The time period of a simple pendulum is $2 \, sec$. If its length is increased $4$ times,then its period becomes ..... $\sec$.

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:

The period of a simple pendulum is measured as $T$ in a stationary lift. If the lift moves upwards with an acceleration of $5g$,the new period will be:

$A$ simple pendulum of length $L$ swings in a vertical plane. The tension in the string when it makes an angle $\theta$ with the vertical and the bob of mass $m$ moves with a speed $v$ is (where $g$ is the gravitational acceleration):

$A$ simple pendulum is suspended from the ceiling of a lift. When the lift is at rest,its period is $T$. With what acceleration $a$ should the lift be accelerated upward in order to reduce the period to $\frac{T}{2}$? (Take $g$ as the acceleration due to gravity.)

The time period of a simple pendulum of length $L$ is $T_1$. The time period of a uniform rod of the same length $L$ suspended from one end and oscillating in a vertical plane is $T_2$. The amplitude of oscillation is small in both cases. Then the ratio $\frac{T_1}{T_2}$ is: