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 $\left.\pi=\frac{22}{7}\right)$
JEE MAIN 2023, Medium
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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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