What is the number of degrees of freedom for an oscillating simple pendulum?

$A$ simple pendulum is executing simple harmonic motion with a time period $T$. If the length of the pendulum is increased by $21\%$,the percentage increase in the time period of the pendulum of increased length is ..... $\%$

$A$ coin is placed on a horizontal plate. The plate performs $S.H.M.$ vertically with an angular frequency $\omega$. The amplitude $A$ of oscillations is gradually increased. The coin will lose contact with the plate for the first time when the amplitude is ($g =$ acceleration due to gravity).

The ratio of the period of oscillation of a conical pendulum to that of a simple pendulum is: (Assume the strings are of the same length in both cases and $\theta$ is the angle made by the string with the vertical in the case of the conical pendulum.)

If a body dropped freely from a height of $20 \,m$ reaches the surface of a planet with a velocity of $31.4 \,ms^{-1}$, then the length of a simple pendulum that ticks seconds on the planet is (in $\,m$)

The bob of a simple pendulum performs $SHM$ with period $T$ in air and with period $T_1$ in water. The relation between $T$ and $T_1$ is (neglect friction due to water,density of the material of the bob is $\frac{9}{8} \times 10^3 \ kg/m^3$,density of water is $1 \ g/cc$):