MHT CET · Physics · Rotational Motion

A solid sphere rolls down from top of inclined plane, 7 m high, without slipping. Its linear speed at the foot of plane is $(g = \frac{10 m}{s^{2}})$

A $\frac{\sqrt{70} m}{s}$
B $\frac{\sqrt{\frac{140}{3}} m}{s}$
C $\frac{\sqrt{\frac{280}{3}} m}{s}$
D $\frac{\sqrt{100} m}{s}$

Verified Solution

Answer & Solution

Correct Answer

(D) $\frac{\sqrt{100} m}{s}$

Step-by-step Solution

Detailed explanation

Given, height of inclined plane, h = 7m
G =

\frac{10 m}{s^{2}}

By conservation of energy,
Potential energy lost by the solid sphere in rolling down the inclined plane = Kinetic energy gained by the sphere.

m g h = \frac{1}{2} m v^{2} + \frac{1}{2} l ω^{2}

= \frac{1}{2} m v^{2} + \frac{1}{2} . \frac{2}{5} m R^{2} {(\frac{v}{R})}^{2}

= \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2}

m g h = \frac{7}{10} m v^{2}

∴ v^{2} = \frac{10}{7} g h

\Rightarrow v = \sqrt{\frac{10}{7}} g h = \sqrt{\frac{10}{7} \times 10 \times 7} = \frac{\sqrt{100} m}{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.

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A solid sphere rolls down from top of inclined plane, 7 m high, without slipping. Its linear speed at the foot of plane is g=10ms2

Step-by-step Solution

A solid sphere rolls down from top of inclined plane, 7 m high, without slipping. Its linear speed at the foot of plane is $(g = \frac{10 m}{s^{2}})$