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 $=0.1 \mathrm{~cm}$ Least count for time $=0.1 \mathrm{~s}$ | 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 $\mathrm{E}_{1}, \mathrm{E}_{11}$ and $\mathrm{E}_{111}$ are the percentage errors in $g$, i.e., $\left(\frac{\Delta g}{g} \times 100\right)$ for students I, II and III, respectively,

(B) $\mathrm{E}_{\mathrm{I}}$ is minimum

(A) $\mathrm{E}_{\mathrm{I}}=0$

(C) $\mathrm{E}_{1}=\mathrm{E}_{11}$

(D) $\mathrm{E}_{\mathrm{II}}$ is maximum

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Units & Measurement

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

E_{1}, E_{11}

and

E_{111}

are the percentage errors in

g

, i.e.,

(\frac{Δ g}{g} \times 100)

for students I, II and III, respectively,

(A) $E_{I} = 0$
(B) $E_{I}$ is minimum✓ Correct
(C) $E_{1} = E_{11}$
(D) $E_{II}$ is maximum

Answer: (B)

Solution

For a simple pendulum,

g = 4 π^{2} \frac{ℓ}{T ^{2}}

. The fractional error is

\frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + 2 \frac{Δ T}{T}

. Since

T = \frac{t}{n}

, we have

\frac{Δ T}{T} = \frac{Δ t}{t}

, so

\frac{Δ g}{g} = \frac{Δ ℓ}{ℓ} + 2 \frac{Δ t}{t}

. For student I:

t = 128.0

s; for student II:

t = 64.0

s; for student III:

t = 36.0

s. Since

Δ t = 0.1

s is constant, the fractional error decreases as

t

increases. Student I has the largest

t

, hence the minimum percentage error.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Single correct option

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JEE Advanced · 2008 · 13 Apr 2008 · Paper 1 · Units & Measurement

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