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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