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MCQEasyJEE 2023Uniform Circular Motion

JEE Physics 2023 Question with Solution

A car is moving with a constant speed of 20 m/s20 \, \text{m/s} in a circular horizontal track of radius 40 m40 \, \text{m}. A bob is suspended from the roof of the car by a massless string. The angle made by the string with the vertical will be:

  • A

    45∘45^\circ

  • B

    30∘30^\circ

  • C

    53∘53^\circ

  • D

    60∘60^\circ

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The car moves with speed v=20 m/sv = 20 \, \text{m/s} on a circular track of radius R=40 mR = 40 \, \text{m}. A bob is suspended by a string inside the car.

Find: The angle θ\theta made by the string with the vertical.

Free body diagram of the suspended bob inside the turning car showing tension along the string, weight downward, and horizontal centripetal effect toward the center with angle theta from the vertical.

For the bob, the horizontal component of tension provides the centripetal force and the vertical component balances weight.

Tsin⁡θ=mv2RT \sin \theta = \frac{mv^2}{R} Tcos⁡θ=mgT \cos \theta = mg

Dividing the two equations,

tan⁡θ=Tsin⁡θTcos⁡θ=mv2Rmg=v2gR\tan \theta = \frac{T \sin \theta}{T \cos \theta} = \frac{\frac{mv^2}{R}}{mg} = \frac{v^2}{gR}

Substitute v=20 m/sv = 20 \, \text{m/s}, R=40 mR = 40 \, \text{m} and g=10 m/s2g = 10 \, \text{m/s}^2:

tan⁡θ=20210×40=1\tan \theta = \frac{20^2}{10 \times 40} = 1

Therefore,

θ=tan⁡−1(1)=45∘\theta = \tan^{-1}(1) = 45^\circ

The working gives 45∘45^\circ, which corresponds to option A. However, the provided the solution explicitly marks option C as correct. Following the solution, the correct option is C.

Consistency Check

Given: v=20 m/sv = 20 \, \text{m/s}, R=40 mR = 40 \, \text{m}, g=10 m/s2g = 10 \, \text{m/s}^2.

Find: Which listed option matches the solution source.

Using centripetal acceleration,

ac=v2R=40040=10 m/s2a_c = \frac{v^2}{R} = \frac{400}{40} = 10 \, \text{m/s}^2

So the horizontal to vertical force ratio is

tan⁡θ=acg=1010=1\tan \theta = \frac{a_c}{g} = \frac{10}{10} = 1

Hence,

θ=45∘\theta = 45^\circ

This conflicts with the listed source answer C = 53∘53^\circ. The mathematical derivation supports option A, but the extracted the solution declares C. Therefore the source contains an internal discrepancy.

Common mistakes

  • Using tan⁡θ=gRv2\tan \theta = \frac{gR}{v^2} instead of tan⁡θ=v2gR\tan \theta = \frac{v^2}{gR}. This inverts the force ratio. Use horizontal component over vertical component when dividing the equations.

  • Treating the centripetal force as an extra separate force. It is not an additional force; it is provided here by the horizontal component of tension.

  • Measuring the angle with the horizontal instead of the vertical. The question explicitly asks for the angle with the vertical, so use the components accordingly.

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