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Practice worksheet · Class 11 Physics

Elasticity

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Class 11 Physics · Mechanical Properties of Solids

Elasticity worksheet for Class 11 Physics

25 practice questions on Elasticity for Class 11 Physics (Mechanical Properties of Solids), from Priya Jain (M.Sc. Physics, B.Ed.). Open the worksheet to view, print or download it, or read the questions below.

21 MCQs 4 numericals Formula sheet Free PDF · 5 pages

Section A · Multiple Choice

Choose the correct option. Questions 1–21

  1. 1

    The possible value of Poisson’s ratio (σ\sigma) for any material can be:

    1. a Between 0 and 0.5
    2. b Between 0.5 and 1.5
    3. c Between −1 and −2
    4. d Between 0.5 and −1
  2. 2

    Shearing strain is equivalent to:

    1. a An extension strain
    2. b A compressive strain
    3. c The sum of mutually parallel expansion and contraction strains
    4. d The sum of mutually perpendicular expansion and contraction strains
  3. 3

    When a wire is twisted, the strain produced in it is:

    1. a Normal strain
    2. b Longitudinal strain
    3. c Lateral strain
    4. d Shearing strain
  4. 4

    If two identical balls, one made of ivory and the other made of clay, are dropped from the same height onto the floor, then:

    1. a The ivory ball will bounce to a higher height
    2. b The clay ball will bounce to a higher height
    3. c Both balls will bounce to the same height
    4. d The ivory ball will not bounce at all
  5. 5

    Which of the following is the most elastic material?

    1. a Glass
    2. b Quartz
    3. c Steel
    4. d Rubber
  6. 6

    The following four wires are made of the same material. On applying the same tension, which one will have the maximum increase in length?

    1. a Length 50 cm, diameter 0.5 mm
    2. b Length 100 cm, diameter 1 mm
    3. c Length 200 cm, diameter 2 mm
    4. d Length 300 cm, diameter 3 mm
  7. 7

    A wire of length LL and cross-sectional area AA is made of a material whose Young’s modulus is YY. If an extension xx is produced in its length, the work done is:

    1. a Yx22\dfrac{Yx^2}{2}
    2. b YAx22L\dfrac{YAx^2}{2L}
    3. c YAx2YAx^2
    4. d Yx2L2\dfrac{Yx^2}{L^2}
  8. 8

    Copper wires are more easily drawn out than iron wires because:

    1. a Copper is a good conductor of heat
    2. b The resistance of copper is less
    3. c The modulus of rigidity of copper is high
    4. d The Young’s modulus of copper is high
  9. 9

    In a metallic wire, if the stress is TT and the Young’s modulus is YY, then the energy stored per unit volume is:

    1. a TY22\dfrac{TY^2}{2}
    2. b TY2\dfrac{TY}{2}
    3. c TYTY
    4. d T22Y\dfrac{T^2}{2Y}
  10. 10

    Elasticity is a property of matter that is:

    1. a Molecular in nature
    2. b Electromagnetic in nature
    3. c Nuclear in nature
    4. d Gravitational in nature
  11. 11

    The relation between YY, η\eta and KK is:

    1. a 19Y=13η+1K\dfrac{1}{9Y} = \dfrac{1}{3\eta} + \dfrac{1}{K}
    2. b 1K=13Y+19η\dfrac{1}{K} = \dfrac{1}{3Y} + \dfrac{1}{9\eta}
    3. c 1η=13Y+19K\dfrac{1}{\eta} = \dfrac{1}{3Y} + \dfrac{1}{9K}
    4. d 1Y=13η+19K\dfrac{1}{Y} = \dfrac{1}{3\eta} + \dfrac{1}{9K}
  12. 12

    The relation between YY, η\eta and σ\sigma is:

    1. a 1η=2(1+2σ)\dfrac{1}{\eta} = 2(1 + 2\sigma)
    2. b Y=2η(1+σ)Y = 2\eta(1 + \sigma)
    3. c η=2Y(1+2σ)\eta = 2Y(1 + 2\sigma)
    4. d Y=3η(1−2σ)Y = 3\eta(1 - 2\sigma)
  13. 13

    If the radius of a wire and the force applied to it are both doubled, then the increase in its length will become:

    1. a Four times
    2. b Two times
    3. c Unchanged
    4. d Half
  14. 14

    The Young’s modulus of elasticity of a wire material is 2×10112 \times 10^{11} N/m². If the strain produced in it is 0.15, then the applied stress (in N/m²) will be:

    1. a 3×10103 \times 10^{10}
    2. b 3×10113 \times 10^{11}
    3. c 2×10112 \times 10^{11}
    4. d 2×10102 \times 10^{10}
  15. 15

    The dimensions of strain are the same as those of which of the following quantities?

    1. a Stress
    2. b Modulus of elasticity
    3. c Angle
    4. d Pressure
  16. 16

    When a weight is suspended from one end of a wire, the wire gets stretched. The potential energy per unit volume of the wire in terms of longitudinal strain σ\sigma and Young’s modulus YY will be:

    1. a Yσ22\dfrac{Y\sigma^2}{2}
    2. b Yσ2\dfrac{Y\sigma}{2}
    3. c 2Yσ22\dfrac{2Y\sigma^2}{2}
    4. d Y2σ2\dfrac{Y^2\sigma}{2}
  17. 17

    The upper end of a wire of radius 4 mm and length 100 cm is clamped, and the other end is twisted by 30∘30^\circ. The shearing angle will be:

    1. a 12∘12^\circ
    2. b 0.12∘0.12^\circ
    3. c 1.2∘1.2^\circ
    4. d 0.012∘0.012^\circ
  18. 18

    The Young’s modulus of a given substance is 2.4 times its modulus of rigidity. Its Poisson’s ratio will be:

    1. a 2.4
    2. b 1.2
    3. c 0.4
    4. d 0.2
  19. 19

    On applying a load to a wire, its length increases by 3 mm. The extension in another wire of the same material and same length, but whose radius is half, under the same load will be:

    1. a 12 mm
    2. b 0.75 mm
    3. c 6 mm
    4. d 15 mm
  20. 20

    How much force is required to stretch a steel wire of cross-sectional area 1 cm² to 1.1 times its length?

    1. a 2×1062 \times 10^{6} N
    2. b 2×1032 \times 10^{3} N
    3. c 2×10−62 \times 10^{-6} N
    4. d 2×10−72 \times 10^{-7} N
  21. 21

    A wire of length LL and cross-sectional area AA has a Young’s modulus of elasticity YY. If the length of the wire is doubled and the cross-sectional area is halved, then the Young’s modulus of elasticity of the wire will become:

    1. a 4Y4Y
    2. b YY
    3. c 2Y2Y
    4. d Y/2Y/2

Section B · Short Answer & Numericals

Show all steps. Questions 22–25

  1. 22

    The edge of a solid copper cube is 10 cm. When a hydraulic pressure of 7.20×1067.20 \times 10^{6} Pa is applied to it, its volume contracts by 0.06 cm³. Find the bulk modulus of copper.

  2. 23

    Two wires A and B of different metals, having the same radius and equal length, are joined together to make a long wire. By fixing one end of it and applying a force on the other end, the elongation in the length of this composite wire is 4 cm. If YB=3YAY_B = 3Y_A, find the change in length of wire B. (YY = Young’s modulus of elasticity)

  3. 24

    Find the work done per unit volume to increase the length of a wire by 1%, having Young’s modulus YY and a cross-sectional area of 1 mm². (Y=9×1011Y = 9 \times 10^{11} N/m²)

  4. 25

    What is the density of water at a depth where the pressure is 80.0 atm, given that its density at the surface is 1.03×1031.03 \times 10^{3} kg m−3^{-3}?

Formula Sheet · Elasticity

Quick revision

Hooke’s law & stress–strain curve

Hooke’s law (within elastic limit) Stress∝Strain\text{Stress} \propto \text{Strain}
Stress–strain curve showing the proportional limit, elastic limit, elastic and plastic regions and breaking strength, with smaller curves for a ductile and a brittle material

Young’s modulus (YY)

Definition Y=Longitudinal stressLongitudinal strain=F/AΔl/lY = \dfrac{\text{Longitudinal stress}}{\text{Longitudinal strain}} = \dfrac{F/A}{\Delta l/l}
Wire of length LL, radius rr, load MgMg Y=Mg/πr2Δl/LY = \dfrac{Mg/\pi r^2}{\Delta l/L}
Elongation due to the wire’s own weight Δl=mgL2AY\Delta l = \dfrac{mgL}{2AY}
Force constant of the wire k=YALk = \dfrac{YA}{L}
Actual length of the wire (LL found from these) L1=L+T1k,L2=L+T2kL_1 = L + \dfrac{T_1}{k}, \quad L_2 = L + \dfrac{T_2}{k}
Work done in stretching the wire W=12 k l2=12 YAL l2W = \tfrac{1}{2}\,k\,l^2 = \tfrac{1}{2}\,\dfrac{YA}{L}\,l^2
Stored energy in a wire U=12×Stress×Strain×VolumeU = \tfrac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}

Bulk modulus (KK)

Definition K=Bulk stressVolume strain=F/A−ΔV/V=−PΔV/VK = \dfrac{\text{Bulk stress}}{\text{Volume strain}} = \dfrac{F/A}{-\Delta V/V} = \dfrac{-P}{\Delta V/V}
Isothermal elasticity modulus Eθ=PE_\theta = P
Adiabatic elasticity modulus Eϕ=γPE_\phi = \gamma P
Density of compressed liquid ρ′=ρ[1+ΔPK]\rho' = \rho\left[1 + \dfrac{\Delta P}{K}\right]
Compressibility C=1KC = \dfrac{1}{K}

Modulus of rigidity (η\eta), Poisson’s ratio (σ\sigma) & relations

Modulus of rigidity η=Shearing stressShearing strain=Ftangential/Aϕ\eta = \dfrac{\text{Shearing stress}}{\text{Shearing strain}} = \dfrac{F_{\text{tangential}}/A}{\phi}
Poisson’s ratio σ=Lateral strainLongitudinal strain=−dr/rdL/L\sigma = \dfrac{\text{Lateral strain}}{\text{Longitudinal strain}} = \dfrac{-dr/r}{dL/L}
Change in volume dVV=(1−2σ) dLL\dfrac{dV}{V} = (1 - 2\sigma)\,\dfrac{dL}{L}
Theoretical value −1<σ<0.5-1 < \sigma < 0.5
Practical value 0<σ<0.50 < \sigma < 0.5
Relation between YY, KK and σ\sigma Y=3K(1−2σ)Y = 3K(1 - 2\sigma)
Relation between YY, η\eta and σ\sigma Y=2η(1+σ)Y = 2\eta(1 + \sigma)

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