TS EAMCET · Physics · Rotational Motion

A solid sphere rolls down without slipping from the top of an inclined plane of height 28 m and angle of inclination \(30^{\circ}\). The velocity of the sphere, when it reaches the bottom of the plane is (Acceleration due to gravity \(=10 \mathrm{~ms}^{-2}\) )

A \(20 \mathrm{~ms}^{-1}\)
B \(28 \mathrm{~ms}^{-1}\)
C \(10 \mathrm{~ms}^{-1}\)
D \(14 \mathrm{~ms}^{-1}\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(20 \mathrm{~ms}^{-1}\)

Step-by-step Solution

Detailed explanation

Velocity of sphere rolls down without slipping \(v=\sqrt{\frac{2 \mathrm{gh}}{1+\frac{\mathrm{k}^2}{\mathrm{R}^2}}}=\sqrt{\frac{2 \times 10 \times 28}{1+\frac{2}{5}}}=\sqrt{\frac{2 \times 10 \times 28 \times 5}{7}}=20 \mathrm{~m} / \mathrm{s}\)

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.

Same subject

Explore more questions on app

From TS EAMCET

Explore more questions on app