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 $\,m$)

$A$ simple pendulum of mass $200\, g$ and length $100\, cm$ is moved aside until the string makes an angle of $60^\circ$ with the vertical. The kinetic and potential energies of the bob,when the string is inclined at $30^\circ$ to the vertical,are

$A$ simple pendulum of length $l$ is made to oscillate with an amplitude of $45^{\circ}$. The acceleration due to gravity is $g$. Let $T_0 = 2 \pi \sqrt{l / g}$. The time period of oscillation of this pendulum will be

The bob of a simple pendulum is hanging vertically down from a fixed identical bob by means of a string of length $l$. If both bobs are charged with a charge $q$ each,what is the time period of the pendulum? (Ignore the radii of the bobs.)

$A$ simple pendulum of length $1\,m$ is allowed to oscillate with an amplitude of $2^o$. It collides elastically with a wall inclined at $1^o$ to the vertical. Its time period will be: (use $g = \pi^2$)

$A$ pendulum has a time period $T$ in air. When it is made to oscillate in water,its time period is $\sqrt{2} T$. Then the relative density of the material of the bob of the pendulum is (neglect damping).