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 a_0. The angle between the inclined plane and ground is \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 \_\_\_\_ . [28 Jan 2026 Shift 2]

Correct Answer is : √4Lgsin2θ−a0(1+cos2θ)

\sqrt{\frac{2 L}{g \sin \theta-a_0 \cos \theta}}

\sqrt{\frac{4 L}{g \sin 2 \theta-a_0(1+\cos 2 \theta)}}

\sqrt{\frac{2 L}{g \sin 2 \theta-a_0(1+\cos 2 \theta)}}

\sqrt{\frac{4 L}{g \cos ^2 \theta-a_0 \sin \theta \cos \theta}}

JEE Main 28 Jan 2026 Shift 2 Paper Solved Paper

Section: Physics

Question : 32 of 75

Marks: +1, -0

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 a0. The angle between the inclined plane and ground is θ 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 ____ .

[28 Jan 2026 Shift 2]

√‌
2L
gsin‌θ−a0‌cos‌θ
√‌
4L
gsin‌2θ−a0(1+cos‌2‌θ)
√‌
2L
gsin‌2θ−a0(1+cos‌2‌θ)
√‌
4L
gcos2θ−a0sin‌θ‌cos‌θ

Validate

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