8.Mechanical Properties of Solids
medium

The diameter of a brass rod is 4 mm and Young's modulus of brass is $9 \times {10^{10}}\,N/{m^2}$. The force required to stretch by $0.1\%$ of its length is

A

$360\,\pi N$

B

$36 \,N$

C

$144\pi \times {10^3}N$

D

$36\pi \times {10^5}N$

Solution

(a)$F = \frac{{YAl}}{L} = \frac{{9 \times {{10}^{10}} \times \pi \times 4 \times {{10}^{ – 6}} \times 0.1}}{{100}} = 360\;\pi \;N$

Standard 11
Physics

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