9. A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross-section of cylindrical container. When a mass m is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, (dr/r) is

1. \(\frac{m g}{3 K a}\)
2. \(\frac{m g}{K a}\)
3. \(\frac{K a}{m g}\)
4. \(\frac{K a}{3 m g}\)

Answer:

Bulk modulus,  K = \(-V \frac{d p}{d V}\)

or, \(-\frac{d V}{V}=\frac{d p}{K}\)

or, \(\frac{-3 d r}{r}=\frac{\frac{m g}{a}}{K}\left[because V=\frac{4}{3} \pi r^3\right]\)

or, \(\frac{d r}{r}=-\frac{m g}{3 K a} \quad therefore\left|\frac{d r}{r}\right|=\frac{m g}{3 K a}\)

The option 1 is correct

Question 10. Copper of fixed volume V is drawn into wire of length l. When this wire is subjected to a constant force F, the extension produced in the wire is Δl. Which of the following graphs is a straight line?

1. Δl versus 1/l
2. Δl versus l²
3. Δl versus 1/l²
4. Δl versus l

Answer:

Y = \(\frac{F l}{A \Delta l} \text { or, } \Delta l=\frac{F l}{A Y}=\frac{F l^2}{V Y}\)

∴ \(\Delta l \propto l^2\)

The option 2 is correct.

Question 11. The approximate depth of an ocean is 2700 m. The compressibility of water is 45.4 x 10^-11 Pal^-1 and density of water is 10³kg/m³. What fractional compression of water will be obtained at the bottom of the ocean?

1. 0.8 x 10^-2
2. 1.0 x 10^-2
3. 1.2 x 10^-2
4. 1.4×10^-2

Answer:

Due to AP amount of increase in pressure, there is AV
amount of compression in volume V.

So, fractional compression = \(\frac{\Delta V}{V}\)

and compressibility, K = \(\frac{1}{V} \frac{\Delta V}{\Delta P}\)

Now consider the magnitude, \(\frac{\Delta V}{V}=K \Delta P\)

Here, ΔP = hρg = 2700 x 10³ x 10 Pa [taking g = 10m/s²]

Hence, fractional compression, \(\frac{\Delta V}{V} =\left(45.4 \times 10^{-11}\right) \times\left(2700 \times 10^3 \times 10\right)\)

= \(1.226 \times 10^{-2}\)

The option 3 is correct.

Question 12. The density of a metal at normal pressure is p. Its density when it is subjected to an excess pressure p is p’ . If B is bulk modulus of the metal, the ratio of \(\frac{e^{\prime}}{\rho}\)

1. \(1+\frac{B}{p}\)
2. \(\frac{1}{1-\frac{p}{B}}\)
3. \(1+\frac{p}{B}\)
4. \(\frac{1}{1+\frac{P}{B}}\)

Answer:

Volume strain = change in pressure = p

Initial volume, V = \(\frac{M}{\rho}\)

Final volume, \(V^{\prime}=\frac{M}{\rho^{\prime}}\)

Change in volume, \(V^{\prime}-V=M\left(\frac{\rho-\rho^{\prime}}{\rho^{\prime} \rho}\right)\)

∴ Volume strain = \(=\frac{V^{\prime}-V}{V}=\frac{\rho-\rho^{\prime}}{\rho^{\prime}}\)

∴ \(B=-\frac{p V}{V^{\prime}-V}=-\frac{p \times \rho^{\prime}}{\rho-\rho^{\prime}}\)

or, \(\underset{B}{p}=-\frac{\rho-\rho^{\prime}}{\rho^{\prime}}=\frac{\rho^{\prime}-\rho}{\rho^{\prime}}\)

or, \(\frac{\rho}{\rho^{\prime}}=1-\frac{p}{B}\)

∴ \(\frac{\rho^{\prime}}{\rho}=\frac{1}{1-\frac{p}{B}}\)

The option 2 is correct.

Question 12. Two wires are made of the same material and have the same volume. The first wire has cross-sectional area A and the second wire has cross-sectional area 3A. If the length of the first wire is increased by Δl. on applying a force F, how much force is needed to stretch the second wire by the same amount?

1. 4F
2. 6F
3. 9F
4. F

Answer:

In case of first wire, Y = \(\frac{F / A}{\Delta l / l_0}=\frac{F l_0}{A \Delta l}\)

or, \(F=\frac{Y A \Delta l}{l_0}\)

In case of second wire,

Y = \(\frac{F^{\prime} / 3 A}{\frac{\Delta l}{l_0 / 3}}=\frac{F^{\prime} l_0}{9 A \Delta l}\)

or, \(F^{\prime}=\frac{9 Y A \Delta l}{l_0}=9 F\)

The option 3 is correct.

Question 13. Which type of substances are called elastomers? Give one example.
Answer:

Elastomers:

Elastomers are those materials for which stress-strain variation is not a straight line within elastic limit. An elastomer is a polymer with viscoelasticity (colloquially elasticity), generally having low Young’s modulus and high failure strain compared with other materials.

Example: Rubber.

Question 14. Bridges are declared unsafe after long use. Why?
Answer:

Bridges are declared unsafe after long use.

A bridge undergoes alternating stress and strain for a large number of times during its use. A bridge loses its elastic strength when it is used for long time. Therefore, the amount of strain for a given stress will become large and ultimately the bridge will collapse. So, they are declared unsafe after long use.

Question 15. What are elastomers? Give two examples for the same.
Answer:

Elastomers:

Elastomers (elastic polymers) are materials of low Young’s modulus but of very high elastic limits. Such a material can withstand high strain but can still develop sufficient stress to bring it back to its initial size and shape.

Examples: natural rubber, thermoplastics.

Question 16. What is the value of rigidity modulus of elasticity for an incompressible liquid?
Answer:

Value of rigidity modulus of elasticity for an incompressible liquid

A liquid, compressible or incompressible, does not have any define shape; it cannot withstand shear. So it can never generate any shearing stress. Hence the rigidity modulus of elasticity of a liquid is zero.

Question 17. Which type of energy is stored in the spring of wrist watch?
Answer:

Potential energy is stored in the spring of wrist watch.

Question 18. The stress-strain graph for materials A and B are as shown in the graphs are drawn to the same scale, which graph represents property of ductile materials? Justify your answer.
Answer:

The graph for material A represents the property of ductile material because of its greater plastic range.

Question 19. A two wires A and B of length l, radius r, and length 21, radius 2 r having same Young’s modulus Y are hung with a weight mg as shown in figure. What is the net elongation in the two wires?
Answer:

Given

A two wires A and B of length l, radius r, and length 21, radius 2 r having same Young’s modulus Y are hung with a weight mg as shown in figure.

The length and radius of wire A are l and r and that of wire B are 2l and 2 r respectively.

If l₁ and l₂ be the individual elonga¬tion of wire A and wire B, then the net elongation,

∴ \(\Delta l =\Delta l_1+\Delta l_2=\frac{m g l}{\pi r^2 Y}+\frac{m g(2 l)}{\pi(2 r)^2 Y}\)

= \(\left(\frac{m g l}{\pi r^2 Y}+\frac{2 m g l}{4 \pi r^2 Y}\right)=\frac{4 m g l+2 m g l}{4 \pi r^2 Y}=\frac{3}{2} \frac{m g l}{\pi r^2 Y}\)

Question 20. Which of the two forces-deforming or restoring is responsible for the elastic behaviour of substance?
Answer:

Restoring force is responsible for the elastic behaviour of substance.

Elasticity Assertion Reason Type Question And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 2 is true; statement D is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, statement 2 is true.

Question 1.

Statement 1: Young’s modulus for a perfectly plastic body is zero.

Statement 2: For a perfectly plastic body, restoring force is zero.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 2.

Statement 1: If length of a rod is doubled the breaking load remains unchanged.

Statement 2: Breaking load is equal to the elastic limit

Answer: 3. Statement 1 is true, statement 2 is false.

Question 3.

Statement 1: Ductile metals are used to prepare thin wires.

Statement 2: In the stress-strain curve of ductile metals, the length between the points representing elastic limit and breaking points is very small.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 4.

Statement 1: The restoring force F on a stretched string at extension x is related to the potential energy U as F = \(=-\frac{d U}{d x}\)

Statement 2: F = -kx and U = 1/2kx², where k is the spring constant.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 5.

Statement 1: Identical springs of steel and copper are equally stretched. More work will be done on the steel spring.

Statement 2: Steel is more elastic than copper.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 6.

Statement 1: A hollow shaft is found to be stronger than a solid shaft made of same material.

Statement 2: The torque required to produce a given twist in hollow cylinder is greater than that required to twist a solid cylinder of same size and same material.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 7.

Statement 1: The bridges are declared unsafe after a long use.

Statement 2: Elastic strength of bridges decreases with time.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 8.

Statement 1: Stress is the internal force per unit area of a body.

Statement 2: Rubber is less elastic than steel.

Answer: 2. Statement 1 is true, statement 2 is true; statement D is not a correct explanation for statement 1.

Elasticity Very Short Answer Type Questions

Question 1. Between rubber and steel—which one is more elastic?
Answer: Steel

Question 2. Which one is more elastic—steel or diamond?
Answer: Diamond

Question 3. State whether the elasticity of a metallic substance increases or decreases with the rise in temperature.
Answer: Decreases

Question 4. ‘Strain has no unit’—state whether the statement is true or false.
Answer: True

Question 5. Write down the dimension of stress.
Answer: ML^-1T^-2

Question 6. Can the length of a steel wire be doubled by hanging a load from its end?
Answer: No

Question 7. Under which kind of stress does a body undergo a change in shape without changing its volume?
Answer: Shearing stress

Question 8. Which property of a spring is represented by its force constant?
Answer: Stiffness of the spring

Question 9. Between steel and copper—which one is usually used to make springs?
Answer: steel

Question 10. Write down the dimension of force constant.
Answer: MT^-2

Question 11. Can liquid and gaseous substances withstand shearing strain?
Answer: No

Question 12. Which property of a metal is manifested when compressional stress more than the yield point is developed?
Answer: Plastic property

Question 13. On what factor does the breaking stress of a wire depend?
Answer: On the material of the wire

Question 14. State whether a body undergoes a change in volume due to shearing stress only.
Answer: No

Question 15. ‘Young’s modulus depends on temperature’—state whether it is true or false?
Answer: True

Question 16. Write down the dimension of elastic limit.
Answer: MLT^-2

Question 17. A wire is halved by cutting it. Would there be any change in the breaking load due to this?
Answer: No

Question 18. All bodies are _______ elastic in reality.
Answer: Partly

Question 19. Strain is a ______ quantity.
Answer: Dimensionless

Question 20. During change in length of a wire or change in volume of a body, ______ stress is developed.
Answer: Normal

Question 21. During change in shape of a body, stress is developed.
Answer: Shearing

Question 22. Compressibilities of solidsand liquids are very _____, but that of a gaseous substance is much ______
Answer: low, higher

Question 23. Modulus of rigidity is a characteristic of ________.
Answer: Solids

Question 24. Poisson’s ratio depends on the of a body.
Answer: Material

Question 25. Loss in elastic ability of a body due to rapid change in the load applied on it is called _____
Answer: Elastic fatigue

Question 26. What is the SI unit of force constant?
Answer: N · m^-1

Question 27. ______ are usually made of not of copper.
Answer: Springs, steel

Question 28. The magnitude of the _______ per unit cross-sectional area on a body is the breaking stress.
Answer: Breaking load

Question 29. Write the name of a substance whose elasticity does not change with the change in temperature.
Answer: Invar

Question 30. What is defined by dividing stress by strain within elastic limit?
Answer: Elastic modulus

Question 31. For which kind of substance is Young’s modulus physically meaningful?
Answer: Solid

Question 32. Two iron wires of equal length are taken. One of them is thick and the other is thin. In which case will Young’s modulus be greater?
Answer: Wall be same in either case

Question 33. What is the Poisson’s ratio of a substance whose volume remains unchanged under elastic strains?
Answer: 1/2

Question 34. A wire is cut into two parts. What will be the change in Young’s modulus of the parts?
Answer: No change occurs

Question 35. What is the range of theoretical values of Poisson’s ratio?
Answer: -1 to 1/2

Question 36. Write down the dimension of Poisson’s ratio with respect to M, L and T.
Answer: M⁰L⁰T⁰

Question 37. What is the value of Young’s modulus for a perfectly rigid body?
Answer: Infinite

Question 38. Name the reciprocal of bulk modulus.
Answer: Compressibility

Question 39. What is the dimension of elastic modulus?
Answer: ML^-1T^-2

Question 40. What is the relation between Y, K and σ?
Answer: Y = 3K(1-2σ)

Question 41. What is the relation between Y, n and <x?
Answer: Y = 2n(1 +σ)

Question 42. What is the value of Young’s modulus for a perfectly plastic body?
Answer: Zero

Question 43. What will be the change in temperature of a stretched wire if it snaps suddenly?
Answer: Temperature increases

Question 44. Which type of energy is stored when an elastic wire is elongated by stretching?
Answer: Potential

Question 45. In a stretched wire, potential energy stored per unit volume = 1/2 x _____ x _____
Answer: stress, strain

Question 46. Force constants of two springs are k₁ and k₂(k₁ >k₂)

1. The springs are elongated by the same amount and
2. The springs are elongated by applying the same force. Then for which of the springs is more work performed?

Answer:

1. The first spring,
2. The second spring

Question 47. Force constants of two springs are k₁ and k₂. What will be the equivalent force constant of the spring system when the springs are joined in a parallel combination?
Answer: k₁ + k₂

Elasticity Match Column 1 With Column 2

Question 1.  

Answer: 1. C, 2. B, 3. A, 4. D

Question 2. 

Answer: 1. A, B, C, 2. B, 3. B

Question 3. 

Answer: 1. A, D, 2. A, D, 3. A, D, 4. C

Question 4.

Answer: 1. A, B, C 2. B, C, 3. A, D

Elasticity Comprehension Type Question And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. According to Hooke’s law, within the elastic limit = stress/strain constant. This constant depends on the type of strain or the type of force acting. Tensile stress might result in compressional or elongative strain however, a tangential stress can only cause a shearing strain. After crossing the elastic limit, the material undergoes elongation and beyond a stage beaks. All modulus of elastically are basically constants for the materials under stress.

1. Two wires of same material have length and radius l, r and 2l, r/2 respectively. The ratio of their Young’s modulus is

1. 1:2
2. 2:3
3. 2:1
4. 1:1

Answer: 4. 1:1

2. After crossing the yield region, the material will have

1. Reduced stress
2. Increased stress
3. Breaking stress
4. Constant stress

Answer: 1. Reduced stress

3. If stress/strain is x in elastic region and y in yield region, then

1. x = y
2. x > y
3. x < y
4. x = 2y

Answer: 2. x > y

Question 2. A sphere of radius 0.1 m and mass 8π kg is attached to the lower end of a steel wire of length 5 m and diameter 10^-3 m. The wire is suspended from 5.22 m high ceiling of a room. When the sphere is made to swing as a simple pendulum, it just grazes the floor at the lowest point. Given, Young’s modulus of steel is 1.994 x 10¹¹ N • m^-2.

1. What is the extension of the wire at the mean position when the sphere is oscillating?

1. 0.01m
2. 0.02 m
3. 0.03 m
4. 0.04 m

Answer: 2. 0.02 m

2. The tension in the wire at the mean position when the sphere is oscillating is

1. 199.4 πN
2. 19.94 πN
3. 1.994 πN
4. 0.1994 πN

Answer: 3. 1.994 πN

Question 3. A light rod of length 2 m is suspended from the ceiling horizontally by means of two vertical wires of equal length tied to its ends. One of the wires is made of steel and is of cross section 10^-3m² and the other is of brass of cross-section 2 x 10^-3 m². Young’s modulus for steel is 2 x 10¹¹ N • m^-2 and for brass is 10¹¹ N • m^-2.

1. Find out the position along the rod at which a weight may be hung to produce equal stress in both wires.

1. 1.39m
2. 1.30m
3. 1.33m
4. 1.24m

Answer: 3. 1.33m

2. Find out the position along the rod at which a weight may be hung to produce equal strains on both wires.

1. 1m
2. 1.2m
3. 0.87m
4. 1.05m

Answer: 1. 1m

Question 4. A bar of cross-section A is subjected to equal and opposite tensile force F at its ends. Consider a plane through the bar making an angle θ with a plane at right angles to the bar as shown in Fig.

1. The tensile stress at this plane in terms of F, A and θ is

1. \(\frac{F \cos ^2 \theta}{A}\)
2. \(\frac{F}{A \cos ^2 \theta}\)
3. \(\frac{F \sin ^2 \theta}{A}\)
4. \(\frac{F}{A \sin ^2 \theta}\)

Answer: 1. \(\frac{F \cos ^2 \theta}{A}\)

2. Find the value of  θ  for which the tensile stress is

1. 0°
2. 90°
3. 45°
4. 60°

Answer: 1. 0°

3. What is the shearing stress at the plane in terms of F, A, and θ?

1. \(\frac{F \cos 2 \theta}{2 A}\)
2. \(\frac{F \sin 2 \theta}{2 A}\)
3. \(\frac{F \sin \theta}{A}\)
4. \(\frac{F \cos \theta}{A}\)

Answer: 2. \(\frac{F \sin 2 \theta}{2 A}\)

Question 5. To measure Poisson’s ratio for rubber, the apparatus shown in Fig is used. T is a piece of thin-walled rubber tubing nearly 1 m long and 2 cm in diameter. The lower end is closed and is joined to a hanger H on which slotted weights can be slipped. The upper end is closed by a rubber bung containing a capillary tube C. The whole of the rubber tube and part of the capillary tube are filled with water. The apparatus is supported on a rigid support. When the weights are inserted in the hanger, the length of the rubber tube increases, and meniscus of water in C is observed to fall. The extension in the rubber tube is measured using a vernier microscope.

Let V, a and l be the inner volume, area iv. The stress in the string is of cross section and length respectively of the tube. Then V = al = πR²l ; R is the inner radius of the tube and r is the radius of the capillary tube. Let dx be the fall in the level of the meniscus of the capillary tube.

1. What is the Poisson’s ratio?

1. \(\sigma=\left(1-\frac{r}{R} \frac{d x}{d V}\right)\)
2. \(\sigma=\frac{1}{2}\left(1-\frac{r}{R} \frac{d x}{d V}\right)\)
3. \(\sigma=\frac{1}{2}\left(1-\frac{r^2}{R^2} \frac{d x}{d \nu}\right)\)
4. \(\sigma=\left(1-\frac{r^2}{R^2} \frac{d x}{d \nu}\right)\)

Answer: 3. \(\sigma=\frac{1}{2}\left(1-\frac{r^2}{R^2} \frac{d x}{d \nu}\right)\)

2. The nature of graph between dx and dl is

Answer: 2.

Question 6. One end of a string of length L and cross-sectional area A is fixed to a support and the other end is fixed to a bob of mass m. The bob is revolved in a horizontal circle of radius r with an angular velocity ω such that the string makes an angle θ with the vertical. Young’s modulus is Y.

1. The angular velocity w is equal to

1. \(\sqrt{\frac{g \sin \theta}{r}}\)
2. \(\sqrt{\frac{g \cos \theta}{r}}\)
3. \(\sqrt{\frac{g \tan \theta}{r}}\)
4. \(\sqrt{\frac{g \cot \theta}{r}}\)

Answer: 3. \(\sqrt{\frac{g \tan \theta}{r}}\)

2. The tension T in the string is

1. \(\frac{m g}{\cos \theta}\)
2. \(\frac{m g}{\sin \theta}\)
3. \(\frac{m g}{\tan \theta}\)
4. \(m\left(g^2+r^2 \omega^4\right)^2\)

Answer: 1. \(\frac{m g}{\cos \theta}\)

3. The increase ΔL in length of the string is

1. \(\frac{T L}{A Y}\)
2. \(\frac{m g L}{A Y \cos \theta}\)
3. \(\frac{m g L}{A Y \sin \theta}\)
4. \(\frac{m g L}{A Y}\)

Answer: 3. \(\frac{m g L}{A Y \sin \theta}\)

4. The stress in the string is

1. \(\frac{m g}{A}\)
2. \(\frac{m g}{A}\left(1-\frac{r}{L}\right)\)
3. \(\frac{m g}{A}\left(1+\frac{r}{L}\right)\)
4. \(\frac{m g}{A}\left(\frac{r}{L}\right)\)

Answer: 1. \(\frac{m g}{A}\left(\frac{r}{L}\right)\)

Elasticity Integer Answer Type Question And Answers

Question 1. In this type, the answer to each of the questions is a single-digit Integer ranging from 0 to 9. 1. A 0.1 kg mass is suspended from a wire of negligible mass. The length of the wire is 1 m and its cross-sectional area is 4.9 x 10^-7 m². If the mass is pulled a little in the vertically downward direction and released it performs simple harmonic motion of angular frequency 140 rad • s^-1. If the Young’s modulus of the material of the wire is n x 10⁹ N • m^-2. Find the value of n.

Answer: 4

Question 2. Two wires A and B have the same length and area of cross section. But Young’s modulus of A is two times the Young’s modulus of B. Then what is the ratio of the force constant of A to that of B?

Answer: 2

Question 3. For a wire of length l, maximum change in length under stress conditions is 2 mm. What is the change in length (in mm) under same conditions when length of wire is halved?

Answer: 1

Question 4. The density of water at the surface is 1030 kg • m^-3 and bulk modulus of water is 2 x 10⁹ N • m^-2. What is the approximate change in density (in kg • m^-3) of water in a lake at a depth of 400 m below the surface?
Answer: 2

Question 5. A body of mass 3.14 kg is suspended from one end of a wire of length 10.0 m. The radius of the wire is changing uniformly from 9.8 x 10^-4 m at one end to 5.0 x 10^-4 m at the other end. Find the change in the length of the wire in mm. Young’s modulus of the material of the wire is 2x 10¹¹ N • m^-2.
Answer: 1

Elasticity Applications of Elasticity in Daily Life

Importance of Elasticity in Daily Products

Most materials used in our daily life undergo some kind of stress. It is therefore important to design things in a way that they continue working under the stress suffered by them. The following examples will illustrate this point.

1. In cranes: The thickness of metallic ropes used in cranes to lift and move heavy weights is decided on the basis of the elastic limit of the material of the rope and the factor of safety.

2. Indesigningabeam: When a transverse load is put across ahorizontal beam, it causes ade pression leading to bending in the beam. In various applications like bridges, buildings, etc., it is important that the beam can with stand the load or the weight and it should not bend too much or break. When the bending is not accompanied by any torsion or shear, it is said to be simple bending.

Elastic Materials Used in Clothing

3. A bridge is declared unsafe after long use: During its long use, a bridge undergoes quick alternating strains repeatedly. It results in the loss of elastic strength of the bridge. After a long period, such a bridge starts developing large strains corresponding to the same usual value of stress and ultimately it may lead to the collapse of the bridge. To avoid such a situation a bridge is declared unsafe after its long use.

Read and Learn More: Class 11 Physics Notes

Elasticity Applications of Elasticity in Daily Life Numerical Examples

Example 1. To increase the length of an elastic string of radius 3.5 mm by 1/20 th of its initial length, within its elastic limit, a 10 N force is required. Calculate the Young’s modulus for the material of the string.
Solution:

Given

To increase the length of an elastic string of radius 3.5 mm by 1/20 th of its initial length, within its elastic limit, a 10 N force is required.

We know that, Y = \(\frac{F L}{\pi r^2 l} .\)

Here, F = 10 N , l = 1/20 L and r = 3.5 mm = 0.0035 m

∴ Y = \(\frac{10 \times L}{3.14 \times(0.0035)^2 \times \frac{L}{20}}=5.2 \times 10^6 \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Example 2. Two wires of the same length but of different materials have diameters of 1 mm and 3 mm respectively. If both of them are stretched by the same force, then the elongation of the first wire becomes thrice that of the second. Compare their Young’s moduli.
Solution:

Given

Two wires of the same length but of different materials have diameters of 1 mm and 3 mm respectively. If both of them are stretched by the same force

Let the initial length of each wire be L and the force applied on each be F.

∴ Y = \(\frac{F L}{\pi r^2 l}\)

So, \(\frac{Y_1}{Y_2}=\left(\frac{r_2}{r_1}\right)^2 \cdot \frac{l_2}{l_1}=\left(\frac{3}{1}\right)^2 \times \frac{1}{3}=\frac{3}{1}\)

∴ \(Y_1: Y_2=3: 1\)

Example 3. If the elastic limit of a typical rock is 3 x 10⁸ N• m^-2 and its mean density is 3 x 10³ kg • m^-3, estimate the maximum height of a mountain on the earth. (g =  10 m•s^-2)
Solution:

Given

If the elastic limit of a typical rock is 3 x 10⁸ N• m^-2 and its mean density is 3 x 10³ kg • m^-3

Let us assume that the maximum height of the mountain is h and g is nearly uniform along this height.

Then the maximum pressure at its bottom = ρg.

According to the problem, hρg = breaking stress = elastic limit.

∴ h = \(\frac{\text { elastic limit }}{\rho g}\)

Here, ρ = 3 x 10³ kg • m^-3 and the elastic limit = 3 x 10⁸ N • m^-2

∴ h = \(\frac{3 \times 10^8}{3 \times 10^3 \times 10}=10^4 \mathrm{~m}\)

Role of Elasticity in Construction Materials

Example 4. Two equal and opposite forces are applied tangentially to two mutually opposite faces of an aluminum cube of side 3 cm to produce a shear of 0.01°. If the modulus of rigidity for aluminum is 7 x 10¹⁰ N • m^-2, then calculate the force applied.
Solution:

Given

Two equal and opposite forces are applied tangentially to two mutually opposite faces of an aluminum cube of side 3 cm to produce a shear of 0.01°. If the modulus of rigidity for aluminum is 7 x 10¹⁰ N • m^-2,

Let the magnitude of applied force be F.

We know that, n = \(\frac{F / A}{\theta} \text { or, } F=n \theta A\)

⇒ \(\theta=0.01^{\circ}=\frac{0.01 \times \pi}{180} \mathrm{rad}\)

A = \(9 \mathrm{~cm}^2=9 \times 10^{-4} \mathrm{~m}^2\)

∴ F = \(7 \times 10^{10} \times \frac{0.01 \times 3.14}{180} \times 9 \times 10^{-4}=1.1 \times 10^4 \mathrm{~N}\)

Example 5. A rubber cord of length 20 m is suspended from—a rigid support by one of Its ends and it hangs vertically. What will be the elongation of the cord due to its own weight? The density of rubber = 1.5 g • cm^-3 and Young’s modulus = 49 x 10⁷ N • m^-2.
Solution:

Given

A rubber cord of length 20 m is suspended from—a rigid support by one of Its ends and it hangs vertically.

Here the downward force,

F = weight of the cord

= volume of the cord x density x acceleration due to gravity

= 20 x Ax 1.5 x 1000 x 9.8 N

[A = area of cross-section of the cord]

The centre of gravity of the rubber cord is at a vertical distance of 10 m from the fixed end. The weight of the cord acts through its centre of gravity.

Hence the length of the upper half, above the centre of gravity, is taken as the initial length to estimate the elongation of the cord. So, L = 10 m.

∴Y = \(\frac{F L}{A l}\)

or, \( l=\frac{F L}{A Y}=\frac{20 \times A \times 1.5 \times 1000 \times 9.8 \times 10}{A \times 49 \times 10^7}=0.006 \mathrm{~m}\)

Example 6. A steel wire of  diameter 0.8 mm and length 1 m is clamped firmly at two points A and B which are 1 m apart in the same horizontal line. A body is hung from the middle point ofthe wire such that the | middle point sags 1 cm from the original position. Calculate the mass of the body. [Y = 2 x 10¹² dyn •cm^-2]
Solution:

Given

A steel wire of  diameter 0.8 mm and length 1 m is clamped firmly at two points A and B which are 1 m apart in the same horizontal line. A body is hung from the middle point ofthe wire such that the | middle point sags 1 cm from the original position

A steel wire of  diameter 0.8 mm and length 1 m is clamped firmly at two points A and B which are 1 m apart in the same horizontal line. A body is hung from the middle point ofthe wire such that the | middle point sags 1 cm from the original position.

C is the mid-point of the wire AB.

When a mass m is hung from C, it sags 1 cm and comes down to the point O.

AB = 1 m = 100 cm , AC = CB = 50 cm,

OC = 1 cm , OA = OB = \(\sqrt{50^2+1^2}\) = 50.01 cm,

cosθ = \(\frac{1}{50.01},\), r = 0.04 cm

Here, tension in the part CM or OB is T. The two horizontal
components of T balance each other and the vertical components balance the weight mg.

∴ 2T cosθ = mg or, T = \(\frac{m g}{2 \cos \theta}\)

The length AC of the wire changes into AO.

∴ Elongation = l = AO – Ac = 50.01-50 = 0.01

Young’s modulus, Y = \(\frac{T L}{A l}=\frac{m g L}{2 \cos \theta \cdot A l}\)

or, \(m=\frac{2 Y A l \cos \theta}{g L}\)

= \(\frac{2 \times\left(2 \times 10^{12}\right) \times 3.14 \times(0.04)^2 \times 0.01}{980 \times 50} \times \frac{1}{50.01}\) = 82.01 g

Elasticity in Everyday Household Items

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Example 7. If the work done in stretching a uniform wire, of cross section 1 mm² and length 2 m, by 1 mm is 0.05 joule, find the Young’s modulus for the material of the wire.
Solution:

Given

If the work done in stretching a uniform wire, of cross section 1 mm² and length 2 m, by 1 mm is 0.05 joule

If the work done in stretching a uniform wire, of cross section 1 mm² and length 2 m, by 1 mm is 0.05 joule

Work done, \(W=\frac{1}{2} \frac{Y \alpha l^2}{L}\).

So, \(Y=\frac{2 W L}{\alpha l^2}\)

Here, \(W=0.05 \mathrm{~J} ; \alpha=1 \mathrm{~mm}^2=10^{-6} \mathrm{~m}^2 ; L=2 \mathrm{~m}\)

⇒ \(l=1 \mathrm{~mm}=10^{-3} \mathrm{~m}\)

∴ \(Y=\frac{2 \times 0.05 \times 2}{10^{-6} \times\left(10^{-3}\right)^2}=2 \times 10^{11} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Example 8. A cylindrical pipe of uniform cross-section and of length 120 cm is closed at one end and is completely filled with water. Kept upright, the cylinder is stretched to increase its length. It elongates by 1 cm, but the length of the water column increases by 0.7 cm. Calculate the Poisson’s ratio of the material of the pipe.
Solution:

Given

A cylindrical pipe of uniform cross-section and of length 120 cm is closed at one end and is completely filled with water. Kept upright, the cylinder is stretched to increase its length. It elongates by 1 cm, but the length of the water column increases by 0.7 cm.

Poisson’s ratio, \(\sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

Here, longitudinal strain = \(\frac{l}{L}=\frac{1}{120}\)

Let the initial diameter of the pipe be D, the decrease in its diameter due to elongation be d.

Initial volume of water = \(\frac{\pi}{4} D^2 \times 120\)

and final volume of water = \(\frac{\pi}{4}(D-d)^2 \times(120+0.7)\)

= \(\frac{\pi}{4}(D-d)^2 \times 120.7\)

Since the volume of water inside the cylinder remains
unchanged,

⇒ \(\frac{\pi}{4} D^2 \times 120=\frac{\pi}{4} \times(D-d)^2 \times 120.7\)

or, \(\frac{D-d}{D}=\sqrt{\frac{120}{120.7}} \quad \text { or, } 1-\frac{d}{D}=\sqrt{0.9942}=0.997\)

or, \(\frac{d}{D}=1-0.997=0.003\)

∴ \(\sigma=\frac{d / D}{l /L}=\frac{0.003}{\frac{1}{120}}=0.003 \times 120=0.36\)

Example 9. A light bar of length 2 m is suspended horizontally by means of two wires of equal lengths connected to its two ends. One wire is of steel having a cross-sectional area of 0.1 cm², the other is of brass with a cross-sectional area of 0.2 cm². Find the point the bar from where a weight must be suspended so that both the wires experience

1. the same stress,
2. the same strain. Given, the Young’s modulus for steel =2x 10¹¹ N • m^-2 and that for brass =10 x 10¹⁰ Nm^-2.

Solution:

Given

A light bar of length 2 m is suspended horizontally by means of two wires of equal lengths connected to its two ends. One wire is of steel having a cross-sectional area of 0.1 cm², the other is of brass with a cross-sectional area of 0.2 cm².

Let the horizontal bar be AB. A load W is hung from the point C. Let AC be x. In this condition, let the tension in the steel wire be T₁ and that in the brass wire be T₂.

Stress on the steel wire = \(\frac{T_1}{\alpha_1}\)

stress on the brass wire = \(\frac{T_2}{a_2}\)

1. If the stresses in both the wires are the same, then

⇒ \(\frac{T_1}{0.1 \times 10^{-4}}=\frac{T_2}{0.2 \times 10^{-4}} \text { or, } \frac{T_1}{T_2}=\frac{1}{2}\)…..(1)

Since the system is in equilibrium, taking moments about the point C, we get,

⇒ \(T_1 x=T_2(2-x) \quad \text { or, } \quad \frac{T_1}{T_2}=\frac{2-x}{x}\)

∴ \(\frac{1}{2}=\frac{2}{x}-1 \quad \text { or, } \frac{2}{x}=\frac{3}{2} \text { or, } x=\frac{4}{3}=1.33 \mathrm{~m}\)

So the weight should be suspended at a distance of 1.33 m from the steel wire.

2. Longitudinal strain = \(\frac{\text { longitudinal stress }}{Y}\)

= \(\frac{\frac{T}{\alpha}}{Y}=\frac{T}{\alpha Y}\)

For the same strain in the two wires, \(\frac{T_1}{0.1 \times 10^{-4} Y_1}=\frac{T_2}{0.2 \times 10^{-4} Y_2}\)

or, \(\frac{T_1}{T_2}=\frac{0.1 \times 10^{-4} \times 2 \times 10^{11}}{0.2 \times 10^{-4} \times 10 \times 10^{10}}\)=1

∴ \(T_1=T_2\)

Since, the system is in equilibrium, taking moments about the point C, we get,

T₁x = T₂(2-x) or, x = 2-x or, x= 1 m

So, the weight should be suspended from the midpoint of the horizontal bar.

Elasticity in Everyday Calculations

Example 10. A sphere of mass 25 kg and radius 0.1 m is hung from the ceiling of a room with the help of a steel wire. The height of the ceiling from the floor is 5.21 m. When the sphere is hung just like a pendulum, its lower surface touches the floor of the room. What will be the velocity of the sphere at the lowest point of its oscillation? The Young’s modulus for steel =2x 10¹¹ N • m²; the initial length of the wire = 5 m and the radius of the wire = 5 x 10⁴ m. 
Solution:

Given

A sphere of mass 25 kg and radius 0.1 m is hung from the ceiling of a room with the help of a steel wire. The height of the ceiling from the floor is 5.21 m. When the sphere is hung just like a pendulum, its lower surface touches the floor of the room.

According to the figure, the elongation of the wire at the lowest position of the sphere (diameter = 0.2m),

l = 5.21-(5+ 0.2) = 0.01 m

If the velocity of the sphere at the lowest point of its oscillation is v, then the tension in the wire,

⇒ \(T-m g=\frac{m v^2}{r}\)

T = \(m g+\frac{m v^2}{r}\) ….(1)

Here, m = mass of the sphere; r = distance of the center of gravity of the sphere from the point of suspension = 5.21 – 0.1 = 5.11m.

Suppose x = radius of the wire.

Then, \(Y=\frac{T}{\pi x^2} \frac{L}{l} \quad \text { or, } \quad T=\frac{Y \pi x^2 l}{L}\)

From equation (1), we get

⇒ \(mg +\frac{m v^2}{r}=\frac{Y \pi x^2 l}{L}\)

or, \(\frac{v^2}{r}=\frac{Y \pi x^2 l}{m L}-g\)

or, \(v^2=\frac{Y \pi x^2 l r}{m L}-r g\)

= \(\frac{\left(2 \times 10^{11}\right) \times 3.14 \times\left(5 \times 10^{-4}\right)^2 \times 0.01 \times 5.11}{25 \times 5}\)

= 14.1036

∴ ν = 3.76 m .s^-1

Example 11. Two blocks A and B are connected to each other by a string and a spring; the string passes over a frictionless pulley as shown. Block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of the blocks is 0. 2. The force constant of the spring Is 1960 N• m^-1. If the mass of the block A is 2 kg, calculate the mass of the block B and the energy stored In the spring.
Solution:

Given

Two blocks A and B are connected to each other by a string and a spring; the string passes over a frictionless pulley as shown. Block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C, both with the same uniform speed. The coefficient of friction between the surfaces of the blocks is 0. 2. The force constant of the spring Is 1960 N• m^-1. If the mass of the block A is 2 kg

Since the block A is descending with uniform velocity, no normal reaction acts on the block A due to the block C.

Considering the motion of the blocks A and B,

T = \(m_A g=\mu m_B g\)

∴ \(m_B=\frac{m_A}{\mu}=\frac{2}{0.2}=10 \mathrm{~kg}\)

If the extension of the spring is xm, then,

T = \(k x \text { or, } x=\frac{T}{k}=\frac{m_A g}{k}=\frac{2 \times 9.8}{1960}=\frac{1}{100} \mathrm{~m}\)

Energy stored in the spring

= \(\frac{1}{2} k x^2=\frac{1}{2} \times 1960 \times\left(\frac{1}{100}\right)^2=0.098 \mathrm{~J}\)

Example 12. On application of a pressure of 21 kg • cm^-2, the volume of 1 litre of an oil decreases by 840 mm³. Calculate the bulk modulus and compressibility of the oil.
Answer:

Given

On application of a pressure of 21 kg • cm^-2, the volume of 1 litre of an oil decreases by 840 mm³.

p = \(21 \mathrm{~kg} \cdot \mathrm{cm}^{-2}=\frac{21 \times 9.8}{(0.01)^2} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

= \(21 \times 98 \times 10^3 \mathrm{~N} \cdot \mathrm{m}^{-2}\)

V = 1 litre 10^-3 m³; ν = 840 mm³ = 840×10^-9 m³

Bulk modulus of the oil,

K = \(\frac{p V}{\nu}=\frac{\left(21 \times 98 \times 10^3\right) \times 10^{-3}}{840 \times 10^{-9}}\)

= \(2.45 \times 10^9 \mathrm{~N} \cdot \mathrm{m}^{-2}\)

= 2.45 x 109 N • m^-2

Compressibility of the oil, \(\frac{1}{K}=\frac{1}{2.45 \times 10^9}=4.1 \times 10^{-10} \mathrm{~m}^2 \cdot \mathrm{N}^{-1}\)

Example 13. A 0.5 kg block Alides from a point A on a horizontal track with an initial speed of 3 m • s^-1 towards a weightless horizontal spring of length 1 in and of force constant 2 N • m^-1. The part AB of the track is frictionless and the part BC has coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2 m and 2.14 m respectively, find the total distance covered by the block before it comes to rest (g = 10 m • s^-2).
Solution:

Given

A 0.5 kg block Alides from a point A on a horizontal track with an initial speed of 3 m • s^-1 towards a weightless horizontal spring of length 1 in and of force constant 2 N • m^-1. The part AB of the track is frictionless and the part BC has coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2 m and 2.14 m respectively

Initial kinetic energy of the block

= \(\frac{1}{2} m v^2=\frac{1}{2} \times 0.5 \times(3)^2=2.25 \mathrm{~J}\)

The part AB of the horizontal track is frictionless and hence during the passage ofthe block over this path no energy is lost.

Loss in energy in the part BD = work done against friction in that path

= μmgxBD = 0.2 x 0.5 x 10×2.14 = 2.14 J

So, the kinetic energy of the block when it reaches the point

D = 2.25-2.14 = 0.11 J

Let us assume that the block compresses the spring through an amount x

According to the principle of conservation of energy,

work done against friction + work done in compressing the spring =0.11

or, \(\mu m g x+\frac{1}{2} k x^2=0.11\)

or, \(0.2 \times 0.5 \times 10 \times x+\frac{1}{2} \times 2 x^2=0.11\)

or, \(x^2+x-0.11=0\)

or, (x + 1.1)(x- 0.1) = 0

x = 0.1 m [as x cannot be -1.1 m]

∴ The distance covered by the block before it comes to rest = 2 + 2.14 + 0.1 = 4.24 m.

Example 14. Two bodies A and B of masses m and 2m respectively are put on a smooth floor. They are connected by a spring. A third body C of mass m moves with a velocity v₀ along the line joining A and B and collides elastically with A as shown in Fig. At a certain instant of time t₀ after the collision, it is found that the instantaneous velocities of A and B are the same. Further, at this instant the compression of the spring is found to be x₀. Find out

1. the common velocity of A and B at time t₀ and
2. the force constant of the spring.

Solution:

Given

Two bodies A and B of masses m and 2m respectively are put on a smooth floor. They are connected by a spring. A third body C of mass m moves with a velocity v₀ along the line joining A and B and collides elastically with A as shown in Fig. At a certain instant of time t₀ after the collision, it is found that the instantaneous velocities of A and B are the same. Further, at this instant the compression of the spring is found to be x₀.

1. After elastic collision, C will come to rest and A will gain a velocity v₀ (as their masses are the same).

Let us assume that the common velocity of A and B is v, at time t₀ after the collision.

According to the principle of conservation of momentum,

⇒ \(m v_0=m \nu+2 m \nu \quad \text { or, } \quad v=\frac{v_0}{3}\)

2. According to the principle of conservation of energy,

⇒ \(\frac{1}{2} m v_0^2=\frac{1}{2} m v^2+\frac{1}{2} \times 2 m v^2+\frac{1}{2} k x_0^2\)

or, \(m v_0^2=3 m v^2+k x_0^2\)

[k = force constant of the spriong]

or, \(m v_0^2=3 m\left(\frac{v_0} {3}\right)^2+k x_0{ }^2 \quad \text { or, } k=\frac{2}{3} \frac{m v_0^2}{x_0^2}\)

Example 15. If the tension in a wire increases gradually to 6 kg, the elongation of the wire becomes 1.13 mm. Calculate the work done.
Solution:

Given

If the tension in a wire increases gradually to 6 kg, the elongation of the wire becomes 1.13 mm.

Work done, W = \(\frac{1}{2} F l=\frac{1}{2} \times 6 \times 9.8 \times \frac{1.13}{1000}=0.033 \mathrm{~J}\)

Elasticity in Everyday Problems And Solutions

Example 16. A body of mass 4 kg and density 2.5 g cm^-3, suspended by a metallic wire of length 1 m and diameter 2 mm, is kept completely immersed in water. What will be the increase in the length of the wire? Young’s modulus of the metal = 2x 10¹¹ N • m^-2 and g = 9.8 m • s^-2.
Solution:

Given

A body of mass 4 kg and density 2.5 g cm^-3, suspended by a metallic wire of length 1 m and diameter 2 mm, is kept completely immersed in water.

Volume of the body, V = \(\frac{4}{2.5 \times 1000}=16 \times 10^{-4} \mathrm{~m}^3\)

Apparent weight of the body when immersed in water,

W’ = mg- V x 1000 x g

= 4 x 9.8 – 16 x 10^-4 x 1000 x 9.8

= 39.2-15.68 = 23.52 N

If the elongation of the wire is l’, then,

Y = \(\frac{F}{A} \cdot \frac{L}{l^{\prime}} \quad\left[\text { Here, } F=W^{\prime}\right]\)

or, \(l^{\prime}=\frac{F}{A} \cdot \frac{L}{Y}\)

= \(\frac{23.52 \times 1}{3.14 \times\left(10^{-3}\right)^2 \times 2 \times 10^{11}}\)

= \(3.75 \times 10^{-5} \mathrm{~m}\)

Example 17. A 0.1kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross sectional area is 4.9 x 10^-7 m². If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad • s^-1. If the Young’s modulus of the material of the wire is n x 10⁹N • m^-2. Calculate the value of n. 
Solution:

Given

A 0.1kg mass is suspended from a wire of negligible mass. The length of the wire is 1m and its cross sectional area is 4.9 x 10^-7 m². If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency 140 rad • s^-1. If the Young’s modulus of the material of the wire is n x 10⁹N • m^-2

Young’s modules of the material of the wire,

Y = \(\frac{\frac{F}{A}}{\frac{\Delta L}{L}}=\frac{F L}{A \Delta L} \quad \text { or, } F=\left(\frac{Y A}{L}\right) \Delta L\)….(1)

If mass m is pulled by a length ΔL then restoring force developed in the wire,

F = kΔL ……(2)

Comparing equations (1) and (2) we get,

k = \(\frac{Y A}{L}\)

Angular frequency, \(\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{Y A}{m L}}\)

or, \(140=\sqrt{\frac{n \times 10^9 \times 4.9 \times 10^{-7}}{0.1 \times 1}}=70 \sqrt{n}\)

∴ \(\sqrt{n}=2\) or, n=4

Example 18. Stress-strain graph of an elastic material is shown. Using the graph And Young’s modulus of the material.

Solution:

Young’s modulus, \(Y=\frac{\text { stress }}{\text { strain }}\)

The strain corresponding to a point P on the graph is

OA = 0.003 and stress – OB = 200 x 10⁶ N • m^-2 .

∴ Young’s modulus of the material,

Y = \(\frac{200 \times 10^6}{0.003}=6.6 \times 10^{10} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Example 19. A 3 kg mass is hanging from one end of a vertical copper wire of length 2 m and diameter 0.5 mm. Due to this the elongation produced in the wire is 2.38 mm. Find Young’s modulus of copper.
Solution:

Given

A 3 kg mass is hanging from one end of a vertical copper wire of length 2 m and diameter 0.5 mm. Due to this the elongation produced in the wire is 2.38 mm

Young’s modulus, Y = \(Y=\frac{m g L}{\pi r^2 l} .\)

Here, m = 3 kg , g = 9.8 m • s^-2 , L = 2 m,

r = \(\frac{0.5}{2} \mathrm{~mm}=\frac{5 \times 10^{-4}}{2} \mathrm{~m}=2.5 \times 10^{-4} \mathrm{~m}\),

l = \(2.38 \mathrm{~mm}=238 \times 10^{-5} \mathrm{~m} \)

∴ Y = \(\frac{3 \times 9.8 \times 2}{3.14 \times\left(2.5 \times 10^{-4}\right)^2 \times 238 \times 10^{-5}}\)

= \(1.26 \times 10^{11} \mathrm{~N} \cdot \mathrm{m}^{-2}\)

Example 20. Six external forces, each of magnitude F, are applied on all the faces of a unit cube. Considering its elastic modulus, calculate the longitudinal strain and the volume strain on the unit cube.
Solution:

Given

Six external forces, each of magnitude F, are applied on all the faces of a unit cube. Considering its elastic modulus

Each side of the unit cube = 1.

∴ Volume of the cube = 1, surface area of each of the six
faces = 1

Let x, y, and z axes be chosen along three mutually perpendicular sides of the cube.

Now, at first, we choose the two opposite faces perpendicular to the x-axis. Due to the forces (F, F) acting on them, the side along the x -direction will have an elongation = l (suppose)

∴ Longitudinal strain along the x -direction

= \(\frac{\text { elongation }}{\text { initial length }}=\frac{l}{1}=l\)

Also, longitudinal stress = \(\frac{\text { applied force }}{\text { surface area }}=\frac{F}{1}=F\)

∴ Young’s modulus, Y = \(\frac{\text { stress }}{\text { strain }}=\frac{F}{l} ; \text { so, } l=\frac{F}{Y}\)

Simultaneously the side along the x-direction will also suffer lateral strains due to the forces acting along y and z directions. We know,

Poisson’s ratio, \(\sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}\)

Lateral strain = σ x longitudinal strain = σl = \(\frac{\sigma F}{Y}\)

Now, an elongation is always associated with a lateral contraction. Therefore, considering the strains along all the three axes, the effective longitudinal strain along the x -axis is

⇒ \(\frac{F}{Y}-\frac{\sigma F}{Y}-\frac{\sigma F}{Y}=\frac{F}{Y}(1-2 \sigma)\)

From symmetry, the effective longitudinal strain along each
of y- and z-axes = \(\frac{F}{Y}(1-2 \sigma)\)

As a result, the final length of each of the three sides of the cube = \(1+\frac{F}{Y}(1-2 \sigma)\)

∴ Final volume of the unit cube

= \(\left[1+\frac{F}{Y}(1-2 \sigma)\right]^3=1+\frac{3 F}{Y}(1-2 \sigma)\) [neglecting higher order terms]

∴ Volume expansion = \(1+\frac{3 F}{Y}(1-2 \sigma)-1\)

= \(\frac{3 F}{Y}(1-2 \sigma)\)

∴ Volume strain = \(\frac{\frac{3 F}{Y}(1-2 \sigma)}{1}=\frac{3 F}{Y}(1-2 \sigma)\)

Example 21. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, find the ratio of the elongation in the thin wire to that in the thick wire.
Solution:

Given

One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends,

Let the elongation produced in wire MN be l₁ and that in PQ be l₂

Since both the wires are made of copper, Young’s moduli will be same for both.

∴ From \(Y=\frac{F L}{A l}\) we have,

⇒ \(\frac{F}{\frac{\pi(2 R)^2}{\frac{l_2}{2 L}}}=\frac{\frac{F}{\pi R^2}}{\frac{l_1}{L}}\)

or, \(\frac{2 F L}{4 \pi R^2 l_2}=\frac{F L}{\pi R^2 l_1}\)

or, \(\frac{l_1}{l_2}=2\)

∴ \(l_1: l_2=2: 1\)