Mechanical Properties of Solids

JEE Main level, single correct, medium

117 of 118

Question 117 of 118

A 3 m long wire of radius 3 mm shows an extension of 0.1 mm when loaded vertically by a mass of 50 kg in an experiment to determine Young's modulus. The value of Young's modulus of the wire as per this experiment is P×1011Nm−2\mathrm{P} \times 10^{11} \mathrm{Nm}^{-2}, where the value of P is: (Take g=3π m/s2)\left.\mathrm{g}=3 \pi \mathrm{~m} / \mathrm{s}^2\right)

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Answer: (A) 5

50 gπr2=y⋅Δℓℓ50×3ππ×(3×10−3)2=P×1011×0.1×10−33⇒P=50×3×332×10−6×1011×0.1×10−3P=5\begin{aligned} & \frac{50 \mathrm{~g}}{\pi \mathrm{r}^2}=\mathrm{y} \cdot \frac{\Delta \ell}{\ell} \\ & \frac{50 \times 3 \pi}{\pi \times\left(3 \times 10^{-3}\right)^2}=\mathrm{P} \times 10^{11} \times \frac{0.1 \times 10^{-3}}{3} \\ & \Rightarrow \mathrm{P}=\frac{50 \times 3 \times 3}{3^2 \times 10^{-6} \times 10^{11} \times 0.1 \times 10^{-3}} \\ & \mathrm{P}=5\end{aligned}

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