A stone of mass m tied to the end of a string, is whirled around in a horizontal circle. (Neglect the force due to gravity). The length of the string is reduced gradually keeping the angular momentum of the stone about the centre of the circle constant. Then, the tension in the string is given by `T = Ar^n` where A is a constant, r is the instantaneous radius of the circle and n=....

Options

A. -3
B. 3
C. 2
D. -4

Answer

a

Explanation

Let r=radius of circle at an instant of time.
r goes on chaning with time but angular momentum of the stone about the center of the circle is constant.

`therefore` Tension in string, `T= mromega^(2)`...........(i)
Angular momenum, L =`Iomega= (mr^(2))omega`.........(ii)
Now, eliminate `omega` between (i) and (ii),
`T=mrL/(mr^(2))^(2) = (mrL^(2))/(m^(2)r^(4)) = L^(2)/(mr^(3))`
Or `T =(L^(2))/((m)r^(-3))`, where `L^(2)/m` = constant
Given `T=Ar^(n)`. Hence `n=-3`.