If the maximum percentage error in the measurement of distance (h) and time ( t ) are ' $e_1$ ' and ' $e_2$ ' respectively, the percentage error in the measurement of acceleration of gravity ( g ) is [use $\mathrm{h}=\frac{1}{2} \mathrm{gt}^2$ ]

The acceleration due to gravity is derived from $h = \frac{1}{2}gt^2$, giving $g = 2ht^{-2}$. For a quantity of form $X = kA^aB^b$, the maximum fractional error is $\frac{\Delta X}{X} = |a|\frac{\Delta A}{A} + |b|\frac{\Delta B}{B}$. Applying this to $g$: $\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t}$. Multiplying by 100 gives the percentage error: $\left(\frac{\Delta g}{g}\right) \times 100 = \left(\frac{\Delta h}{h}\right) \times 100 + 2\left(\frac{\Delta t}{t}\right) \times 100$. With percentage errors $e_1$ for height and $e_2$ for time: $\text{Percentage error in } g = e_1 + 2e_2$. Final Answer: $\boxed{\text{D}}$