A particle is moving with a uniform spee...

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the n th power of R, If the period of rotation of the particle is T, then

`T prop R^((n + 1))""^(/2)`

`T prop R^(n//2)`

`T prop R^(3//2)` for any n

`T prop R^((n)/(2)+1)`

Text Solution

Verified by Experts

According to the given condition, `F prop = (1)/(R^(n))`
or, `F= (k)/(R^(n)) " " or, m omega^(2) R = (k)/(R^(n))`
or, `" "m((2pi)/(T))^(2) = (k)/(R^(n+1)) " " or, (4pi^(2)m)/(T^(2)) = (k)/(R^(n + 1))`
So, `(1)/(T^(2)) prop (1)/(R^(n + 1)) " "or, T prop R^((n+1)/(2))`
The option A is correct.

Topper's Solved these Questions

CIRCULAR MOTION
CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE WITH SOLUTIONS (AIPMT)|1 Videos
CIRCULAR MOTION
CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE WITH SOLUTIONS (NEET)|2 Videos
CIRCULAR MOTION
CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE WITH SOLUTIONS (WBJEE)|2 Videos
CALORIMETRY
CHHAYA PUBLICATION|Exercise EXAMINATION ARCHIVE|5 Videos
DOPPLER EFFECT IN SOUND
CHHAYA PUBLICATION|Exercise CBSE SCANNER|2 Videos

If a particle of mass is moving in a horizontal circle of radius r with a centripetal force (k/r^2) the total energy is

(-k/2r)

(-k/r)

(-2k/r)

(-4k/r)

A particle moves with a constant speed v along a circular path of radius r and completes the circle in time T. What is the acceleration of the particle?

`(2pir)/T`

`(2piv)/T`

`(pir^2)/T`

`(piv^2)/T`

A particle is moving uniformly in a circular path of radius r. when it moves through an annular displacement theta then the magnitude of the corresponding linear displacement will be

`2rcos(theta/2)`

`2rcot(theta/2)`

`2rtan(theta/2)`

`2rsin(theta/2)`