$A$ uniform thin ring of radius $R$ and mass $m$ is suspended in a vertical plane from a point on its circumference. Its time period of oscillation is ........

$A$ 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 $s$)

Under the influence of force $F_1$,a body oscillates with a period $T_1$,and due to another force $F_2$,the body oscillates with a period $T_2$. If both forces act simultaneously,what is the resultant period? (Consider the displacement is the same in all three cases.)

$A$ particle of mass $m$ undergoes oscillations about $x=0$ in a potential given by $V(x) = \frac{1}{2} k x^2 - V_0 \cos \left(\frac{x}{a}\right)$,where $V_0, k, a$ are constants. If the amplitude of oscillation is much smaller than $a$,the time period is given by

$A$ light hollow cube of side length $10 \ cm$ and mass $10 \ g$ is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y \pi \times 10^{-2} \ s$,where the value of $y$ is (Acceleration due to gravity,$g = 10 \ m/s^2$,density of water $\rho = 10^3 \ kg/m^3$)

If a body is released into a tunnel dug across the diameter of the Earth,it executes simple harmonic motion with a time period of: