Q: A student measures the distance traversed in free fall of a body, initially at rest, in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage error in measurement of the distance and the time are e_{1} and e_{2} respectively, the percentage error in the estimation of g is

(a) e_{2} – e_{1}

(b) e_{1} + 2 e_{2}

(c) e_{1} + e_{2}

(d) e_{1} – 2 e_{2}

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Ans: (b)

Sol: $\displaystyle s = u t + \frac{1}{2}g t^2 $

$\displaystyle s = 0 + \frac{1}{2}g t^2 $

$\displaystyle g = \frac{2 s}{t^2} $

$\displaystyle \frac{\Delta g}{g} \times 100 = \frac{\Delta s}{s} \times 100 + 2 \frac{\Delta t}{t} \times 100$

$\displaystyle \frac{\Delta g}{g} \times 100 = e_1 + 2 e_2 $