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JEE Mains · Physics · STD 11- 8. mechanical properties of solids

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 \(\left.\pi=\frac{22}{7}\right)\)

  1. A \(10\)
  2. B \(9\)
  3. C \(11\)
  4. D \(7\)
Verified Solution

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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