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

$A$ lift is descending with an acceleration of $g/3$. What will be the time period of a simple pendulum suspended from its ceiling if its time period in a stationary lift is $T$?

If the length of a simple pendulum is increased by $44\%$,then what is the percentage change in the time period of the pendulum (in $\%$)?

$A$ 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 $s$)

In an experiment to determine the period of a simple pendulum of length $1\, m$,it is attached to different spherical bobs of radii $r_1$ and $r_2$. The two spherical bobs have uniform mass distribution. If the relative difference in the periods is found to be $5 \times 10^{-4}$,the difference in radii,$|r_1 - r_2|$,is best given by .... $cm$.