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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Work and Energy

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Work and Energy

The figure shows a smooth track which consists of a straight...

The figure shows a smooth track which consists of a straight inclined part of length ll joining smoothly with the circular part. A particle of mass mm is projected up the incline from its bottom. (a) Find the minimum projection-speed v0v_0 for which the particle reaches the top of the track. (b) Assuming that the projection-speed is 2v02v_0 and that the block does not lose contact with the track before reaching its top, find the force acting on it when it reaches the top. (c) Assuming that the projection-speed is only slightly greater than v0v_0, where will the block lose contact with the track?
The figure shows a smooth track which consists of a straight inclined part of length l joining smoothly with the

Options

  1. A(a) 2g[R(1−cos⁡θ)+lsin⁡θ]\sqrt{2g[R(1 - \cos\theta) + l\sin\theta]}
    (b) 6mg(1−cos⁡θ+lRsin⁡θ)6mg\left(1 - \cos\theta + \dfrac{l}{R}\sin\theta\right)
    (c) The radius through the particle makes an angle cos⁡−1(2/3)\cos^{-1}(2/3) with the vertical.
  2. B(a) 2g[R(1−cos⁡θ)+lsin⁡θ]\sqrt{2g[R(1 - \cos\theta) + l\sin\theta]}
    (b) 3mg(1−cos⁡θ+lRsin⁡θ)3mg\left(1 - \cos\theta + \dfrac{l}{R}\sin\theta\right)
    (c) The radius through the particle makes an angle cos⁡−1(1/3)\cos^{-1}(1/3) with the vertical.
  3. C(a) g[R(1−cos⁡θ)+lsin⁡θ]\sqrt{g[R(1 - \cos\theta) + l\sin\theta]}
    (b) 6mg(1−cos⁡θ+lRsin⁡θ)6mg\left(1 - \cos\theta + \dfrac{l}{R}\sin\theta\right)
    (c) The radius through the particle makes an angle cos⁡−1(1/2)\cos^{-1}(1/2) with the vertical.
  4. D(a) 2g[R(1+cos⁡θ)+lsin⁡θ]\sqrt{2g[R(1 + \cos\theta) + l\sin\theta]}
    (b) 4mg(1+cos⁡θ+lRsin⁡θ)4mg\left(1 + \cos\theta + \dfrac{l}{R}\sin\theta\right)
    (c) The radius through the particle makes an angle cos⁡−1(2/3)\cos^{-1}(2/3) with the vertical.

Correct answer

A. (a) 2g[R(1−cos⁡θ)+lsin⁡θ]\sqrt{2g[R(1 - \cos\theta) + l\sin\theta]}
(b) 6mg(1−cos⁡θ+lRsin⁡θ)6mg\left(1 - \cos\theta + \dfrac{l}{R}\sin\theta\right)
(c) The radius through the particle makes an angle cos⁡−1(2/3)\cos^{-1}(2/3) with the vertical.

Step-by-step solution

(a) 2g[R(1−cos⁡θ)+lsin⁡θ]\sqrt{2g[R(1 - \cos\theta) + l\sin\theta]}
(b) 6mg(1−cos⁡θ+lRsin⁡θ)6mg\left(1 - \cos\theta + \dfrac{l}{R}\sin\theta\right)
(c) The radius through the particle makes an angle cos⁡−1(2/3)\cos^{-1}(2/3) with the vertical.
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