A cup of water is placed in a car moving at a constant acceleration $a$ to the left. Inside the water is a small air bubble. The figure correctly shows the shape of the water surface and the direction of motion of the air bubble is-

A 

B 

C 

D 

A stick of square cross-section $(5 \mathrm{cm} \times 5 \mathrm{cm})$ and length $‘4 \mathrm{m}’$ weighs $2.5 \mathrm{kg}$ is in equilibrium as shown in the figure below. Determine its angle of inclination (in degree) in equilibrium when the water surface is $1 \mathrm{m}$ above the hinge. 

A non-viscous liquid of constant density $1000 \mathrm{kg} {\mathrm{m}}^{-3}$ flows in a streamlined motion along a tube of the variable cross-section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross-section of the tube at two points $P$ and $Q$ at heights of $2 \mathrm{m}$ and $5 \mathrm{m}$ are respectively $4 \times {10}^{–3} {\mathrm{m}}^2 \& 8 \times {10}^{–3} {\mathrm{m}}^2.$ The velocity of the liquid at point $P$ is $1 \mathrm{m} {\mathrm{s}}^{-1}$. If the work done per unit volume by the pressure is $\left(1161\right)\mathrm{\alpha } \mathrm{joule} {\mathrm{m}}^{-3}$as the liquid flows from point $P$ to $\mathrm{Q}$.then $\mathrm{\alpha }$ will be $(g=9.8 \mathrm{m} {\mathrm{s}}^{-2})$

Water shoots out of a pipe and nozzle as shown in the figure. The cross-sectional area for the tube at point $A$ is four times that of the nozzle. The pressure of water at point $A$ is $41\times {10}^3 \mathrm{N} {\mathrm{m}}^{-2}$ (gauge). If the height $‘h’$ above the nozzle to which water jet will shoot is$\frac{x}{10}$meter than $x$ is- (Neglect all the losses that occurred in the above process) $[g=10 \mathrm{m} {\mathrm{s}}^{-2 }]$

The vessel shown in the figure has two sections of the area of cross-section $A_1$ and $A_2$. A liquid of density $\mathrm{\rho }$ fills both the sections, up to height $h$ in each. Neglecting atmospheric pressure.