Mechanical Properties of Solids

JEE Main level, numerical answer, medium

103 of 118

Question 103 of 118

A cube has side length 55 cm and modulus of rigidity 10510^5 N/m2^2. The displacement produced by a force of 1010 N in the upper face of cube is _____ mm.

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Answer: 2

Given:
Side length of the cube, L=5L = 5 cm =0.05= 0.05 m
Modulus of rigidity, η=105\eta = 10^5 N/m2^2
Force applied on the upper face, F=10F = 10 N

Area of the upper face, A=L2=(0.05)2=2.5×10−3A = L^2 = (0.05)^2 = 2.5 \times 10^{-3} m2^2

Shear stress is given by:
Stress=FA=102.5×10−3=4000\text{Stress} = \dfrac{F}{A} = \dfrac{10}{2.5 \times 10^{-3}} = 4000 N/m2^2

Shear strain is given by:
Strain=ΔxL\text{Strain} = \dfrac{\Delta x}{L}

Modulus of rigidity is defined as the ratio of shear stress to shear strain:
η=StressStrain=F/AΔx/L\eta = \dfrac{\text{Stress}}{\text{Strain}} = \dfrac{F/A}{\Delta x/L}

Rearranging for displacement Δx\Delta x:
Δx=F⋅LA⋅η\Delta x = \dfrac{F \cdot L}{A \cdot \eta}

Substituting the values:
Δx=10×0.052.5×10−3×105\Delta x = \dfrac{10 \times 0.05}{2.5 \times 10^{-3} \times 10^5}

Δx=0.5250=0.002\Delta x = \dfrac{0.5}{250} = 0.002 m

Converting to millimeters:
Δx=0.002×1000=2\Delta x = 0.002 \times 1000 = 2 mm

Answer: 22

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