An experiment is performed to… - JEE Main 2014 Physics PYQ

JEE Main 2014 — Physics Mechanics

2014mcqeasy

An experiment is performed to obtain the value of acceleration due to gravity g by using a simple pendulum of length L. In this experiment time for 100 oscillations is measured by using a watch of 1 second least count and the value is 90.0 seconds. The length L is measured by using a meter scale of least count 1 mm and the value is 20.0 cm. The error in the determination of g would be :

Official previous-year question

Held on 9 Apr 2014 · Verified 6 Jul 2026.

Options

A
4.4%
B
2.27%
C
1.7%
D
2.7%

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Solution

$\text{T} = 2 \pi \sqrt{ \frac{ \text{L} }{ \text{g} } }$

$\therefore {\text{T}}^{2} = 4 {\pi }^{2} \frac{ \text{L} }{ \text{g} }$

$\therefore \text{g} = 4 {\pi }^{2} \frac{ \text{L} }{ {\text{T}}^{2} }$

% Error in g $= ( \frac{ \Delta \text{L} }{ {\text{L}}_{0} } \times 1 0 0 ) + 2 ( \frac{ \Delta \text{T} }{ {\text{T}}_{0} } \times 1 0 0 )$

$= ( \frac{ 1 \times 1 {0}^{ - 3 } }{ \text{0.20} } \times 1 0 0 ) + 2 { \frac{ 1 }{ 9 0 } \times 1 0 0 }$

$= \text{0.5} + 2 ( \frac{ 1 0 }{ 9 } )$

$= \text{0.5} + \frac{ 2 0 }{ 9 }$

$= \text{0.5} + \text{2.22}$

= 2.7%

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Solution

$\text{T} = 2 \pi \sqrt{ \frac{ \text{L} }{ \text{g} } }$

$\therefore {\text{T}}^{2} = 4 {\pi }^{2} \frac{ \text{L} }{ \text{g} }$

$\therefore \text{g} = 4 {\pi }^{2} \frac{ \text{L} }{ {\text{T}}^{2} }$

% Error in g $= ( \frac{ \Delta \text{L} }{ {\text{L}}_{0} } \times 1 0 0 ) + 2 ( \frac{ \Delta \text{T} }{ {\text{T}}_{0} } \times 1 0 0 )$

$= ( \frac{ 1 \times 1 {0}^{ - 3 } }{ \text{0.20} } \times 1 0 0 ) + 2 { \frac{ 1 }{ 9 0 } \times 1 0 0 }$

$= \text{0.5} + 2 ( \frac{ 1 0 }{ 9 } )$

$= \text{0.5} + \frac{ 2 0 }{ 9 }$

$= \text{0.5} + \text{2.22}$

= 2.7%

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics