The time period of a simple pendulum inside a stationary lift is $\sqrt{3} \ s$. When the lift moves upwards with an acceleration $g/3$,the new time period will be ($g=$ acceleration due to gravity).

An oscillating pendulum suspended from the roof of a lift which is at rest has time period $T_1$. When the lift moves up with acceleration $a$, its time period is $T_2$. When the lift moves down with acceleration $a$, its time period is $T_3$. The relation between $T_1$, $T_2$, and $T_3$ is:

$A$ 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 $\text{ s}$)

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 $\,m$)

Which of the following plots represents schematically the dependence of the time period of a pendulum,if measured and plotted as a function of the amplitude of its oscillations? (Note: amplitude need not be small)

If the length of a clock pendulum increases by $0.2 \%$ due to atmospheric temperature rise,then the loss in time of the clock per day is ........... $s$.