JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2023
- Medium
- Numerical
A metal block of mass \(m\) is suspended from a rigid support through a metal wire of diameter \(14\,mm\). The tensile stress developed in the wire under equilibrium state is \(7 \times 10^5\,Nm ^{-2}\). The value of mass \(m\) is \(......kg\). (Take, \(g =9.8\,ms ^{-2}\) and \(\left.\pi=\frac{22}{7}\right)\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(11\)
Step-by-step Solution
Detailed explanation
Tensile stress, \(\sigma=\frac{F}{A}=\frac{4 m g}{\pi D^2}\)
\(\therefore m=\frac{\pi D^2 \sigma}{4 g}\)
\(=\frac{22}{7} \times \frac{\left(14 \times 10^{-3}\right)^2 \times 7 \times 10^5}{4 \times 9.8}\)
\(=11\,kg\)
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