Question

The density $\rho$ of a liquid varies with depth $h$ from the free surface as $\rho =kh$. A small body of density ${\rho }_1$ is released from the surface of liquid. The body will

• A. Come to a momentary rest at a depth $\left(2{\rho }_1/k\right)$ from the free surface.
• B. Executes simple harmonic motion about a point at a depth $\left(\rho 1/k\right)$ from the surface.
• C. Executes simple harmonic motion of amplitude ${\rho }_1/k$.
• D. Does not execute simple harmonic motion.

Answer 1

Answer: (A), (B), (C)

The correct options are
A Come to a momentary rest at a depth $\left(2{\rho }_1/k\right)$ from the free surface.
B Executes simple harmonic motion about a point at a depth $\left(\rho 1/k\right)$ from the surface.
C Executes simple harmonic motion of amplitude ${\rho }_1/k$.


FBD for ${\rho }_1$ body $=$

At equilibrium;
$mg=F_T$ $(\because mg={\rho }_{1}Vg,F_T=(Kh)Vg$
${\rho }_{1}Vg=\left(Kh\right)Vg$
$h=\frac{{\rho }_1}{K}$
$\therefore$ At $h$, no force will act on the body.
$\Rightarrow F_{net}=0$
$\Rightarrow$ Mean position of SHM
Energy relation:
Loss in P.E. = work done by up thrust force.
${\rho }_{1}Vg{h}^{\prime }=\underset{0}{\overset{h^{\prime }}{\int }}{F}_T.dx$
${\rho }_1{h}^{\prime }Vg=\underset{0}{\overset{h^{\prime }}{\int }}(Kh.Vg)dh$
${\rho }_1{h}^{\prime }Vg=KVg\frac{(h^{\prime }{)}^2}{2}$
$\Rightarrow {\rho }_1{h}^{\prime }=\frac{K}{2}(h^{\prime }{)}^2$
$\Rightarrow h^{\prime }=\frac{2{\rho }_1}{K}$
So, the body cannot go beyond this depth $h^{\prime }$.
i.e. zero velocity point :
Amplitude of SHM $=$Extreme Position$-$Mean position
$\frac{2{\rho }_1}{K}=\frac{{\rho }_1}{K}=\frac{{\rho }_1}{K}$
$\therefore$ A,B,C.