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MCQMediumJEE Main 2026 · 28 January, Shift 2Newton's Second Law & Force

Physics Question from JEE Main 2026 · 28 January, Shift 2

A small block of mass mm 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 _.

A block marked m is at the top of an inclined plane of angle theta, base length L, with the wedge accelerating left with acceleration a0.
  • A

    4Lgsin⁡2θ−a0(1+cos⁡2θ)\displaystyle \sqrt{\frac{4L}{g\sin 2\theta-a_0(1+\cos 2\theta)}}

  • B

    2Lgsin⁡θ−a0cos⁡θ\displaystyle \sqrt{\frac{2L}{g\sin\theta-a_0\cos\theta}}

  • C

    4Lgcos⁡2θ−a0sin⁡θcos⁡θ\displaystyle \sqrt{\frac{4L}{g\cos^2\theta-a_0\sin\theta\cos\theta}}

  • D

    2Lgsin⁡2θ−a0(1+cos⁡2θ)\displaystyle \sqrt{\frac{2L}{g\sin 2\theta-a_0(1+\cos 2\theta)}}

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