JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Units & Measurement

In an experiment to find acceleration due to gravity

(g)

using simple pendulum, time period of 0.5 s is measured from time of 100 oscillations with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy, the accuracy in the determination of

g

is found to be

x %

. The value of

x

is ____

Answer: 5

Solution

From

T = 2 π \frac{ℓ}{g}

, we get

g = \frac{4 π ^{2} ℓ}{T ^{2}}

. The fractional error is

\frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + 2 \frac{Δ T}{T}

. For 100 oscillations with 1 s resolution,

Δ T = \frac{1 s}{100 \times 0.5 s} = 0.02

. For length,

\frac{Δ ℓ}{ℓ} = \frac{1 mm}{10 cm} = \frac{0.1}{10} = 0.01

. Therefore

\frac{Δ g}{g} = 0.01 + 2 (0.02) = 0.05 = 5%

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Integer answer

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