Verified Solution • 2025–26 Session moderate

QJEEM15Ph4463: JEE Main 2015 Physics - Class 11 JEEM15Ph4463 Solution: The period of oscillation of a simple pendulum is T = 2 π L g . …

The period of oscillation of a simple pendulum is

T = 2 π \sqrt{\frac{L}{g}}

. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using wrist watch of 1 s resolution. The accuracy in the determination of g is:

Multiple Choice Options

A 1%
B 5%
C 2%
D 3%

Step-by-Step Verified Solution

Given

T = 2 π \sqrt{\frac{L}{g}}

\Rightarrow g = \frac{4 π^{2} L}{T^{2}}

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

[ as

T = \frac{t}{n}

]

So, percentage error in

g

\frac{Δ g}{g} \times 100 = \frac{Δ L}{L} \times 100 + 2 \frac{Δ t}{t} \times 100

\frac{0.1}{20.0} \times 100 + 2 \times \frac{1}{90} \times 100

= 2.72 % = 3 %

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