Question

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 errors 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. $e_1+2e_2$
• C. $e_1+e_2$
• D. $e_1-2e_2$

Answer 1

The Correct Option is B

To solve this problem, we need to understand how the measurement errors propagate in the context of calculating gravitational acceleration \( g \) from the distance \( s \) traversed and the time \( t \) taken. The formula for calculating \( g \) using these measurements in a free-fall scenario is derived from the equation of motion:

s = \frac{1}{2}gt^2

Solving for \( g \), we get:

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

Next, we'll consider the percentage errors in the measurements of distance \( s \) and time \( t \). Let:

• e_1 be the percentage error in measuring the distance \( s \).
• e_2 be the percentage error in measuring the time \( t \).

The formula for percentage error in a derived quantity is given by adding the absolute value of the relative errors of the measurements, considering their respective powers:

\frac{\Delta g}{g} \times 100\% = \left|\frac{\Delta s}{s}\right| \times 100\% + 2\left|\frac{\Delta t}{t}\right| \times 100\%

Therefore, the percentage error in \( g \) is:

=\ e_1 + 2e_2

This implies that both the error in measuring the distance and twice the error in measuring the time contribute to the overall error in estimating \( g \).

Thus, the correct answer is e_1 + 2e_2.