IIT-JEE 2008 · Paper 1 · Physics Q24

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Students I, II and III perform an experiment for measuring the acceleration due to gravity ( $g$ ) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table.

Least count for length = 0.1 cm. Least count for time = 0.1 s.

Student	Length of the pendulum (cm)	Number of oscillations ( $n$ )	Total time for ( $n$ ) oscillations (s)	Time period (s)
I	64.0	8	128.0	16.0
II	64.0	4	64.0	16.0
III	20.0	4	36.0	9.0

If $E_I$ , $E_{II}$ and $E_{III}$ are the percentage errors in $g$ , i.e., $\left(\dfrac{\Delta g}{g}\times100\right)$ for students I, II and III, respectively,

E_I=0

E_I

is minimum

E_I=E_{II}

E_{II}

is maximum

Answer and worked solution

Official answer: B

Error in g. $g=\dfrac{4\pi^2L}{T^2}$ , so
$\frac{\Delta g}g=\frac{\Delta L}L+2\frac{\Delta T}T,\qquad\frac{\Delta T}T=\frac{\Delta t}{t}\ (t=\text{total time}).$

Student	$\frac{\Delta L}L$	$2\frac{\Delta t}t$	total
I	$\frac{0.1}{64}=0.0016$	$\frac{0.2}{128}=0.0016$	0.0031
II	0.0016	$\frac{0.2}{64}=0.0031$	0.0047
III	$\frac{0.1}{20}=0.005$	$\frac{0.2}{36}=0.0056$	0.0106

$E_I$ is the smallest (and not zero); $E_{III}$ , not $E_{II}$ , is the largest.

(B)

Topics:Simple pendulum experiment Errors in measurement

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IIT-JEE 2008 · Paper 1, Physics question 24