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Physics · Work, Power & Energy

JEE Main 2024 — 29 January, Shift 2 — Question 36

The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 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:[Use, g:10 ms−2\mathrm{g}: 10 \mathrm{~ms}^{-2} ]

  1. Option A:

    65 ms−16 \sqrt{5} \mathrm{~ms}^{-1}

    Correct
  2. Option B:

    56 ms−15 \sqrt{6} \mathrm{~ms}^{-1}

  3. Option C:

    55 ms−15 \sqrt{5} \mathrm{~ms}^{-1}

  4. Option D:

    25 ms−12 \sqrt{5} \mathrm{~ms}^{-1}

Answer: A

Step-by-step solution

ℓ=10 m\ell=10 \mathrm{~m},

Initial energy =mg⁡ℓ=\operatorname{mg} \ell

So, 910mgℓ=12mv2\frac{9}{10} \mathrm{mg} \ell=\frac{1}{2} \mathrm{mv}^{2}

⇒910×10×10=12v2\Rightarrow \frac{9}{10} \times 10 \times 10=\frac{1}{2} \mathrm{v}^{2}

v2=180\mathrm{v}^{2}=180

v=180=65 m/s\mathrm{v}=\sqrt{180}=6 \sqrt{5} \mathrm{~m} / \mathrm{s}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 2
Subject
Physics
Chapter
Work, Power & Energy
Topic
Applications of Conservation of Mechanical Energy

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