JEE Main 2018 · Physics · Units and Measurements · medium

To find g, a student times 20 oscillations of a simple pendulum. The length is measured as 1.00 m with a metre rule of least count 1 cm, and the…

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To find

g

, a student times

20

oscillations of a simple pendulum. The length is measured as

1.00\ \mathrm{m}

with a metre rule of least count

1\ \mathrm{cm}

, and the total time for

20

oscillations is

40.0\ \mathrm{s}

with a stopwatch of least count

0.1\ \mathrm{s}

. The maximum percentage error in the value of

g

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Answer: (D) $1.5\%$

From

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

we get

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

, so

\dfrac{\Delta g}{g}=\dfrac{\Delta L}{L}+2\dfrac{\Delta T}{T}

. The length error is

\dfrac{1\ \mathrm{cm}}{100\ \mathrm{cm}}=1\%

. The period is found from the total time

t

20

oscillations, so

\dfrac{\Delta T}{T}=\dfrac{\Delta t}{t}=\dfrac{0.1}{40.0}=0.25\%

(the count

20

is exact and does not change the fractional error). Hence

\dfrac{\Delta g}{g}=1\%+2(0.25\%)=1.5\%

. A common slip is to omit the square on

T

, which gives

1.25\%

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake