$F_B=m g \Rightarrow 3 \rho A y\,g=\rho A\left(60 \times 10^{-2}\right)\,g$
$y=20\,cm$
$\Rightarrow \text { Maximum amplitude }=20\,cm$
Restoring force when it is slightly depressed by an amount of $x$.
$F =-(\Delta V \sigma g )=-( A \sigma g ) x$
$T =2 \pi \sqrt{\frac{ m }{ A \sigma g }}=2 \pi \sqrt{\frac{\rho A h}{3 A \rho g}}=2 \pi \sqrt{\frac{ h }{3 g}}$
$=2 \pi \sqrt{\frac{60 \times 10^{-2}}{3 \times 9.8}}=\frac{2 \pi}{7}$
$1.$ If the total energy of the particle is $E$, it will perform periodic motion only if
$(A)$ $E$ $<0$ $(B)$ $E$ $>0$ $(C)$ $\mathrm{V}_0 > \mathrm{E}>0$ $(D)$ $E > V_0$
$2.$ For periodic motion of small amplitude $\mathrm{A}$, the time period $\mathrm{T}$ of this particle is proportional to
$(A)$ $\mathrm{A} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(B)$ $\frac{1}{\mathrm{~A}} \sqrt{\frac{\mathrm{m}}{\alpha}}$ $(C)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$ $(D)$ $\mathrm{A} \sqrt{\frac{\alpha}{\mathrm{m}}}$
$3.$ The acceleration of this particle for $|\mathrm{x}|>\mathrm{X}_0$ is
$(A)$ proportional to $\mathrm{V}_0$
$(B)$ proportional to $\frac{\mathrm{V}_0}{\mathrm{mX}_0}$
$(C)$ proportional to $\sqrt{\frac{\mathrm{V}_0}{\mathrm{mX}_0}}$
$(D)$ zero
Give the answer qustion $1,2$ and $3.$

