In a physics experiment to determine the acceleration due to gravity g using a free-fall a… — JAMB UTME

In a physics experiment to determine the acceleration due to gravity $g$ using a free-fall apparatus, the distance of fall is measured as $h = (2.00 \pm 0.04)\text{ m}$ and the duration of fall is measured as $t = (0.50 \pm 0.01)\text{ s}$ . Given that $g = \frac{2h}{t^2}$ , what is the percentage error in the calculated value of $g$ ?

Answer: 6 %

Answer

The percentage error in the calculated value of

g

is 6%.

For a physical quantity defined by

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

, the maximum percentage error is determined by adding the percentage error in

h

to twice the percentage error in

t

. The percentage error in

h

\frac{0.04}{2.00} \times 100\% = 2\%

and in

t

\frac{0.01}{0.50} \times 100\% = 2\%

. Therefore, the total percentage error in

g

2\% + 2(2\%) = 6\%

Step-by-Step Solution

Calculate the percentage error in the distance measurement

h

The fractional error in height is

\frac{\Delta h}{h} = \frac{0.04}{2.00} = 0.02

, which corresponds to

2\%

Calculate the percentage error in the time measurement

t

The fractional error in time is

\frac{\Delta t}{t} = \frac{0.01}{0.50} = 0.02

, which corresponds to

2\%

Apply the power-law error propagation rule to find the maximum percentage error in

g

For

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

, the total relative error is

\frac{\Delta g}{g} = \frac{\Delta h}{h} + 2\frac{\Delta t}{t} = 2\% + 2(2\%) = 6\%

, since the exponent of

t

is 2.

Key Concept

Error propagation in physical formulas involving powers and quotients

Estimated Time:2m 0s

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