Q59 / 75 · JEE Main 2025 (8 Apr Shift 2) · Physics

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 $\mathrm{P} \times 10^{11} \mathrm{Nm}^{-2}$, where the value of P is: (Take $\left.\mathrm{g}=3 \pi \mathrm{~m} / \mathrm{s}^2\right)$

  • Correct answer
    A
    5
  • B
    10
  • C
    25
  • D
    2.5

Solution

$\begin{aligned} & \frac{50 \mathrm{~g}}{\pi \mathrm{r}^2}=\mathrm{y} \cdot \frac{\Delta \ell}{\ell} \\ & \frac{50 \times 3 \pi}{\pi \times\left(3 \times 10^{-3}\right)^2}=\mathrm{P} \times 10^{11} \times \frac{0.1 \times 10^{-3}}{3} \\ & \Rightarrow \mathrm{P}=\frac{50 \times 3 \times 3}{3^2 \times 10^{-6} \times 10^{11} \times 0.1 \times 10^{-3}} \\ & \mathrm{P}=5\end{aligned}$

Hence, the correct answer is option: 'A'.

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