JEE Main 2019 · Physics · Units and Measurements · easy

A student finds g from the time period of a simple pendulum, g = 4π^2 L/T^2. The length is measured as L = 1.00 m with an uncertainty of 0.01 m. The…

Single correct option

A student finds

g

from the time period of a simple pendulum,

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

. The length is measured as

L = 1.00\ \mathrm{m}

with an uncertainty of

0.01\ \mathrm{m}

. The time for

20

complete oscillations is measured as

40.0\ \mathrm{s}

with a stopwatch of least count

0.1\ \mathrm{s}

. The maximum percentage error in the computed value of

g

Show answer and solution

Answer: (B) $1.5\%$

Since

g\propto L\,T^{-2}

, the maximum fractional error is

\Delta g/g = \Delta L/L + 2\,\Delta T/T

. Here

\Delta L/L = 0.01/1.00 = 1\%

. The stopwatch error applies to the total time of

20

oscillations, and

T = t/20

, so

\Delta T/T = \Delta t/t = 0.1/40.0 = 0.25\%

. Hence

\Delta g/g = 1\% + 2(0.25\%) = 1.5\%

. The common slip is to divide the least count by the single period

T=2\ \mathrm{s}

instead of using the total time, which would give a far larger error.

Original question written in the JEE Main 2019 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake