A small block of mass `4kg` is attached to a cord passing through a hole in a horizontal frictionless surface. The block is originally revolving in a circle of radius `0.5 m` about the hole with a tangential velocity of `4m//s`. The cord is then pulled slowly from below, shorteing the radius of the circle in which the block revolves. The breaking strength of the cord is `600 N`. What will be radius of the circle when the cord breaks?

Options

A. `4.0m`
B. `1.0m`
C. `3.0m`
D. `2.0m`

Answer

D

Explanation

The tension of the rope is the only net force on the block and it does not exert torque about the axis of rotation. Hence, the angular momentum of the block about the axis should remain conserved.

`impliesmvr=`constant
Let `r_(1)=5m` and `v_(1)=4m//s`
Let `r_(2), v_(2), T_(2)` be the radius, velocity and tension when the string breaks.
`implies T_2=200N`
`mv_(1)r_(1)=mv_(2)r_(2)` and `T_(2)=(mv_(2)^(2))/(r_(2))`.....i
`implies mv_(1)r_(1)=msqrt((r_(2)T_(2))/mr_(2))`
`r_(2)=((mv_(1)^(2)r_(1)^(2))/(T_(2)))^(1/3)=((4x15xx25)/200)^(1/3)=2m`