A particle of mass m moves in a circular path radius r under the action of a force (m v2/r). The work done during its motion over half of the circumference of the circular path will be

Question

A particle of mass m moves in a circular path radius r under the action of a force (m v2/r). The work done during its motion over half of the circumference of the circular path will be (A) ((m v2/r)) × 2 π r (B) ((m v2/r)) × π r (C) ((2 π r)/((m v2)r)) (D) zero

Tardigrade Admin · Accepted Answer

Work done by centripetal force is always zero.

Q. A particle of mass $m$ moves in a circular path radius $r$ under the action of a force $\frac{m v ^{2}}{r}$ . The work done during its motion over half of the circumference of the circular path will be

$(\frac{m v ^{2}}{r}) \times 2 π r$

$(\frac{m v ^{2}}{r}) \times π r$

$\frac{( 2 π r )}{( \frac{m v ^{2}}{r} )}$

zero

Work done by centripetal force is always zero.

Q. A particle of mass m moves in a circular path radius r under the action of a force rmv2​. The work done during its motion over half of the circumference of the circular path will be

(rmv2​)×2πr

(rmv2​)×πr

(rmv2​)(2πr)​

zero

Work done by centripetal force is always zero.

Q. A particle of mass $m$ moves in a circular path radius $r$ under the action of a force $\frac{m v ^{2}}{r}$ . The work done during its motion over half of the circumference of the circular path will be

$(\frac{m v ^{2}}{r}) \times 2 π r$

$(\frac{m v ^{2}}{r}) \times π r$

$\frac{( 2 π r )}{( \frac{m v ^{2}}{r} )}$