JEE Mains · Physics · STD 11- 8. mechanical properties of solids
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 \(\left.\mathrm{g}=3 \pi \mathrm{~m} / \mathrm{s}^2\right)\)
- A 5
- B 10
- C 25
- D 2.5
Answer & Solution
Correct Answer
(A) 5
Step-by-step Solution
Detailed explanation
\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…
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