JEE Main 2020 · Physics · Units and Measurements · medium

In an experiment on the simple pendulum, the length is measured as L = 100.0 cm with a metre scale of least count 0.1 cm, and the total time for 20…

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In an experiment on the simple pendulum, the length is measured as

L = 100.0\ \mathrm{cm}

with a metre scale of least count

0.1\ \mathrm{cm}

, and the total time for

20

oscillations is measured as

40.0\ \mathrm{s}

with a stopwatch of least count

0.2\ \mathrm{s}

. The maximum percentage error in the value of

g

obtained from

T = 2\pi\sqrt{L/g}

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Answer: (B) $1.1\%$

g = 4\pi^2 L/T^2

with

T = t/20

, so

\frac{\Delta g}{g} = \frac{\Delta L}{L} + 2\frac{\Delta t}{t}

(the factor 20 is exact).

\frac{\Delta L}{L} = \frac{0.1}{100.0} = 0.1\%

and

\frac{\Delta t}{t} = \frac{0.2}{40.0} = 0.5\%

. Hence

\frac{\Delta g}{g} = 0.1 + 2(0.5) = 1.1\%

. Common slip: forgetting that

T

enters squared gives

0.6\%

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