A particle tied to a string of negligible weight and length l is swung in a horizontal circular path with constant angular velocity having time period T. If the string length is shortened by `l//2` while the particle is in motion , the time period is .

Options

A. 4T
B. 2T
C. `T//2`
D. `T//4`

Answer

D

Explanation

`omega = (2pi)/(T)=(v)/(l)` According to the law of conservation of angular momentum , `m_(1)v_(1)r_(1)=m_(2)v_(2)r_(2)`
`therefore r_(1)^(2)omega_(1)=r_(2)^(2) omega _(2)implies (r_(1)^(2))/(T_(1))=(r_(2)^(2))/(T_(2))`
`therefore T_(2)=((r_(2)^(2))/(r_(1)^(2)))T_(1)((l^(2)//4)/(l^(2)))T=(T)/(4)`