A simple pendulum has a time period T_(1). When its point of suspension is moved vertically upwards according as y=kt^(2),where y is vertical covered and k=1 ms^(-2) , its time period becomes T_(2) then(T_(1)^(2))/(T_(2)^(2)) is (g=10 ms^(-2))

A simple pendulum has a time period T_(1). When its point of suspensio

1.	A simple pendulum has a time period T_(1). When its point of suspension is moved vertically upwards according as y=kt^(2),where y is vertical covered and k=1 ms^(-2) , its time period becomes T_(2) then(T_(1)^(2))/(T_(2)^(2)) is (g=10 ms^(-2))
Answer» `(5)/(6)` `(11)/(10)` `(6)/(5)` `(5)/(4)` SOLUTION :`T=T_(1)= 2pi sqrt((l)/(G))` `:.T_(1)^(2)= 4PI^(2)((l)/(g))` `y= kt^(2)` `K= 1 MS^(-1)` `:. Y= t^(2)` `:. (y_(1))/(y_(2))=(T_(1)^(2))/(T_(2)^(2))=(6)/(5)`