WBCHSE Class 11 Physics On Elasticity Short Questions And Answers

WBCHSE Class 11 Physics On 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:

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:

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:

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 increases by 0.04%. The pressure applied to 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.

Question 5. The stress along the length of a rod (with a rectangular cross section) is 1% of the Young’s modulus of its material. What is the approximate percentage of change in 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, the 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, \(\sigma=\frac{\text { lateral strain }}{\text { longitudinal strain }}=\frac{\frac{\Delta y}{y}}{\frac{\Delta x}{x}}=\frac{\frac{\Delta z}{z}}{\frac{\Delta x}{x}}\)]

The negative symbol in equation (1) implies that, as length increases due to stress, the 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 the 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 it. 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.

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 floats on the surface of the liquid, covering the entire cross-section of the 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. The copper of fixed volume V is drawn into a 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 the 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 the 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 the case of the 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 are those materials for which stress-strain variation is not a straight line within the 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:

A bridge undergoes alternating stress and strain a large number of times during its use. A bridge loses its elastic strength when it is used for a 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 (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:

A liquid, compressible or incompressible, does not have any defined 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 wristwatch?
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 drawn to the same scale, which graph represents a 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. Two wires A and B of length l, radius r, and length 21, radius 2 r having the same Young’s modulus Y are hung with a weight of mg as shown in the figure. What is the net elongation in the two wires?
Answer:

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 are 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 behavior of a substance?
Answer: Restoring force is responsible for the elastic behavior of a substance.