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JEE Main 2026 (23 Jan, Shift 1) · Simple Pendulum

Simple Pendulum question from JEE Main 2026 (23 Jan, Shift 1)

Q4JEE Main 2026 · 23 JAN · SHIFT 1Easy

Simple Pendulum

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.]

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Answer: (b)

  1. T1=1020=0.5T_1 = \frac{10}{20} = 0.5 s and T2=1040=0.25T_2 = \frac{10}{40} = 0.25 s.
  2. l∝T2⇒l2=l1(T2T1)2=30×14l \propto T^2 \Rightarrow l_2 = l_1\left(\frac{T_2}{T_1}\right)^2 = 30 \times \frac{1}{4}
  3. l2=7.5l_2 = 7.5 cm
  4. So the answer is (b) 7.5 cm.
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