In an experiment to find g with a…

In an experiment to find g with a simple pendulum, the length is measured as L=100.0±0.1 cm and the time for 20 oscillations as 40.0±0.2 s. Use…

One or more options correct

In an experiment to find

g

with a simple pendulum, the length is measured as

L=100.0\pm0.1\ \mathrm{cm}

and the time for

20

oscillations as

40.0\pm0.2\ \mathrm{s}

. Use

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

. Which of the following statements is/are correct?

A. The value of $g$ is $9.87\ \mathrm{m\,s^{-2}}$ .
B. The maximum fractional error in $g$ is $1.1\%$ .
C. The maximum fractional error in $g$ is $1.6\%$ .
D. Timing $40$ oscillations (about $80.0\ \mathrm{s}$ ) with the same absolute error $0.2\ \mathrm{s}$ would reduce the fractional error in $g$ to $0.35\%$ .

Show answer and solution

Answer: A, B

The period is

T=40.0/20=2.000\ \mathrm{s}

, so

g=4\pi^2(1.000)/4.000=9.87\ \mathrm{m\,s^{-2}}

; (A) is true. Since

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

, the maximum fractional error is

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

. Here

\Delta L/L=0.1/100=0.1\%

, and

\Delta T/T=0.2/40=0.5\%

because the error in the total time carries over to the period by the same fraction. Thus

\Delta g/g=0.1+2(0.5)=1.1\%

, so (B) is true and (C) is false. With

40

oscillations the timing fraction becomes

0.2/80=0.25\%

and the total is

0.1+2(0.25)=0.6\%

, not

0.35\%

(the length error stays), so (D) is false. Correct options: (A) and (B).

Original question written in the JEE Advanced 2019 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake