A stone of mass `1 kg` tied to a light inextensible string of length `L = (10)/(3) m` is whirling in a circular path of radius `L` is a vertical plane . If the ratio in the string is `4` and if `g` is taken to be `10 m//sec^(2)`, the speed of the stone at the highest point of the circle is

Options

A. `5sqrt(2) m//s`
B. `20 m//s`
C. `5sqrt(3) m//s`
D. `10 m//s`

Answer

D

Explanation

At bighest position :`T_(1) + mg = (mv_(1)^(2))/(l)`
or`T_(2) = (mv_(b)^(2))/(l)- mg` …..(i)
Similarly at lowest position
`T_(b) - m_(g) = (mv_(b)^(2))/(l)`
`or T_(b) = (mv_(b)^(2))/(l) + mg`…(ii)
Given `(T_(b))/(T_(t)) = 4 rArr((mvb^(2))/(l) +mg)/((mvb^(2))/(l) - mg) =4`....(iii)
Using C.O.M.E`: Delta E+Delta U = 0`
`((1)/(2) mv_(b)^(2) - (1)/(2) mv_(b)^(2)) + mg 2s= 0 `...(iv) From (ii) and (iv) we get `u_(t) = 10m//s`.