The period of oscillation of a simple pendulum is T = 2π (L)/(g) . Measured value of L is 1 . 0m from meter scale having a minimum division of 1mm and time of one complete oscillation is 1 . 95s measured from stopwatch

Correct answer: D. 1 . 13 % T = 2π (ℓ)/(g) g = 4 π ^ 2 ℓ T ^ 2 ( )/(g) = (Δℓ)/(ℓ) + (2 )/(T) ( )/(g) = 1 × 10 ^ - 3 1 + 2 × (0 . 01)/(1 . 95) ( )/(g) = 0 . 0113 or 1 . 13 %

JEE Main 202124 FebruaryEvening Shift

Exam: JEE Main
Year: 2021
Date: 24 February
Shift: Evening Shift
Subject: Physics
Chapter: Mathematics in Physics

JEE Main 2021 · 24 February · Evening Shift · Physics · Mathematics in Physics

The period of oscillation of a simple pendulum is T = 2π (L)/(g) ....

The period of oscillation of a simple pendulum is

T = 2\pi\sqrt{\dfrac{L}{g}} .

Measured value of

L

1 . 0m

from meter scale having a minimum division of

1mm

and time of one complete oscillation is

1 . 95s

measured from stopwatch of

0 . 01s

resolution. The percentage error in the determination of

' g '

will be:

Options

A $1 . 03 %$
B $1 . 33 %$
C $1 . 30 %$
D $1 . 13 %$

Correct answer

D. $1 . 13 %$

Step-by-step solution

$T = 2\pi\sqrt{\dfrac{ℓ}{g}}$

$g = \dfrac{4{\pi}^{2}ℓ}{{T}^{2}}$

$\dfrac{\Deltag}{g} = \dfrac{\Deltaℓ}{ℓ} + \dfrac{2\DeltaT}{T}$

$\dfrac{\Deltag}{g} = \dfrac{1 \times {10}^{- 3}}{1} + 2 \times \dfrac{0 . 01}{1 . 95}$

$\dfrac{\Deltag}{g} = 0 . 0113$ or $1 . 13 %$

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