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JEE Physics 2024 Question with Solution

A ball suspended by a thread swings in a vertical plane so that its acceleration in the extreme and lowest position are equal. The angle (θ\theta) of deflection in the extreme position is:

  • A

    tan⁡−1(2)\tan^{-1}(\sqrt{2})

  • B

    2tan⁡−1(1/2)2\tan^{-1}(1/2)

  • C

    tan⁡−1(1/2)\tan^{-1}(1/2)

  • D

    2tan⁡−1(1/5)2\tan^{-1}(1/\sqrt{5})

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A pendulum of length ll swings with extreme angular deflection θ\theta. The magnitudes of acceleration at the extreme position and at the lowest position are equal.

Find: The value of θ\theta and hence the correct option.

At the extreme position, the bob has zero speed, so there is no centripetal acceleration. The acceleration is only tangential:

aextreme=gsin⁡θa_{\text{extreme}} = g\sin\theta

At the lowest position, the tangential acceleration is zero, and the acceleration is purely centripetal:

alowest=v2la_{\text{lowest}} = \frac{v^2}{l}

Given that these are equal in magnitude,

gsin⁡θ=v2lg\sin\theta = \frac{v^2}{l}

Using conservation of mechanical energy between the extreme position and the lowest position,

mgl(1−cos⁡θ)=12mv2mgl(1-\cos\theta) = \frac{1}{2}mv^2

So,

v2=2gl(1−cos⁡θ)v^2 = 2gl(1-\cos\theta)

Substituting into the acceleration condition,

gsin⁡θ=2gl(1−cos⁡θ)lg\sin\theta = \frac{2gl(1-\cos\theta)}{l}

Hence,

sin⁡θ=2(1−cos⁡θ)\sin\theta = 2(1-\cos\theta)

Using

sin⁡θ=2sin⁡θ2cos⁡θ2,1−cos⁡θ=2sin⁡2θ2\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}, \qquad 1-\cos\theta = 2\sin^2\frac{\theta}{2}

we get

2sin⁡θ2cos⁡θ2=4sin⁡2θ22\sin\frac{\theta}{2}\cos\frac{\theta}{2} = 4\sin^2\frac{\theta}{2}

Dividing by 2sin⁡θ22\sin\frac{\theta}{2},

cos⁡θ2=2sin⁡θ2\cos\frac{\theta}{2} = 2\sin\frac{\theta}{2}

Therefore,

tan⁡θ2=12\tan\frac{\theta}{2} = \frac{1}{2}

So,

θ=2tan⁡−1(12)\theta = 2\tan^{-1}\left(\frac{1}{2}\right)

Therefore, the correct option is B.

Direct Energy-Acceleration Relation

Given: Equal acceleration magnitudes at the extreme and lowest positions.

Find: The extreme angular deflection θ\theta.

At the lowest point,

a=v2la = \frac{v^2}{l}

and from energy conservation,

v2l=2g(1−cos⁡θ)\frac{v^2}{l} = 2g(1-\cos\theta)

At the extreme point,

a=gsin⁡θa = g\sin\theta

Equating them directly,

2g(1−cos⁡θ)=gsin⁡θ2g(1-\cos\theta) = g\sin\theta

which gives

sin⁡θ=2(1−cos⁡θ)\sin\theta = 2(1-\cos\theta)

Using the half-angle identity immediately,

tan⁡θ2=12\tan\frac{\theta}{2} = \frac{1}{2}

Hence,

θ=2tan⁡−1(12)\theta = 2\tan^{-1}\left(\frac{1}{2}\right)

Therefore, the correct option is B.

Common mistakes

  • At the extreme position, taking the acceleration as centripetal is incorrect because the speed there is zero. The acceleration is purely tangential, equal to gsin⁡θg\sin\theta. Always check whether the bob has speed before using v2/lv^2/l.

  • At the lowest position, using gsin⁡θg\sin\theta again is wrong because the tangent is horizontal there, so tangential acceleration is zero. The acceleration at that point is centripetal, equal to v2/lv^2/l.

  • While using energy conservation, writing the height change as lcos⁡θl\cos\theta instead of l(1−cos⁡θ)l(1-\cos\theta) gives a wrong expression for v2v^2. The vertical rise of the bob from the lowest point to the extreme point is l(1−cos⁡θ)l(1-\cos\theta).

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