$A$ mass falls from a height $h$ and its time of fall $t$ is recorded in terms of time period $T$ of a simple pendulum. On the surface of the Earth,it is found that $t = 2T$. The entire setup is taken to the surface of another planet whose mass is half that of the Earth and radius is the same. The same experiment is repeated and corresponding times are noted as $t'$ and $T'$.

Let $l_1$ be the length of a simple pendulum. Its length changes to $l_2$ to increase the periodic time by $20 \%$. The ratio $\frac{l_2}{l_1}$ is:

$A$ rubber ball filled with water,having a small hole,is used as the bob of a simple pendulum. The time period of such a pendulum

$A$ simple pendulum with a metallic bob has a time period $T$. The bob is now immersed in a non-viscous liquid and oscillated. If the density of the liquid is $1/4$ that of the metal,the time period of the same pendulum will be ...........

$A$ block of mass $m$ is sliding on a fixed frictionless concave surface of radius $R$. It is released from rest at point $P$ which is at a height of $H \ll R$ from the lowest point $Q$.
$(a)$ What is the potential energy as a function of $\theta$,taking the lowest point $Q$ as the reference level for potential energy?
$(b)$ What is the kinetic energy as a function of $\theta$?
$(c)$ What is the time taken for the particle to reach from point $P$ to the lowest point $Q$?
$(d)$ How much force is exerted by the block on the concave surface at the point $Q$?

For a simple pendulum, the graph of time period $(T)$ versus length $(l)$ is depicted by: