The length of a cylinder is measured with a meter rod having least count 0.1text{ cm}. Its diameter is measured with vernier calipers having least co…

Volume of a cylinder is $V = \pi R^2 L$. The maximum fractional error is $\frac{\Delta V}{V} = \frac{\Delta L}{L} + 2\frac{\Delta R}{R}$. Here $\Delta L = 0.1\text{ cm}$, $L = 5.0\text{ cm}$, $\Delta R = 0.01\text{ cm}$, and $R = 2.0\text{ cm}$. So $\frac{\Delta V}{V} = \frac{0.1}{5.0} + 2\left(\frac{0.01}{2.0}\right) = 0.02 + 0.01 = 0.03$, meaning the percentage error is $3\%$? Wait, let's recalculate: $0.02 + 0.01 = 0.03$, which is $3\%$. Let's check option C: $3.0\%$. Let's ensure option C is marked correct.