$A$ pendulum clock is running fast. To correct its time,we should

The mass and diameter of a planet are twice those of Earth. What will be the period of oscillation of a pendulum on this planet if it is a seconds pendulum on Earth?

The ratio of frequencies of two oscillating pendulums are $3: 2$. Their lengths are in the ratio:

The length of a seconds pendulum is $1 \,m$ on the earth. If the mass and diameter of a planet are double that of the earth, then the length of the seconds pendulum on the planet will be: (in $\,m$)

$A$ pendulum clock is set to give correct time at sea level. This clock is moved to a hill station at an altitude of $2500 \ m$ above sea level. In order to keep correct time at the hill station,the length of the pendulum:

$A$ simple pendulum is hanging from a peg inserted in a vertical wall. Its bob is stretched in a horizontal position from the wall and is left free to move. The bob hits the wall,and the coefficient of restitution is $\frac{2}{\sqrt{5}}$. After how many collisions will the amplitude of vibration become less than $60^\circ$?