The bob of a pendulum has a mass of 200g.The string of the pendulum is initially kept in the horizontal position and then released,the tension (in N ) in the string in the lowest position is `(g=10m/s^(2))` 1) 6 2) 8 3) 2 4) 4

The string of a pendulum is horizontal. The mass of the bob is m. Now the string is released. The tension in the string in the lowest position is -

A simple pendulum is oscillating with an angular amplitude 60^@ . If mass of bob is 50 gram, then the tension in the string at mean position is (g = 10 ms^(-2))

The bob of a simple pendulum of length L, is displaced through 90^(@) , from its mean position and then released. What will be the tension in the string when the bob of mass m will be its lowest position ?

The bob of a simple pendulum of length L, is displaced through 90^(@) , from its mean position and then released. What will be the tension in the string when the bob of mass m will be at its lowest position?

The length of a simple pendulum of 0.4 m. The bob of the pendulum is drawn aside till the string is horizontal, it is then released from this horizontal position, What will be its velocity when it reached the lowest point in its swing ? (g = 9.8 m//s^(2)) [Hint : Loss in potential energy = gain in kinetic energy]

If a simple pendulum of mass m is released from a horizontal position, find the tension in the string as a function of the angle theta with the horizontal.

A mass of 2 kg tied to a string 1 m length is rotate in a verticle circle with a uniform speed of 4 m//s . The tension in the string will be 52 N, when the mass is at (g=10 m//s^(2))

A 1kg stone at the end of 1m long string is whirled in a vertical circle at a constant speed of 4m//s . The tension in the string is 6N , when the stone is at (g=10m//s^(2))

The bob of a simple pendulum of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the stirng at this instant.