Question

Match column I with column II

Answer 1

$Case:A$ lets assume volume of cube is $V$.

Forces in Y-direction:

$F_y=\left(\sigma Vg\right)-\left(\rho Vg\right)=\sigma V{a}_x$

$a_y=\frac{(\sigma -\rho )g}{\sigma }$

Forces in x- direction: Pseudo force will be applied i $-vex-direction$ as system is moving with acceleration, $g/2$.

Pseudo force $=-\left(\rho Vg/2\right)$

$F_x=\sigma V{a}_x=\left(\sigma Vg\right)-\left(\rho Vg/2\right)\Rightarrow a_x=\frac{(-\rho +2\sigma )g}{2\sigma }$

Now, $\frac{a_x}{a_y}=\frac{(2\sigma -\rho )}{2(\sigma -\rho )}$

$Case:B$

$F_y=\sigma V{a}_y=\left(\rho Vg\right)-\left(\sigma Vg\right)$

$a_y=\frac{(\rho -\sigma )g}{\sigma }$

$a_x=g$

$\frac{a_x}{a_y}=\frac{\sigma }{(\rho -\sigma )}$

$Case:C$

$F_y=\sigma V{a}_y=\left(\rho Vg\right)-\left(\sigma Vg\right)$

$a_y=\frac{(\rho -\sigma )g}{\sigma }$

As system is rotating with $\omega =\sqrt{\frac{2g}{3x}}$. Hence Pseudo force on tube is $F_{pseudo}=\rho V{\omega }^2\left(3x/2\right)$

$F_x=F_{pseudo}=\sigma V{a}_x=\rho V{\omega }^2\left(3x/2\right)$

$\Rightarrow a_x=\frac{\rho }{\sigma }.g$

Now, $\frac{a_x}{g}=\frac{\rho }{\sigma }$

$Case:D$

When there is air then, $T\sin \theta =M.a$ $T\cos \theta =Mg$

$\Rightarrow \tan \theta =a/g$

When there is liquid of density, $\rho <\sigma$, then

Pseudo force $=\left(\rho Va\right)-\left(\sigma Va\right)$

$T\sin \theta =\left(\rho Va\right)-\left(\sigma Va\right)$ $T\cos \theta =\left(\rho Vg\right)-\left(\sigma Vg\right)$

From above, we have, $\tan {\theta }^{{}^{\prime }}=a/g$

Hence $\frac{\tan \theta }{\tan {\theta }^{{}^{\prime }}}=1:1$