$1.$  The piston is now pulled out slowly and held at a distance $2 \mathrm{~L}$ from the top. The pressure in the cylinder between its top and the piston will then be

$(A)$ $\mathrm{P}_0$ $(B)$ $\frac{\mathrm{P}_0}{2}$  $(C)$ $\frac{P_0}{2}+\frac{M g}{\pi R^2}$  $(D)$ $\frac{\mathrm{P}_0}{2}-\frac{\mathrm{Mg}}{\pi \mathrm{R}^2}$

$2.$  While the piston is at a distance $2 \mathrm{~L}$ from the top, the hole at the top is sealed. The piston is then released, to a position where it can stay in equilibrium. In this condition, the distance of the piston from the top is

$(A)$ $\left(\frac{2 \mathrm{P}_0 \pi \mathrm{R}^2}{\pi \mathrm{R}^2 \mathrm{P}_0+\mathrm{Mg}}\right)(2 \mathrm{~L})$  $(B)$ $\left(\frac{\mathrm{P}_0 \pi R^2-\mathrm{Mg}}{\pi R^2 \mathrm{P}_0}\right)(2 \mathrm{~L})$ 

$(C)$ $\left(\frac{\mathrm{P}_0 \pi \mathrm{R}^2+\mathrm{Mg}}{\pi \mathrm{R}^2 \mathrm{P}_0}\right)(2 \mathrm{~L})$  $(D)$ $\left(\frac{\mathrm{P}_0 \pi \mathrm{R}^2}{\pi \mathrm{R}^2 \mathrm{P}_0-\mathrm{Mg}}\right)(2 \mathrm{~L})$

$3.$  The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in the figure. The density of the water is $\rho$. In equilibrium, the height $\mathrm{H}$ of the water column in the cylinder satisfies

$(A)$ $\rho g\left(\mathrm{~L}_0-\mathrm{H}\right)^2+\mathrm{P}_0\left(\mathrm{~L}_0-\mathrm{H}\right)+\mathrm{L}_0 \mathrm{P}_0=0$

$(B)$ $\rho \mathrm{g}\left(\mathrm{L}_0-\mathrm{H}\right)^2-\mathrm{P}_0\left(\mathrm{~L}_0-\mathrm{H}\right)-\mathrm{L}_0 \mathrm{P}_0=0$

$(C)$ $\rho g\left(\mathrm{~L}_0-\mathrm{H}\right)^2+\mathrm{P}_0\left(\mathrm{~L}_0-\mathrm{H}\right)-\mathrm{L}_0 \mathrm{P}_0=0$

$(D)$ $\rho \mathrm{g}\left(\mathrm{L}_0-\mathrm{H}\right)^2-\mathrm{P}_0\left(\mathrm{~L}_0-\mathrm{H}\right)+\mathrm{L}_0 \mathrm{P}_0=0$

Give the answer question $1,2$ and $3.$

Figure: $Image$

$1.$ As the bubble moves upwards, besides the buoyancy force the following forces are acting on it

$(A)$ Only the force of gravity

$(B)$ The force due to gravity and the force due to the pressure of the liquid

$(C)$ The force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the liquid

$(D)$ The force due to gravity and the force due to viscosity of the liquid

$2.$ When the gas bubble is at a height $\mathrm{y}$ from the bottom, its temperature is

$(A)$ $\mathrm{T}_0\left(\frac{\mathrm{P}_0+\rho_0 \mathrm{gH}}{\mathrm{P}_0+\rho_t \mathrm{gy}}\right)^{2 / 5}$

$(B)$ $T_0\left(\frac{P_0+\rho_t g(H-y)}{P_0+\rho_t g H}\right)^{2 / 5}$

$(C)$ $\mathrm{T}_0\left(\frac{\mathrm{P}_0+\rho_t \mathrm{gH}}{\mathrm{P}_0+\rho_t \mathrm{gy}}\right)^{3 / 5}$

$(D)$ $T_0\left(\frac{P_0+\rho_t g(H-y)}{P_0+\rho_t g H}\right)^{3 / 5}$

$3.$ The buoyancy force acting on the gas bubble is (Assume $R$ is the universal gas constant)

$(A)$ $\rho_t \mathrm{nRgT}_0 \frac{\left(\mathrm{P}_0+\rho_t \mathrm{gH}\right)^{2 / 5}}{\left(\mathrm{P}_0+\rho_t \mathrm{gy}\right)^{7 / 5}}$

$(B)$ $\frac{\rho_{\ell} \mathrm{nRgT}_0}{\left(\mathrm{P}_0+\rho_{\ell} \mathrm{gH}\right)^{2 / 5}\left[\mathrm{P}_0+\rho_{\ell} \mathrm{g}(\mathrm{H}-\mathrm{y})\right]^{3 / 5}}$

$(C)$ $\rho_t \mathrm{nRgT} \frac{\left(\mathrm{P}_0+\rho_t g \mathrm{H}\right)^{3 / 5}}{\left(\mathrm{P}_0+\rho_t g \mathrm{~g}\right)^{8 / 5}}$

$(D)$ $\frac{\rho_{\ell} \mathrm{nRgT}_0}{\left(\mathrm{P}_0+\rho_{\ell} \mathrm{gH}\right)^{3 / 5}\left[\mathrm{P}_0+\rho_t \mathrm{~g}(\mathrm{H}-\mathrm{y})\right]^{2 / 5}}$

Give the answer question $1,2,$ and $3.$

[Given: $\pi=22 / 7, g=10 ms ^{-2}$, density of water $=1 \times 10^3 kg m ^{-3}$, viscosity of water $=1 \times 10^{-3} Pa$-s.]

$(A)$ The work done in pushing the ball to the depth $d$ is $0.077 J$.

$(B)$ If we neglect the viscous force in water, then the speed $v=7 m / s$.

$(C)$ If we neglect the viscous force in water, then the height $H=1.4 m$.

$(D)$ The ratio of the magnitudes of the net force excluding the viscous force to the maximum viscous force in water is $500 / 9$.