The ratio of the period of oscillation of a conical pendulum to that of a simple pendulum is: (Assume the strings are of the same length in both cases and $\theta$ is the angle made by the string with the vertical in the case of the conical pendulum.)

$A$ simple pendulum of length $l$ has a brass bob attached at its lower end. Its period is $T$. $A$ steel bob of the same size,having density $x$ times that of brass,replaces the brass bob. Its length is then changed such that the period becomes $2T$. What is the new length?

$A$ cylindrical block of wood (density $= 650 \ kg \ m^{-3}$),of base area $30 \ cm^2$ and height $54 \ cm$,floats in a liquid of density $900 \ kg \ m^{-3}$. The block is depressed slightly and then released. The time period of the resulting oscillations of the block would be equal to that of a simple pendulum of length ..... $cm$ (nearly).

Two simple pendulums $A$ and $B$ are made to oscillate simultaneously. It is found that $A$ completes $10$ oscillations in $20 \ s$ and $B$ completes $8$ oscillations in $10 \ s$. The ratio of the lengths of $A$ and $B$ is

$A$ man measures the time period of a pendulum $(T)$ in a stationary lift. If the lift moves upward with an acceleration of $\frac{g}{4}$,then the new time period will be:

$A$ $metre$ scale is suspended vertically from a horizontal axis passing through one end of it. Its time period would be ....... $\sec$