The acceleration due to gravity on the surface of the moon is `1.7ms^(-2)`. What is the time perioid of a simple pendulum on the surface of the moon, if its time period on the surface of earth is `3.5s ?` Take `g=9.8ms^(-2)` on the surface of the earth.

The time period of a simple pendulum on the surface of the earth is 4s. Its time period on the surface of the moon is

A simple pendulum is set up at an unknown planet.The acceleration due to gravity on the surface is 6.2ms^(-2) .Find the time period of the pendulum on the surface of the planet,if its time period on the surface of the earth is 3.5s .Take g=9.8ms^(-2)

A simple pendulum is set up at an unknown planet.The acceleration due to gravity on the surface is 6.2ms^(-2) .Find the time period of the pendulum on the surface of the planet,if its time period on the surface of the earth is 3.5s ,Take g =9.8ms^(-2)

If acceleration due to gravity on the moon is one-sixth that on the earth, then the length and time period of seconds pendulum on the surface of moon are

The time period of a simple pendulum when it is made to oscillate on the surface of moon

The value of acceleration due to gravity 'g' on the surface of the moon with radius 1/2 that of the earth and same mean density as that of the earth

Assume that the acceleration due to gravity on the surface of the moon is 0.2 times the acceleration due to gravity on the surface of the earth. If R_e is the maximum range of a projectile on the earth's surface, what is the maximum range on the surface of the moon for the same velocity of projection

The acceleration due to gravity on the surface of the moon = (1)/(6) the acceleration due to gravity on the surface of the earth. What will be the mass of a steel ball on the surface of the moon, if its mass on the surface of the earth is 6 kg ?