Question

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

Answer 1

Answer: (A)

$360\pi N$

$Y=\frac{FL}{A\Delta l}$

$F=\frac{Y\Delta lA}{L}=\frac{YA\times 0.1L}{L}=0.1YA=0.1\times 9\times {10}^{10}\times \pi \times (\frac{4\times {10}^{-3}}{2}{)}^2=360\pi$