Mechanical Properties of Solids

JEE Main level, single correct, medium

38 of 118

Question 38 of 118

The Young's modulus of a steel wire of length 6m6m and cross-sectional area 3mm23{mm}^{2}, is 2×1111N/m22\times {11}^{11}N/{m}^{2} . The wire is suspended from its support on a given planet. A block of mass 4kg4kg is attached to the free end of the wire. The acceleration due to gravity on the planet is 14\dfrac{1}{4} of its value on the earth. The elongation of wire is (Take gg on the earth =10m/s2=10m/{s}^{2} ):

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Answer: (C) 0.1mm0.1mm

The external force or tension (F)\left(F\right) acting on the wire is its own weight. It is given by
F=mg=4kg×104ms−2=10N\begin{matrix}F & = & mg \\ & = & 4kg\times \dfrac{10}{4}m{s}^{-2} \\ & = & 10N \\ & & \end{matrix}
The formula to calculate the Young's modulus of the material of the wire is given by
Y=FA∆LL=FL∆LA..........................(1)\begin{matrix}Y & = & \dfrac{\dfrac{F}{A}}{\dfrac{∆L}{L}} \\ & = & \dfrac{FL}{∆LA}..........................(1)\end{matrix}
where, AA is the cross-sectional area, LL is the original length and ∆L∆L is the elongation of the wire.
Substitute the given values of the known parameters into equation (1) and solve to calculate the required elongation of the wire.
2×1011N/m2=10N×6m∆L×3mm2×1m2106mm2∆L=10N×6m2×1011N/m2×3mm2×1m2106mm2=0.0001m×103mm1m=0.1mm\begin{matrix}2\times {10}^{11}N/{m}^{2} & = & \dfrac{10N\times 6m}{∆L\times 3{mm}^{2}\times \dfrac{1{m}^{2}}{{10}^{6}{mm}^{2}}} \\ ∆L & = & \dfrac{10N\times 6m}{2\times {10}^{11}N/{m}^{2}\times 3{mm}^{2}\times \dfrac{1{m}^{2}}{{10}^{6}{mm}^{2}}} \\ & = & 0.0001m\times \dfrac{{10}^{3}mm}{1m} \\ & = & 0.1mm\end{matrix}

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