MHTCET 13 Apr 2026 Shift 2 Paper

Concept:The percentage error in a calculated quantity is found by summing the relative errors of each measured factor, multiplied by their powers.Explanation:Volume of a cylinder is given by:$$V = \pi r^{2} l$$Therefore, the relative error in volume is:$$\frac{\Delta V}{V} = \frac{2\Delta r}{r} + \frac{\Delta l}{l}$$Here, length is measured with a meter rod of least count $0.1\ \text{cm}$, so:$$\Delta l = 0.1\ \text{cm}, \quad l = 8.0\ \text{cm}$$$$\frac{\Delta l}{l} = \frac{0.1}{8.0}$$Diameter is measured with vernier calipers of least count $0.01\ \text{cm}$.Given radius $r = 4.0\ \text{cm}$, so diameter $d = 8.0\ \text{cm}$.Since $r = \frac{d}{2}$, the relative error in radius equals the relative error in diameter:$$\frac{\Delta r}{r} = \frac{\Delta d}{d} = \frac{0.01}{8.0}$$Now substitute values:$$\frac{\Delta V}{V} = 2\left(\frac{0.01}{8.0}\right) + \frac{0.1}{8.0}$$$$\frac{\Delta V}{V} = \frac{0.02}{8.0} + \frac{0.1}{8.0} = \frac{0.12}{8.0} = 0.015$$Percentage error in volume:$$0.015 \times 100 = 1.5\%$$Answer:The percentage error in the calculated volume is $1.5\%$, which corresponds to Option C.