JEE Main 2017 · Physics · Units and Measurements · easy

In an experiment to find g with a simple pendulum, the length is measured as L = 1.00 m with an uncertainty of 1 mm. The time for 20 oscillations is…

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In an experiment to find

g

with a simple pendulum, the length is measured as

L = 1.00\ \mathrm{m}

with an uncertainty of

1\ \mathrm{mm}

. The time for 20 oscillations is found to be

40.0\ \mathrm{s}

with an uncertainty of

0.2\ \mathrm{s}

in this total time. The percentage error in the value of

g

obtained from

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

is nearly

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Answer: (B) $1.1\%$

The period is

T = 40.0/20 = 2.00\ \mathrm{s}

. An error of

0.2\ \mathrm{s}

in the total time of 20 oscillations gives

\Delta T = 0.2/20 = 0.01\ \mathrm{s}

, so

\Delta T/T = 0.005

. Since

g \propto L T^{-2}

\dfrac{\Delta g}{g} = \dfrac{\Delta L}{L} + 2\dfrac{\Delta T}{T} = 0.001 + 0.010 = 0.011

, i.e.

1.1\%

. The common slip is to use

0.2\ \mathrm{s}

as the error in one period, which gives

2.1\%

, or to forget the factor

2

on the time term, which gives

0.6\%

Original question written in the JEE Main 2017 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake