MHTCET 11 Apr 2026 Shift 1 Paper

Concept:Percentage error in $g$ is found by combining the relative errors of length and period using error propagation.Explanation:The formula for the period is $T=2 \pi \sqrt{\frac{l}{g}}$.Rearranging for $g$ gives $g=\frac{4 \pi^2 l}{T^2}$.Since $4 \pi^2$ is a constant, it does not contribute to the error.So the percentage error in $g$ is $\frac{\Delta g}{g} \times 100 = \frac{\Delta l}{l} \times 100 + 2 \frac{\Delta T}{T} \times 100$.Given $l = 80 \, \text{cm}$ and $\Delta l = 0.1 \, \text{cm}$:$\frac{\Delta l}{l} \times 100 = \frac{0.1}{80} \times 100 = 0.125\%$.Time for 50 oscillations is $t = 50 \times 1.5 = 75 \, \text{s}$.Least count of the stopwatch is $\Delta t = 0.1 \, \text{s}$.Since $T = \frac{t}{50}$, the fractional error in $T$ equals the fractional error in $t$:$\frac{\Delta T}{T} = \frac{\Delta t}{t} = \frac{0.1}{75}$.Therefore, $2 \frac{\Delta T}{T} \times 100 = 2 \times \frac{0.1}{75} \times 100 = 0.267\%$.Total percentage error: $0.125\% + 0.267\% = 0.392\% \approx 0.4\%$.Answer:The percentage error in $g$ is nearly $0.4\%$, which corresponds to Option B.