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Physics · Newton's Laws of Motion

JEE Main 2026 — 28 January, Evening Shift — Question 40

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 a0\mathrm{a}_{0}. The angle between the inclined plane and ground is θ\theta and its base length is L . 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 ____\_\_\_\_ .

Question figure
  1. Option A:

    2 L gsin⁡2θ−a0(1+cos⁡2θ)\sqrt{\frac{2 \mathrm{~L}}{\mathrm{~g} \sin 2 \theta-\mathrm{a}_{0}(1+\cos 2 \theta)}}

  2. Option B:

    4 L gsin⁡2θ−a0(1+cos⁡2θ)\sqrt{\frac{4 \mathrm{~L}}{\mathrm{~g} \sin 2 \theta-\mathrm{a}_{0}(1+\cos 2 \theta)}}

    Correct
  3. Option C:

    4 L gcos⁡2θ−a0sin⁡θcos⁡θ\sqrt{\frac{4 \mathrm{~L}}{\mathrm{~g} \cos ^{2} \theta-\mathrm{a}_{0} \sin \theta \cos \theta}}

  4. Option D:

    2Lgsin⁡θ−a0cos⁡θ\sqrt{\frac{2 L}{g \sin \theta-a_{0} \cos \theta}}

Answer: B

Step-by-step solution

mgsin⁡θ−ma0cos⁡θ=m g \sin \theta-m a_{0} \cos \theta= ma a=gsin⁡θ−a0cos⁡θ\mathrm{a}=\mathrm{g} \sin \theta-\mathrm{a}_{0} \cos \theta Now using, S=ut+12adown t2\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{a}_{\text {down }} \mathrm{t}^{2} Lcos⁡θ=12( gsin⁡θ−a0cos⁡θ)t2\frac{\mathrm{L}}{\cos \theta}=\frac{1}{2}\left(\mathrm{~g} \sin \theta-\mathrm{a}_{0} \cos \theta\right) \mathrm{t}^{2} t=2 L gsin⁡θcos⁡θ−a0cos⁡2θ\mathrm{t}=\sqrt{\frac{2 \mathrm{~L}}{\mathrm{~g} \sin \theta \cos \theta-\mathrm{a}_{0} \cos ^{2} \theta}} t=4 L gsin⁡2θ−a0(1+cos⁡2θ)\mathrm{t}=\sqrt{\frac{4 \mathrm{~L}}{\mathrm{~g} \sin 2 \theta-\mathrm{a}_{0}(1+\cos 2 \theta)}}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Evening Shift
Subject
Physics
Chapter
Newton's Laws of Motion
Topic
Non-Inertial Frame of Reference
A small block of mass m slides down from the top of a frictionless… | JEE Main 2026 PYQ with Solution · DhiX AI