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MCQEasyJEE 2023Rolling Motion & Rotational Kinematics

JEE Physics 2023 Question with Solution

A solid cylinder is released from rest from the top of an inclined plane of inclination 30∘30^\circ and length 60 cm60 \, \text{cm}. If the cylinder rolls without slipping, its speed upon reaching the bottom of the inclined plane is _____ ms−1\text{ms}^{-1}. (Given g=10 ms−2g = 10 \, \text{ms}^{-2})

A solid cylinder at the top of a 30 degree inclined plane of length 60 cm, rolling down the incline.Diagram of an inclined plane marked 30 degree at the base and 60 cm along the slope, with a cylinder at the top.
  • A

    11

  • B

    22

  • C

    33

  • D

    44

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: A solid cylinder rolls without slipping down an inclined plane of length 60 cm60 \, \text{cm} and inclination 30∘30^\circ. Also, g=10 ms−2g = 10 \, \text{ms}^{-2}.

Find: The speed at the bottom.

For rolling motion,

v=2gh1+k2R2v = \sqrt{\frac{2gh}{1 + \frac{k^2}{R^2}}}

where the vertical height is

h=60sin⁡30∘=30 cmh = 60 \sin 30^\circ = 30 \, \text{cm}

and for a solid cylinder,

k2=R22k^2 = \frac{R^2}{2}

Substituting in the given relation,

v=2 ms−1v = 2 \, \text{ms}^{-1}

Therefore, the speed of the cylinder at the bottom is 2 ms−12 \, \text{ms}^{-1}. The correct option is B.

Common mistakes

  • Using the full incline length as the vertical height is incorrect because gravitational potential energy depends on vertical drop, not distance along the plane. First find h=lsin⁡θh = l \sin \theta.

  • Treating the cylinder like a sliding block is wrong because rolling without slipping includes rotational kinetic energy. Use the rolling expression with k2R2\frac{k^2}{R^2}, not only translational motion.

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