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 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?

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?

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?

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))

A simple pendulum with a bob of mass 40g and charge +2mu C makes 20 oscillation in 44 s. A vertical electric field magnitude 4.2xx10^(4) NC^(-1) pointing downward is applied. The time taken by the pendulum to make 15 oscillation in the electric field is (acceleration due to gravity =10 ms^(-2) )

The bob of a simple pendulum is oscillating with a frequency of 25 Hz. How much time will the pendulum take to make 100 oscillations?

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 bob of a simple pendulum of mass 40 mg with a positive charge 4 xx 10^(-6) C is oscilliating with time period "T_(1). An electric field of intensity 3.6 xx 10^(4) N/c is applied vertically upwards now time period is T_(2) . The value of (T_(2))/(T_(1)) is (g = 10 m/s^(2))