[" Figure shows the kinetic energy "K" of a simple pendulum versus its angle of "],[" pendulum bob has mass "0.2kg" .The length of the pendulum is equal to "(9=10n],[qquad 15],[10],[(100)/((B)1.8m)(100)/(0)(100)/((C)1.5m)f(mrad)]

Figure shown the kinetic energy K of a pendulum versus. its angle theta from the verticle. The pendulum bob has mass 0.2kg The length of the pendulum is equal to (g = 10m//s^(2)) ,

Figure shown the kinetic energy K of a pendulum versus. its angle theta from the vertical. The pendulum bob has mass 0.2kg The length of the pendulum is equal to (g = 10m//s^(2)) ,

Figure shows the kinetic energy K of a simple pendulum versus its angle theta from the verticle. The pendulum bob has mass 0.2 kg . If the length of the pendulum is equal to (n)/(g) meter. Then find n (g = 10 m//s^(2))

The period of a simple pendulum is double, when its length is increased by 1.2 m. The original length of the simple pendulum is

The given figure shows the variation of the kinetic energy of a simple pendulum with its angular displacement (theta) from the vertical. Mass of the pendulum bob is m = 0.2 kg. Find the time period of the pendulum. Take g = 10 ms^(-2) .

If the inertial mass m_(1) of the bob of a simple pendulum of length 'l' is not equal to the gravitationalmass m_(g) , then its period is

If the inertial mass m_(1) of the bob of a simple pendulum of length 'l' is not equal to the gravitationalmass m_(g) , then its period is

A simple pendulum is released from A as shown. If m and 1 represent the mass of the bob and length of the pendulum, the gain kinetic energy at B is

A simple pendulum is released from A as shown. If m and 1 represent the mass of the bob and length of the pendulum, the gain kinetic energy at B is