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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Physics
Chapter
Oscillations

JEE Main 2026 · 23 January · Morning Shift · Physics · Oscillations

A simple pendulum of string length 30 cm performs 20 oscillations...

A simple pendulum of string length 30 cm performs 20 oscillations in 10 s. The length of the string required for the pendulum to perform 40 oscillations in the same time duration is ____\_\_\_\_ cm. [Assume that the mass of the pendulum remains same.]

Options

  1. A7.5
  2. B0.75
  3. C120
  4. D15

Correct answer

A. 7.5

Step-by-step solution

Period of simple pendulum: T=timenumber of oscillationsT = \frac{\text{time}}{\text{number of oscillations}}
Initial period: T1=1020=0.5T_1 = \frac{10}{20} = 0.5 s
Required period: T2=1040=0.25T_2 = \frac{10}{40} = 0.25 s
Using T=2πLgT = 2\pi\sqrt{\frac{L}{g}}, the ratio is: T2T1=L2L1\frac{T_2}{T_1} = \sqrt{\frac{L_2}{L_1}}
0.250.5=L230⇒0.5=L230\frac{0.25}{0.5} = \sqrt{\frac{L_2}{30}} \Rightarrow 0.5 = \sqrt{\frac{L_2}{30}}
0.25=L230⇒L2=7.50.25 = \frac{L_2}{30} \Rightarrow L_2 = 7.5 cm
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