In both figures shown below, there is a hole along the diameter of the Earth. In the first figure, a particle is released from point $A$ and it oscillates with a time period $T_1$. In the second figure, the same particle is released from point $B$ and it oscillates with a time period $T_2$. Then, [$O$ is the centre of the Earth]

$A$ rectangular block of mass $m$ and cross-sectional area $A$ floats on a liquid of density $\rho$. It is given a small vertical displacement from equilibrium,and it starts oscillating with frequency $n$. The frequency $n$ is equal to (where $g$ is the acceleration due to gravity):

Two light identical springs of spring constant $k$ are attached horizontally at the two ends of a uniform horizontal rod $AB$ of length $l$ and mass $m$. The rod is pivoted at its centre $O$ and can rotate freely in a horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in the figure. The rod is gently pushed through a small angle $\theta$ and released. The frequency of the resulting oscillation is

$A$ ball is rolling without slipping in a spherical shallow bowl (radius $R$) as shown in the figure and is executing simple harmonic motion. If the radius of the ball is doubled,then the time period of oscillation

$A$ test tube of mass $6 \ g$ and uniform area of cross-section $10 \ cm^2$ is floating in water vertically when $10 \ g$ of mercury is in the bottom. The tube is depressed by a small amount and then released. The time period of oscillation is (Acceleration due to gravity $= 10 \ m/s^2$) (in $s$)

$A$ particle executes $S.H.M.$ along a straight line. The amplitude of oscillation is $2 \,cm$. When the displacement of the particle from the mean position is $1 \,cm$,the magnitude of its acceleration is equal to the magnitude of its velocity. The time period of oscillation is ........