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NEET PHYSICSWORK, ENERGY AND POWEREasy

Question

A sphere of mass mm is tied to the end of a string of length ll and rotated through the other end along a horizontal circular path with speed vv. The work done by the centripetal force in a full horizontal circle is:

A

0

B

(mv2l)2πl\left(\frac{mv^2}{l}\right) \cdot 2\pi l

C

mg2πlmg \cdot 2\pi l

D

(mv2l)l\left(\frac{mv^2}{l}\right) \cdot l

Step-by-Step Solution

When a sphere moves in a horizontal circular path, the centripetal force (provided by the tension in the string) is continuously directed towards the centre of the circle . At any instant, the displacement of the sphere is tangential to the circular path. Since the radius and the tangent are perpendicular to each other, the angle θ\theta between the centripetal force vector and the displacement vector is 9090^\circ . The work done by a force is given by W=FdcosθW = Fd \cos \theta. Thus, W=Fdcos90=0W = Fd \cos 90^\circ = 0. Therefore, the work done by the centripetal force over a full circle (or any part of the circle) is always zero.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

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