The kinetic energies of two similar cars A and B are 100 J and 225 J respectively. On applying breaks, car A stops after 1000 m and car B stops after 1500 m. If Fₐ and Fb are the forces applied by the breaks on cars A and B, respectively, then the ratio Fₐ/Fb is:

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NEET 2025 Official Paper (Held On: 04 May, 2025)
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  1. \(\frac{3}{2} \)
  2. \(\frac{2}{3} \)
  3. \(\frac{1}{3} \)
  4. \(\frac{1}{2} \)

Answer (Detailed Solution Below)

Option 2 : \(\frac{2}{3} \)
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Correct option is : (2) 2/3

From work energy theorm

W . D = ΔKE

⇒ F . S = ΔKE

⇒ (ΔKE)A / (ΔKE)B = - FA SA / FB SB

⇒ (-100) / (225) = - FA(1000) / FB(1500)

FA / FB = 2 / 3

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