If $R$ is the radius of the earth and the acceleration due to gravity on the surface of the earth is $g = \pi^2 \ m/s^2$,then the length of the second's pendulum at a height $h = 2R$ from the surface of the earth will be:

The bob of a pendulum of mass $m$, suspended by an inextensible string of length $L$ as shown in the figure, carries a small charge $q$. An infinite horizontal plane conductor with uniform surface charge density $\sigma$ is placed below it. What will be the time period of the pendulum for small amplitude oscillations?

$A$ small sphere of radius $r$ is placed on a concave surface of radius of curvature $R$ a little away from the center. When the sphere is released, it oscillates. Assuming the oscillation to be simple harmonic motion and $r << R$, then the time period is

$A$ pendulum performs $S.H.M.$ with period $\sqrt{3} \ s$ in a stationary lift. If the lift moves up with acceleration $\frac{g}{3}$,the period of the pendulum is $[g=$ acceleration due to gravity $]$. (in $s$)

$A$ simple pendulum of length $l_1$ has a periodic time of $2.4 \text{ s}$. Another simple pendulum of length $l_2 < l_1$ has a periodic time of $1.8 \text{ s}$. The periodic time of a simple pendulum of length $(l_1 - l_2)$ is nearly: (in $\text{ s}$)

$A$ simple pendulum has a time period $T$ in air. Its time period when it is completely immersed in a liquid of density one-eighth the density of the material of the bob is: