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Physics · Newton's Laws of Motion

JEE Main 2024 — 30 January, Shift 1 — Question 42

A spherical body of mass 100 g is dropped from a height of 10 m from the ground. After hitting the ground, the body rebounds to a height of 5 m . The impulse of force imparted by the ground to the body is given by : (given g=9.8 m/s2\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^{2} )

  1. Option A:

    4.32 kg ms−14.32 \mathrm{~kg} \mathrm{~ms}^{-1}

  2. Option B:

    43.2 kg ms−143.2 \mathrm{~kg} \mathrm{~ms}^{-1}

  3. Option C:

    23.9 kg ms−123.9 \mathrm{~kg} \mathrm{~ms}^{-1}

  4. Option D:

    2.39 kg ms−12.39 \mathrm{~kg} \mathrm{~ms}^{-1}

    Correct

Answer: D

Step-by-step solution

I⃗=ΔP⃗=P⃗f−P⃗i\vec{I}=\Delta \vec{P}=\vec{P}_{f}-\vec{P}_{i}

M=0.1 kg\mathrm{M}=0.1 \mathrm{~kg}

I=ΔP=0.1(2×9.8×5−(−2×9.8×10))I=\Delta P=0.1(\sqrt{2 \times 9.8 \times 5}-(-\sqrt{2 \times 9.8 \times 10}))

=0.1(14+72)≈2.39 kg ms−1=0.1(14+7 \sqrt{2}) \approx 2.39 \mathrm{~kg} \mathrm{~ms}^{-1}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
30 January, Shift 1
Subject
Physics
Chapter
Newton's Laws of Motion
Topic
Application of NLM and Impulse
A spherical body of mass 100 g is dropped from a height of 10 m from… | JEE Main 2024 PYQ with Solution · DhiX AI