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JEE Main 2026 (28 Jan, Shift 2) · Connected Bodies & Pulleys

Connected Bodies & Pulleys question from JEE Main 2026 (28 Jan, Shift 2)

Q6JEE Main 2026 · 28 JAN · SHIFT 2Difficult

Connected Bodies & Pulleys

A small block of mass m slides down from the top of a frictionless inclined surface, while the inclined plane is moving towards left with constant acceleration a0a_0. The angle between the inclined plane and ground is θ\theta and its base length is LL. Assuming that initially the small block is at the top of the inclined plane, the time it takes to reach the lowest point of the inclined plane is ________.

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Answer: (b)

  1. In the incline frame the block feels a pseudo force ma0ma_0 to the right, which has a component ma0cos⁡θma_0\cos\theta up the slope.
  2. arel=gsin⁡θ−a0cos⁡θa_{rel} = g\sin\theta - a_0\cos\theta; slope length =Lcos⁡θ= \dfrac{L}{\cos\theta}
  3. t=2Lcos⁡θ(gsin⁡θ−a0cos⁡θ)=2Lgsin⁡θcos⁡θ−a0cos⁡2θt = \sqrt{\dfrac{2L}{\cos\theta(g\sin\theta - a_0\cos\theta)}} = \sqrt{\dfrac{2L}{g\sin\theta\cos\theta - a_0\cos^2\theta}}
  4. Using sin⁡θcos⁡θ=12sin⁡2θ\sin\theta\cos\theta = \tfrac{1}{2}\sin 2\theta, cos⁡2θ=12(1+cos⁡2θ)\cos^2\theta = \tfrac{1}{2}(1+\cos 2\theta): t=4Lgsin⁡2θ−a0(1+cos⁡2θ)t = \sqrt{\dfrac{4L}{g\sin 2\theta - a_0(1+\cos 2\theta)}}
  5. So the answer is (b).
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