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Physics · PYQ NEET UG 2025

हिन्दी

The kinetic energies of two similar cars A and B are \( 100\text{ J} \) and \( 225\text{ J} \) respectively. On applying breaks, car A stops after \( 1000\text{ m} \) and car B stops after \( 1500\text{ m} \). If \( F_A \) and \( F_B \) are the forces applied by the breaks on cars A and B, respectively, then the ratio \( F_A / F_B \) is

  1. A
    \( \frac{3}{2} \)
  2. B
    \( \frac{2}{3} \)
  3. C
    \( \frac{1}{3} \)
  4. D
    \( \frac{1}{2} \)
View Answer & Solution

Correct answer · Option B

\( \frac{2}{3} \)

Explanation

Work done by braking force equals initial kinetic energy: \( F_A d_A = K_A \implies F_A (1000) = 100 \) and \( F_B (1500) = 225 \). Thus \( \frac{F_A}{F_B} = \frac{100/1000}{225/1500} = \frac{0.1}{0.15} = \frac{2}{3} \).
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