Three students $S_{1}, S_{2}$ and $S_{3}$ perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table. | Student No. | Length of Pendulum (cm) | No. of oscillations (n) | Total for n oscillations | Time period (s) | |-------------|------------------------|------------------------|--------------------------|----------------| | 1 | 64.0 | 8 | 128.0 | 16.0 | | 2 | 64.0 | 4 | 64.0 | 16.0 | | 3 | 20.0 | 4 | 36.0 | 9.0 | (Least count of length $=0.1 \mathrm{~cm}$ least count for time $=0.1 \mathrm{~s}$ ) If $E_{1}, E_{2}$ and $E_{3}$ are the percentage errors ' $g$ ' for students 1,2 and 3 respectively, then the minimum percentage error is obtained by student no. ______.

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Units & Measurement

Three students

S_{1}, S_{2}

and

S_{3}

perform an experiment for determining the acceleration due to gravity (g) using a simple pendulum. They use different lengths of pendulum and record time for different number of oscillations. The observations are as shown in the table.

Student No.	Length of Pendulum (cm)	No. of oscillations (n)	Total for n oscillations	Time period (s)
1	64.0	8	128.0	16.0
2	64.0	4	64.0	16.0
3	20.0	4	36.0	9.0

(Least count of length

= 0.1 cm

least count for time

= 0.1 s

)
If

E_{1}, E_{2}

and

E_{3}

are the percentage errors '

g

' for students 1,2 and 3 respectively, then the minimum percentage error is obtained by student no. ______.

Answer: 1

Solution

With

T = \frac{t}{n}

, where

t

is the single stopwatch reading for

n

oscillations,

g = \frac{4 π ^{2} L}{T ^{2}} = \frac{4 π ^{2} L n ^{2}}{t ^{2}} \Rightarrow \frac{Δ g}{g} = \frac{Δ L}{L} + 2 \frac{Δ t}{t}

The oscillations are counted exactly, so the whole timing error sits in the single reading

t

, with

Δ t = 0.1

s and

Δ L = 0.1

cm.

E_{1} = \frac{0.1}{64.0} + 2 \times \frac{0.1}{128.0} = 0.31%

E_{2} = \frac{0.1}{64.0} + 2 \times \frac{0.1}{64.0} = 0.47%

E_{3} = \frac{0.1}{20.0} + 2 \times \frac{0.1}{36.0} = 1.06%

The smallest percentage error is obtained by student number

1

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Integer answer

Practise more: all JEE Main PYQs on Units & Measurement →

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Units & Measurement

Solution

More JEE Main questions on Units & Measurement