The diameter of a wire measured by a screw gauge of least count $0.001$ cm is $0.08$ cm. The length measured by a scale of least count $0.1$ cm is $150$ cm. When a weight of $100$ N is applied to the wire, the extension in length is $0.5$ cm, measured by a micrometer of least count $0.001$ cm. The error in the measured Young's modulus is $\alpha\times10^{9}\ \text{N/m}^2$. The value of $\alpha$ is ______.
(Ignore the contribution of the load to Young's modulus error calculation)
Answer: (B) $1.65$
$$Y=\frac{FL}{A\,\Delta l}=\frac{4FL}{\pi d^2\,\Delta l}$$
$Y=\dfrac{4\times100\times1.5}{\pi\times(8\times10^{-4})^2\times(5\times10^{-3})}=\dfrac{600}{1.005\times10^{-8}}\approx5.97\times10^{10}\ \text{N/m}^2$
Relative error (the load is taken as exact):
$$\frac{\Delta Y}{Y}=2\frac{\Delta d}{d}+\frac{\Delta L}{L}+\frac{\Delta(\Delta l)}{\Delta l}=2\times\frac{0.001}{0.08}+\frac{0.1}{150}+\frac{0.001}{0.5}=0.025+0.00067+0.002=0.02767$$
$\Delta Y=0.02767\times5.97\times10^{10}\approx1.65\times10^{9}\ \text{N/m}^2$, so $\alpha=1.65$.
Solution by Sreeraj P, M.Sc Physics