MCQMediumJEE 2024Kinetic Energy & Work-Energy Theorem

JEE Physics 2024 Question with Solution

The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10m10 \, \text{m}. If it dissipates 10%10\% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is:

  • A

    65m/s6\sqrt{5} \, \text{m/s}

  • B

    56m/s5\sqrt{6} \, \text{m/s}

  • C

    55m/s5\sqrt{5} \, \text{m/s}

  • D

    25m/s2\sqrt{5} \, \text{m/s}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The pendulum length is =10m\ell = 10 \, \text{m} and 10%10\% of the initial energy is dissipated against air resistance.

Find: The speed of the bob at the lowest point.

When the bob is released from the horizontal position, its initial potential energy relative to the lowest point is

Einitial=mgE_{\text{initial}} = mg\ell

Using g=10m/s2g = 10 \, \text{m/s}^2 and =10m\ell = 10 \, \text{m},

Einitial=m1010=100mE_{\text{initial}} = m \cdot 10 \cdot 10 = 100m

Common mistakes

  • Using height as 22\ell instead of \ell is incorrect because the bob is released from the horizontal position, so the vertical drop to the lowest point is exactly \ell. Use h=h = \ell.

  • Subtracting 10%10\% from the final speed instead of from the initial energy is wrong because the question states that energy is dissipated. First reduce the energy to 90%90\% of its initial value, then compute the speed from kinetic energy.

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