Question

The real force 'F' acting on a particle of mass ' m' performing circular motion acts along the radius of circle 'r' and is directed towards the centre of circle. The square root of the magnitude of such force is (T = periodic time).

विकल्प

• `(2pi)/"T" sqrt"mr"`

• `"Tmr"/(4pi)`

• `(2pi"T")/sqrt"mr"`

• `("T"^2"mr")/(4pi)`

Answer 1

`(2pi)/"T" sqrt"mr"`

Explanation:

Force F acting on a body of mass m performing circular motion of radius r,

F = `"mv"^2/"r"`   (centripetal force) .....(i)

where, v = velocity of the particle

The time period of one complete cycle,

T = `"perimeter of a circular path"/"velocity of body" = (2pi"r")/"v"`

`=> "v" = (2pi"r")/"T"`     ...(ii)

From Eqs. (i) and (ii), we get

F = `"m"/"r"((2pi"r")/"T")^2`

`= "mr"((2pi)/"T")^2` or

`sqrt"F" = (2pi)/"T" sqrt"mr"`