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

A stone of mass 1 kg tied to a light inextensible string of length L=

1.	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
Answer» `5 m//s` `5sqrt(3) m//s` `5sqrt(2) m//s` `10 m//s` Solution :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`.