A small mass attached to a string rotates on a frictionless table top as shown in Fig. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will

Options

A. remain constant
B. increase by a factor of `2`
C. increase by a factor of `4`
D. decrease by a foctor of `2`

Answer

C

Explanation

`K.E`. of rotation,
`K = (1)/(2)I omega^(2) = (1)/(2) ((I omega)^(2))/(I) = (1)/(2)(L^(2))/(I) = (1)/(2) (L^(2))/((mr^(2))`
`:. K prop (1)/(r^(2))` (As `(L^(2))/(2m)` is constant during rotation)
`:. (K')/(K) = (r^(2))/((r//2)^(2)) = 4 or K' = 4K`
Thus, `K.E.` of mass increases by a factor of `4`.