AP EAMCET · PHYSICS · Laws of Motion

A sphere of mass $' m'$ is attached to a spring of spring constant $' k'$ and is held in unstretched position over an inclined plane as shown in the figure. After letting the sphere go, find the maximum length by which the spring extends, given the sphere only rolls.

A $\frac{2 m g \sin θ}{k}$
B $\frac{k}{2 m g \sin θ}$
C $\frac{2 \sin θ}{k m g}$
D $\frac{2 m g \cos θ}{k}$

Verified Solution

Answer & Solution

Correct Answer

(A) $\frac{2 m g \sin θ}{k}$

Step-by-step Solution

Detailed explanation

By applying conservation of energy, loss in gravitational P.E = Gain in P.E of spring. ⇒mgxsinθ=12kx2 ⇒2mgsinθ=kx ⇒x=2mgsinθk

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 sphere of mass 'm' is attached to a spring of spring constant 'k' and is held in unstretched position over an inclined plane as shown in the figure. After letting the sphere go, find the maximum length by which the spring extends, given the sphere only rolls.

Step-by-step Solution

A sphere of mass $' m'$ is attached to a spring of spring constant $' k'$ and is held in unstretched position over an inclined plane as shown in the figure. After letting the sphere go, find the maximum length by which the spring extends, given the sphere only rolls.