ELASTICITY AND VISCOSITY_JEE MAIN_080425_S2 - Competishun – Live JEE & NEET Courses and Exam Prep

JEE MAIN 2025

08-04-2025 S2

Question

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

Select the correct option:

2.5

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

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

Question Tags

JEE Main

Physics

Medium

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