MHT CET · Physics · Rotational Motion

A solid cylinder of mass 'M' and radius 'R' rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity ' \(\mathrm{v}^{\prime}\). The height of the inclined plane
is \((\mathrm{g}=\) acceleration due to gravity \()\)

A \(\frac{3 v^{2}}{4 g}\)
B \(\frac{4 v^{2}}{5 g}\)
C \(\frac{7 v^{2}}{9 g}\)
D \(\frac{2 v^{2}}{3 g}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{3 v^{2}}{4 g}\)

Step-by-step Solution

Detailed explanation

\(Mgh = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2\) \(Mgh = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v}{R}\right)^2\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A solid cylinder of mass 'M' and radius 'R' rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity ' \(\mathrm{v}^{\prime}\). The height of the inclined planeis \((\mathrm{g}=\) acceleration due to gravity \()\)

Step-by-step Solution

A solid cylinder of mass 'M' and radius 'R' rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity ' \(\mathrm{v}^{\prime}\). The height of the inclined plane
is \((\mathrm{g}=\) acceleration due to gravity \()\)