Q 11-08-061JEE MainJEE Main 2023 (6 Apr, Shift 2)Easy
A metal block of mass $m$ is suspended from a rigid support through a metal wire of diameter $14\ \text{mm}$. The tensile stress developed in the wire under equilibrium state is $7\times10^5\ \text{N m}^{-2}$. The value of mass $m$ is ______ kg. (Take $g=9.8\ \text{m s}^{-2}$ and $\pi=\frac{22}{7}$)
Numerical value type. Enter your answer.
Answer: 11
$A=\dfrac{22}{7}\times(7\times10^{-3})^2=1.54\times10^{-4}\ \text{m}^2$.
$m=\dfrac{\sigma A}{g}=\dfrac{7\times10^5\times1.54\times10^{-4}}{9.8}=11\ \text{kg}$.
Solution by Sreeraj P, M.Sc Physics