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A particle of mass m is initially situated at point P inside a hemispherical surface of radius r as shown in the figure. A horizontal acceleration of magnitude α₀ is suddenly produced acceleration on the particle in the horizontal direction. If gravitational acceleration is neglected, then time taken by the particle to touch the sphere again is

A. √4r sin α/α₀
B. √4r tan α/α₀
C. √4r cos α/α₀
D. None of these

1 Answer

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Final answer:

The time taken by the particle to touch the hemispherical surface again, with gravity neglected, is given by √4r tan α/α0, which is based on the kinematic equation for uniformly accelerated motion along a chord of the hemisphere. Therefore, the correct option is B.

Step-by-step explanation:

You're tasked with finding the time it takes for a particle, which is subject to a horizontal acceleration, to touch a hemispherical surface again after starting from an initial point inside the hemisphere. Neglecting gravitational acceleration, the particle will follow a linear trajectory within the hemisphere due to the horizontal force applied.

The answer to the question is based on the geometry of the situation and the kinematic equation for uniformly accelerated motion. The correct option is B. √4r tan α/α0, where α is the angle made by the radius vector with the vertical at the starting point and α0 is the horizontal acceleration.

Since the particle is moving horizontally, the trajectory will be a chord of the hemisphere with length 2r sin α, as projected onto the base of the hemisphere. The time taken to travel this distance with constant acceleration α0 is given by the kinematic formula t = √(2s/a), where s is the distance traveled. Substituting s = 2r sin α and a = α0 gives the correct answer.

User Al Belmondo
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