WBCHSE Class 11 Physics Expansion Of Solid And Liquids Long Answer Questions

Expansion Of Solid And Liquids Long Answer Type Questions

Question 1. Is it possible to maintain a constant difference in the lengths of a brass rod and a steel rod, at all temperatures?
Answer:

It is possible. If the initial lengths of the brass rod and the steel rod are such that their expansion is the same for the same rise in temperature, a constant difference in length can be maintained.

Let at a certain temperature, l_b and l_s be the lengths of the brass rod and the steel rod, and a_b and a_s be their coefficients of linear expansion, respectively.

For a rise t in temperature, the expansion of brass rod = \(l_b \times \alpha_b \times t\)

and the expansion of steel rod = \(l_s \times alpha_s \times t\)

To preserve the difference in length we must have, \(l_b \times \alpha_b \times t=l_s \times \alpha_s \times t\)

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or, \(\frac{l_b}{l_s}=\frac{\alpha_s}{\alpha_b}\)

i. e., if the initial length is inversely proportional to the coef¬ficient of linear expansion, the difference in length of the two rods remains constant at all temperatures.

Example 2. A brass disc is stuck in a hole of a steel plate. Should you heat the system to free the disc from the hole? Given, α for brass = 19 x 10^-6 °C^-1 and α for steel I = 12 x 10^{-6 °}C^-1.
Answer:

To free the disc from the hole, the system should cooled.

Since \(\alpha_{\text {brass }}>\alpha_{\text {steel }}\), and initially the radii of the disc and the hole were the same, the disc expands more on heating and gets stuck more tightly.

On cooling, the disc contracts more than the hole, and thus becomes free.

Question 3. An isosceles triangle is made of three zinc rods. Will there be any change in its base angles with the rise in temperature? Give reasons for your answer.
Answer:

Let the initial length of each equal arm of the triangle be a and the length of the base be b. Hence, initial ratio of the lengths of the arms = a: a: b.

Let the rise in temperature be t and the coefficient of linear expansion of zinc be a .

Now, the ratio of the lengths of the arms

= a(1 + αt): a(1 + αt): b(1 + αt) = a: a: b.

Hence, the ratio of the lengths of the sides remains unchanged with the rise in temperature. Thus, the base angles too, do not change.

Example 4. A copper plate and an iron plate of the same volume are riveted together. Explain the likely observations, when the system is heated.
Answer:

On heating, the strip will bend, instead of remaining straight.

Since the coefficient of linear expansion of copper \(\left(\alpha_{\mathrm{Cu}}\right)\) is more than that of iron \(\left(\alpha_{\mathrm{Fe}}\right)\), the plates expand differently and thus the system bends.

The copper plate forms the outer surface due to a greater expansion, and the iron plate forms the inner surface of the arc. If the coefficients of linear expansion would have been the same for copper and iron, then the strip would remain straight.

Question 5. An iron rod is fitted along the diameter of a circular iron ring. Explain, whether the ring remains circular or not, when the system is heated.
Answer:

The ring will continue to be circular even when the system is heated.

Let the initial diameter of the ring be d and the coefficient of linear expansion of iron be α.

∴ Circumference of the ring =πd and length of the rod = d.

∴ \(\frac{\text { length of the circumference of the ring }}{\text { length of the rod }}=\frac{\pi d}{d}=\pi\)

When the system is heated to a temperature θ,

the new circumference = πd(1 + αθ) and the new length = d(1 + αθ)

∴ The new ratio = \(\frac{\pi d(1+\alpha \theta)}{d(1+\alpha \theta)}=\pi\)

Hence, the ring continues to be circular.

Question 6. Two rods of different materials are of the same length and cross-sectional area. They are joined lengthwise and rigidly fixed between two walls. The materials of the two rods have different thermal and mechanical properties. What should be the relation between the Young’s moduli and the coefficients of linear expansion of the two materials so that the position of the junction point of the rods does not alter even when heated?
Answer:

The rods are rigidly fixed. If the thermal stresses developed remains the same for all temperatures, the position of the junction point will not change.

Let Young’s moduli and coefficients of linear expansion for the two materials be Y₁, α₁ and Y₂, α₂ respectively and the temperature be increased by t.

Hence, the thermal stress in the 1st rod =Y₁α₁t and that in the 2nd rod = Y₂α₂t.

For no movement of the junction point, \(Y_1 \alpha_1 t=Y_2 \alpha_2 t\)

or, \(\frac{Y_1}{Y_2}=\frac{\alpha_2}{\alpha_1}\) is the required relation.

Question 7. If the other factors remain unaltered, does the change in the volume of a solid, with the change in temperature, depend on the presence of a hole in the solid? Answer with reason.
Answer:

The change in volume of a solid with the change in temperature, does not depend on the presence of a hole in the solid. The hole size will change byotheosame amount as the solid that would have exactly fitted in the hole.

Question 8. A solid and a hollow cylinder of the same size and made of the same material ore heated for the same change of temperature. Will the expanisons be the same? Explain your answer.
Answer:

The volume expansion, in both cases, will be the same.

• Let us imagine a solid cylinder made of the same material inserted inside tire hollow cylinder to fit exactly in the inner gap.
• Then the hollow cylinder is identical to the initially supplied solid cylinder. They will expand equally for the same rise in temperature.
• When their external dimensions are observed, we shall see that their volumes are the same. So, their expansions are equal.

Question 9. The difference in lengths of two rods, made up of different materials, remains constant at all temperatures. Show that their lengths at 0°C are inversely proportional to their coefficients of linear expansion.
Answer:

The lengths of the rods at 0°C are l₁ and l₂, and their respective coefficients of linear expansion are α₁ and α₂

Linear expansion of the first rod for t°C rise in temperature = l₁α₁t and that of the second rod = l₂α₂t

For the difference in lengths to be constant at all temperatures, \(l_1 \alpha_1 t=l_2 \alpha_2 t \text { or, } l_1 \alpha_1=l_2 \alpha_2 \quad \text { or, } \frac{l_1}{l_2}=\frac{\alpha_2}{\alpha_1} \text {. }\)

Question 10. While casting concrete roofs, the meshwork is made only using iron rods and no other metal. Why?
Answer:

Coefficients of thermal expansion of concrete and iron are almost the same. Thus on increase or decrease of temperature, iron and concrete expand or contract almost equally and concrete joints do not get separated from the rods. No thermal stress develops between them.

If rods of other metals were used, the two expansions or contractions would not match and cracks would develop.

Question 11. A platinum wire can easily be sealed to a glass bulb while a copper wire cannot Why?
Answer:

A copper wire cannot be sealed to glass as their coefficients of thermal expansion are different. After being sealed at a high temperature, and then being bolted, glass cracks or a gap develops between the wire and the glass because of unequal contraction.

Glass and platinum, having almost the same coefficients of expansion, contract or expand equally on cooling or heating, keeping the seal intact.

Question 12. Electric wires are hung between two consecutive poles with a slight sag, instead of stretching them out fully. Give reasons.
Answer:

This arrangement takes into account the expansion or contraction of the wire with the change in seasonal temperatures. Wires are fitted to two consecutive poles such that during winter, due to the fall in temperature, the contraction does not produce a thermal stress to bend the poles or to snap the wire.

Question 13. An aluminium rod of length l₁ and an iron rod of length I₂ are joined lengthwise to form a single rod of length \(\left(l_1+l_2\right)\). The coefficients of linear expansion of aluminium and iron are aa and as respectively. If both the rods expand equally for a rise in temperature t find the ratio between and \(\left(l_1+l_2\right)\).
Answer:

Expansion of aluminium rod for rise t in temperature = l₁α_at

and that of iron rod for the same rise in temperature =l₂α_st.

According to the problem, \(l_1 \alpha_a t=l_2 \alpha_s t\)

∴ \(\frac{l_1}{l_2}=\frac{\alpha_s}{\alpha_a}\)

∴ \(\frac{l_1}{l_1+l_2}=\frac{\alpha_s}{\alpha_a+\alpha_s}\)

Question 14. On being heated, the length of each side of a cube increases by 2%. What is the percentage increase in the volume of the cube?
Answer:

If l is the length of the cube of volume V, then \( V=l^3\)….(1)

∴ \(dV = 3 l^2 d l\)….(2)

or, \(\frac{d V}{V}=\frac{3 l^2 d l}{l^3}=\frac{3 d l}{l} . \text { Given, } \frac{d l}{l}=\frac{2}{100}\)

∴ \(\frac{d V}{V}=\frac{3}{l} \cdot \frac{2 l}{100}=\frac{6}{100}\)

∴ Increase in volume = \(\frac{6}{100} \times 100 \%=6 \% .\)

Example 15. Coefficient of linear expansion along one edge of a cube-shaped crystal is α₁, but that along a particular direction is α₂ find the coefficient of volume expansion of the crystal.
Answer:

Let l₀ = length of a side of the crystal at 0°C, and t = rise in temperature

∴ Final volume of tire crystal = \(I_0\left(1+a_1 t\right) \times I_0^2\left(1+a_2 t\right)^2\)

= \(l_0^3\left(1+a_1 t\right)\left(1+a_2 t\right)^2\)

= \(V_0\left(1+a_1 t\right)\left(1+a_2 t\right)^2 \quad\left[because l_0^3=V_0\right]\)

Now, V = V₀(l + γt), where γ is the coefficient of volume expansion of the crystal.

∴ \(1+\gamma t =\left(1+\alpha_1 t\right)\left(1+\alpha_2 t\right)^2=\left(1+\alpha_1 t\right)\left(1+2 \alpha_2 t\right)\)

= \(1+t\left(\alpha_1+2 \alpha_2\right)\) [neglecting higher powers]

or, \(\gamma=\alpha_1+2 \alpha_2.\)

Example 16. The length of each side of a skeleton cube (made up of rods) at 0°C is I₀. Taking any vertex of the cube as the origin, the coefficients of linear expansion of the rods along x, y, z-axes or parallel to them are α₁, α₂ and α₃ respectively. Show that the equivalent coefficient of volume expansion of the skeleton cube is α₁ + α₂ + α₃.
Answer:

Volume of the skeleton cube at 0°C = \(l_0^3\)

For a rise r in temperature, the final volume of the cube

= \(l_0\left(1+\alpha_1 t\right) \times l_0\left(1+\alpha_2 t\right) \times l_0\left(1+\alpha_3 t\right)\)

= \(l_0^3\left(1+\alpha_1 t\right)\left(1+\alpha_2 t\right)\left(1+\alpha_3 t\right)\)

= \(l_0^3\left(1+\alpha_1 t+\alpha_2 t+\alpha_3 t\right) \text { [neglecting higher powers] }\)

= \(l_0^3+l_0^3\left(\alpha_1+\alpha_2+\alpha_3\right) t\)

∴ Increase in volume = \(l_0^3+l_0^3\left(\alpha_1+\alpha_2+\alpha_3\right) t-l_0^3\)

= \(l_0^3\left(\alpha_1+\alpha_2+\alpha_3\right) t\)

Hence, equivalent coefficient of volume expansion of the skeleton cube

= \(\frac{\text { increase in volume }}{\text { initial volume } \times \text { rise in temperature }}\)

= \(\frac{l_0^3\left(\alpha_1+\alpha_2+\alpha_3\right) t}{l_0^3 \times t}=\alpha_1+\alpha_2+\alpha_3 .\)

Example 17. When the temperature of a body increases from t to t + Δt, the moment of inertia of it increases from I to I+ ΔI. If the coefficient of linear expansion of the body is α, then show that \(\frac{\Delta I}{I}=2 \alpha \Delta t\)
Answer:

We know, moment of inertia I = mk²

∴ ΔI = 2mkΔk

∴ \(\frac{\Delta I}{I}=\frac{2 m k \Delta k}{m k^2}=\frac{2 \Delta k}{k}=\frac{2 \alpha k \Delta t}{k} \quad[\Delta k=\alpha k \Delta t]\)

= 2 αΔt

Question 18. The apparent coefficient of expansion of a liquid when heated in a copper vessel is C and when heated in a silver vessel is S. If A is the linear coefficient of expansion of copper, then show that the linear coefficient of expansion of silver Is \(\frac{C+3 A-S}{3}\)
Answer:

The relation between the apparent coefficient of expansion γ_a and real coefficient of expansion γ_r is \(\gamma_r=\gamma_a+3 \alpha\)

or, \(\gamma_a=\gamma_r-3 \alpha. \) [α is the linear coefficient of expansion of the material of the vessel] In case of copper vessel]

In case of silver vessel, \(C=\gamma_r-3 A\)

[α’ is the linear coefficient of expansion of silver (say)]

∴ C+3A = \(S+3 \alpha^{\prime} \quad \text { or, } \alpha^{\prime}=\frac{C+3 A-S}{3}\)

Question 19. A liquid in a container is heated suddenly. At first the liquid level goes down a little and then the level begins to rise. Explain why.
Answer:

As heating starts, the container receives the heat first and expands in volume. No significant change occurs in the temperature of the liquid in the container. So, initially, the volume of the liquid remains approximately unchanged.

• For this, the liquid level in the container goes down a little. But on prolonged heating, the liquid gradually becomes hot and begins to increase in volume.
• Generally, the expansion of a liquid is greater than that of a solid. Hence, after some time, the liquid level rises above its initial level.

Question 20. What will be the effect of using water in place of mercury, in a thermometer for measuring the temperatures in the range of 0°C to 10°C?
Answer:

From 0°C to 4°C water contracts, but then expands above 4°C. So, for a water thermometer, the markings on the thermometer stem will be confusing. Volume of water, both at 0°C and 8°C, may be the same.

Then the same marking will correspond to two temperatures. Hence, water is not suitable as a thermometric liquid for the range of 0°C to 10°C.

Question 21. If mercury is kept in a glass vessel and in a copper! vessel separately, will the coefficient of apparent expansion in both cases be the same? If not, In which case will it be greater? Coefficients of volume expansion of glass and copper are 25 x 10^-6  °C^-1 and 50 x 10^-6 °C^-1 respectively.
Answer:

The coefficient of apparent expansion of mercury in a glass vessel will not have the same value as in a copper vessel.

The coefficient of apparent expansion = the coefficient of real expansion – the coefficient of volume expansion of the material of the container. The value of coefficient of real expansion of a liquid is fixed. So, coefficients of apparent expansion obtained in the two cases are

⇒ \(\gamma_{\text {apparent }}=\gamma-\gamma_{\text {glass }}\), when glass vessel is used,

and \(\gamma_{\text {apparent }}^{\prime}=\gamma-\gamma_{\text {copper }}\), when copper vessel is used.

Here, \(\gamma_{\text {copper }}>\gamma_{\text {glass }}, \gamma_{\text {apparent }}>\gamma_{\text {apparent }}^{\prime}\).

Therefore, the coefficient of apparent expansion of mercury in a glass vessel will be higher than that in a copper vessel.

Example 22. A wooden block floats in water keeping a volume V of it above the surface, at 0°C. If the temperature of water is slowly raised from 0°C to 20°C, what will be the change in the value of V?
Answer:

The upthrust of a liquid buoyant force on a floating body depends on the density of the liquid. The part of a floating object above the liquid surface [V, in this case) increases with the increase in density of the liquid.

The density of water at 20°C is less than that at 0°C. But, at 4° C, it is the highest. So, from 0°C to 4°C, V will increase, but after that V will decrease gradually. Eventually, V₂₀ < V₀

Question 23. What will be the difference observed when one heats mercury and water from 0°C?
Answer:

Mercury expands continuously from 0°C when heat is supplied to it. But when water is heated from 0°C, it contracts up to 4°C, attains a minimum volume, and then expands like other liquids from 4°C.

Question 24. Ice is formed on the top surface of a lake. Temperature of air above ice is -15°C.

1. What is the tern- perature of the water layer just below ice?
2. Find the maximum possible temperature at the bottom of the lake.

Answer:

1. The temperature of the water layer just below the fro¬zen surface is 0°C. A thick ice layer acts as a good insulator that prevents heat from escaping to the outer atmosphere. So the water below does not freeze easily.
2. Maximum temperature can be 4°C at the bottom of the lake. As the density of water is maximum at 4°C, water of this temperature occupies the bottom.

Question 25. A beaker is filled up to its brim with a liquid of denity 1.5 g cm^-3. A piece of ice is floating on the liquid. Does the liquid overflow due to melting of ice?
Answer:

The liquid in the beaker overflows due to melting of ice.

Let the mass of ice =mg.

Hence, while floating, the ice displaces = \(\frac{m}{1.5} \mathrm{~cm}^3=V_1 \text { (say) }\) of the liquid.

Volume of water produced on melting of ice = m cm³ = V₂ (say)

As V₂ > V₁ i.e., volume of water produced is greater than the space occupied by floating ice, the excess water rises above the liquid level and overflows.

Question 26. A container is filled up to its brim with water with a piece of ice floating on it. State the effect on the level of water in the container when ice melts completely, if the initial temperature of water was

1. 0°C
2. 4°C and
3. above 4°C.

Answer:

Volume expansion of the container at the tempera¬tures mentioned is neglected.

1. There will be no change in the water level. Here mass of ice is equal to the mass of the displaced water as per the condition of floatation.

Ice, being at 0°C, produces water at 0°C on melting. Volume of that water is equal to the space formerly occupied by the Immersed portion of ice. Hence, water thus formed on melting of ice fills the space.

2. Some water will overflow from this container. Density of water is maximum at 4°C so water formed on melting of ice at 0°C is of lower density. Therefore the volume of this water will be greater than the volume of water displaced by floating ice at 4°C.

3. The water level will come down. Mass of the ice floating is equal to the mass of water displaced by the ice. According to the question, temperature of the water is above 4°C. Receiving heat from the surroundings and water, the ice produces water at 0°C on melting.

Density of water at 0°C is higher than that of water at 4°C. Hence, volume of the water displaced is more than the volume of the water produced on melting. Also, the temperature of the water falls as it supplies heat to the ice. With decrease in temperature, volume of the water also decreases.

Question 27. A body, immersed in a liquid, is suspended from the left arm of a hydrostatic balance, and is balanced by putting measuring weights on the scale pan at the right arm. If the liquid with the body in it is heated, will the equilibrium be disturbed?
Answer:

The left arm of the balance will go down.

• With rise in temperature, density of the body as well as that of the liquid decreases. But the decrease in density of the liquid is greater than that of the solid body, because the volume expansion coefficient of a liquid is much higher.
• Hence, the value of the upthrust decreases, i.e., the apparent weight of the body increases. Now more counterpoising weights should be put on the right pan to balance the beam.

Question 28. State the factors on which the coefficients of expansion of a liquid depend.
Answer:

The coefficient of real expansion of a liquid depends on the nature of the liquid only.

But, the coefficient of apparent expansion depends on the nature of the liquid, and also on the coefficient of volume expansion of the material of its container.

Question 29. Is the apparent expansion coefficient of a liquid a constant?
Answer:

The apparent expansion coefficient of a liquid is not a constant it depends on the material of the container in which the liquid is kept.

Question 30. Which one of the following graphs in Fig. correctly represents the change of density of water with temperature?

Answer:

Graph (1) represents the correct nature of variation of density of water with temperature. The convex nature is due to the anomalous expansion of water. The maximum point corresponds to the maximum density of water at 4°C.

Question 31. If the coefficient of volume expansion of glass had been equal to the coefficient of real expansion of mercury, a mercury thermometer would not be active.
Answer:

A thermometer shows the apparent expansion of the thermometric liquid.

But, the coefficient of apparent expansion of mercury = the coefficient of real expansion of mercury – the coefficient of volume expansion of glass = 0, if γ_glass = γ_mercury (given).

Therefore, the mercury within the tube Would remain at the same level, no matter what the change in temperature is.

Expansion Of Solid And Liquids Conclusion

Thermal expansion of solid are of three kinds

1. Linear Expansion,
2. Surface Expansion And
3. Volume Expansion.

The increase in length for unit rise in temperature for a unit length of a solid is called the coefficient of linear expansion (α) of that solid.

The value of α does not depend on the unit of length but depends on the unit of temperature.

The unit of α is °C^-1 or °F^-1.

• The increase in surface area for unit rise in temperature for a unit surface area of a solid is called the coefficient of surface expansion (β) of that solid.
• The value of ft does not depend on the unit of area, but depends on the unit of temperature.

The unit of β is °C^-1 or °F^-1.

• The increase in volume for unit rise in temperature for a unit volume of a solid is called the coefficient of volume expansion (γ) of that solid.
• The value of γ does not depend on the unit of volume, but depends on the unit of temperature.

The unit of γ is °C^-1 or °F^-1.

• With the increase in temperature, the volume of a solid increases and hence its density decreases, and vice-versa.
• With the increase in temperature of a rod, its length increases and with the decrease in temperature its length decreases.
• But, if the two ends of the rod are rigidly fixed, then expansion or contraction of the rod faces some obstruction when its temperature changes.
• As a result, a force develops inside the rod. This force, per unit area of the rod, is called thermal stress.
• Thermal stress is independent of the length or area of cross-section of a rod or a wire.

Liquids undergo expansion in volume due to rise in tent perature. This expansion is of two kinds

1. Real expansion and
2. Apparent expansion.

Apparent expansion: The expansion of a Uqnid me a sured by ignoring the expansion of the container is called apparent expansion of the liquid.

Real expansion: The sum of the apparent expansion of the liquid and the expansion of the container is called real expansion of the liquid.

Coefficient of apparent expansion: The apparent expansion of unit volume of a liquid for a temperature rise of 1° is called the coefficient of apparent expansion of the liquid.

The coefficient of apparent expansion of a liquid depends on the coefficient of expansion of the material of the container. This is not a characteristic property of the liquid.

Coefficient of real expansion: The real expansion of unit volume of a liquid for a temperature rise of 1° is called the die coefficient of real expansion of the liquid.

• This coefficient is a characteristic property of the liquid.
• With the increase in temperature, the apparent weight of a solid body immersed in a liquid will increase.
• With the increase in temperature of a liquid, its density decreases.

Exception: Density of water increases with the increase in temperature from 0°C to 4°C. Beyond this temperature density of water decreases with the rise of temperature like other liquids.

This exceptional or peculiar expansion of water from 0°C to 4°C is called anomalous expansion of water.

Volume of some amount of water at 4 ° C is the minimum (volume of 1g of water is 1 cm³) and the density is the maximum (1 g ⋅ cm^-3).