A stone of mass 1 kg tied to a light inextensible string of length `L= 5/3m`, is whirling in a vertical circle of radius L. If the ratio of the maximum and minimum tension in the string is 4 and `g = 10 m//s^2`, the speed of stone at the highest point of the circle is 

Options

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

Answer

C

Explanation

In the vertical circular motion, the tension in the string is maximum at A and minimum at B.
`T_A = (mv_A^2)/(L) + mg = T_("max")`
`T_B = (mv_B^2)/(L) - mg = T_("min")`
`(T_("max"))/(T_("min")) = 4 ((mv_A^2)/L + mg)/((mv_B^2)/L - mg)`
or `(v_A^2) + gL)/(v_B^2 - gL) = 4/1 " " ....(i)`
But `v_A^2 = v_B^2 + 2g(2L) = v_B^2 + 4gL`
(from the law of conservation of enrergy)
From (i), `(v_B^2 + 5gL)/(v_B^2 - gL) = 4/1`
On solving, we get , `v_B = sqrt(3gL) = sqrt(3 xx 10 xx 5/3) = sqrt(50) = 5sqrt(2) m//s`.