Question:

If the kinetic energy of a moving body reduces to \(49\%\), then its velocity is reduced to

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If KE becomes \(k\) times, velocity becomes \(\sqrt{k}\) times.
Updated On: Apr 24, 2026
  • \(49\%\)
  • \(70\%\)
  • \(30\%\)
  • \(25\%\)
  • \(51\%\)
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Kinetic energy \(KE = \frac{1}{2}mv^2\). So \(KE \propto v^2\).

Step 2:
Detailed Explanation:
\(\frac{KE_2}{KE_1} = \frac{49}{100} = 0.49\)
\(\frac{v_2^2}{v_1^2} = 0.49 \Rightarrow \frac{v_2}{v_1} = \sqrt{0.49} = 0.7 = 70\%\)

Step 3:
Final Answer:
Velocity is reduced to \(70\%\).
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