Errors in Measurement question from MHT CET 2026 (15 Apr, Shift 1)

Answer: (b)

1. $V = \pi r^2 l = \frac{\pi d^2 l}{4}$, so $\frac{\Delta V}{V} = 2\frac{\Delta d}{d} + \frac{\Delta l}{l}$
2. Diameter $d = 4.0$ cm, so $\frac{\Delta d}{d} = \frac{0.01}{4.0} = 0.25\%$; $\frac{\Delta l}{l} = \frac{0.1}{5.0} = 2\%$
3. $\frac{\Delta V}{V} = 2(0.25\%) + 2\% = 2.5\%$
4. So the answer is (b) $2.5\%$.