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JEE Main 2026 (23 Jan, Shift 1) · Conservation of Energy

Conservation of Energy question from JEE Main 2026 (23 Jan, Shift 1)

Q1JEE Main 2026 · 23 JAN · SHIFT 1Medium

Conservation of Energy

An object is projected with kinetic energy KK from a point AA at an angle 60∘60^\circ with the horizontal. The ratio of the difference in kinetic energies at points BB and CC to that at point AA (see figure), in the absence of air friction is:

Figure

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Answer: (a)

  1. A and C are at the same level, so KC=KA=KK_C = K_A = K.
  2. At the top B only ucos⁡60∘u\cos 60^\circ remains: KB=Kcos⁡260∘=K4K_B = K\cos^2 60^\circ = \frac{K}{4}.
  3. KC−KB=K−K4=3K4K_C - K_B = K - \frac{K}{4} = \frac{3K}{4}
  4. Ratio to KAK_A: 34\frac{3}{4}. So the answer is (a) 3:43 : 4.
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