A bob of a simple pendulum of mass `40gm` with a positive charge `4xx10^(-6)C` is oscillating with a time period `T_(1)`.An electric field of intensity `3.6xx10^(4)N//C` is applied vertically upwards.Now the time period is `T_(2)` the value of `(T_(2))/(T_(1))` is `(g=10//s^(2))`

A simple pendulum has a bob of mass m = 40 gm and a positive charge q =4 xx 10^(-6) C . It makes 20 oscillation in 4.5 s. A vertical upward electric field of magnitude E = 2.5 xx 10^(4) N//C is switched on in space. How much time will the simple pendulum will now take to complete 20 oscillation.

The bob of a simple pendulum has a mass of 60 g and a positive charge of 6xx10^(-6) C . It makes 30 oscillations in 50 s above earth's surface. A vertical electric field pointing upward and of magnitude 5xx10^(4) N//C is switched on. How much time will it now take to complete 60 oscillations ? (g=10 m//s^(2))

The bob of a simple pendulum has a mass of 40 g and a positive charge of 4.0 xx 10 ^(-5) C . It makes 20 oscillations in 45 s. A vertical electric field pointing upward and of magnitude 2.5 xx 10^ 4 NC^(-1) is switched on. How much time will it now take to complete 20 oscillations?

A simple pendulum has a time period T_(1) on the surface of earth of radius R. When taken to a height of R above the earth's surface, its time period is T_(2) , ratio (T_(2))/(T) is

A simple pendulu has a time period T_(1) . The point of suspension of the pendulum is moved upword according to the relation y= (3)/(2)t^(2) , Where y is the vertical displacement . If the new time period is T_(2) , The ratio of (T_(1)^(2))/(T_(2)^(2)) is ( g=10m//s^(2))

The bob of a simple pendulum of mass 100g is oscillating with a time period of 1.42s . If the bob is replaced by another bob of mass 150g but of same radius, the new time period of oscillation

A simple pendulum has time period T_(1) / The point of suspension is now moved upward according to the realtion y = kt^(2)(k = 1 m//s^(2)) where y is vertical displacement, the time period now becomes T_(2) . The ratio of ((T_(1))/(T_(2)))^(2) is : (g = 10 m//s^(2))

The bob of a pendulum of mass 8 mu g carries an electric charge of 39.2xx10^(-10) coulomb in an electric field of 20xx10^(3) volt/meter and it is at rest. The angle made by the pendulum with the vertical will be

A simple pendulum is oscillating in a lift. If the lift is going down with constant velocity, the time period of the simple pendulum is T_(1) . If the lift is going down with some retardation its time period is T_(2) , then