Errors in Measurement question from JEE Main 2021 (1 Sep, Shift 2)

Answer: (b)

1. $Y = \frac{MgL^3}{4bd^3\delta} \Rightarrow \frac{\Delta Y}{Y} = \frac{\Delta M}{M} + 3\frac{\Delta L}{L} + \frac{\Delta b}{b} + 3\frac{\Delta d}{d} + \frac{\Delta \delta}{\delta}$
2. $\frac{\Delta M}{M} = \frac{1}{2000} = 0.0005$, $3\frac{\Delta L}{L} = 3\left(\frac{1}{1000}\right) = 0.003$
3. $\frac{\Delta b}{b} = \frac{0.1}{40} = 0.0025$, $3\frac{\Delta d}{d} = 3\left(\frac{0.01}{4}\right) = 0.0075$, $\frac{\Delta \delta}{\delta} = \frac{0.01}{5} = 0.002$
4. Sum $= 0.0155$
5. So the answer is (b) 0.0155.