$A$ hollow sphere is filled with water through a small hole in it. It is then hung by a long thread and made to oscillate. As the water slowly flows out of the hole at the bottom,the period of oscillation will

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:

$A$ simple pendulum with a bob of mass $40 \ g$ and charge $+2 \ \mu C$ makes $20$ oscillations in $44 \ s$. $A$ vertical electric field of magnitude $4.2 \times 10^4 \ NC^{-1}$ pointing downward is applied. The time taken by the pendulum to make $15$ oscillations in the electric field is (acceleration due to gravity $= 10 \ ms^{-2}$) (in $s$)

$A$ plate oscillates with a time period $T$. If another plate is suddenly placed on the first plate,what happens to the time period?

$A$ simple pendulum of length $l$ is made to oscillate with an amplitude of $45^{\circ}$. The acceleration due to gravity is $g$. Let $T_0 = 2 \pi \sqrt{l / g}$. The time period of oscillation of this pendulum will be

$A$ simple pendulum consisting of a mass $M$ attached to a string of length $L$ is released from rest at an angle $\alpha$. $A$ pin is located at a distance $l$ below the pivot point. When the pendulum swings down,the string hits the pin as shown in the figure. The maximum angle $\theta$ which the string makes with the vertical after hitting the pin is: