There is a circular tube in a vertical plane. Two liquids which do not mix and of densities $d_1$ and $d_2$ are filled in the tube. Each liquid subtends $90^o$ angle at centre. Radius joining their interface makes an angle $\alpha$ with vertical. Ratio $\frac{{{d_1}}}{{{d_2}}}$ is
$\frac{{1 + cos\alpha }}{{1 - cos\alpha }}$
$\;\frac{{1 + tan\alpha }}{{1 - tan\alpha }}$
$\;\frac{{1 + sin\alpha }}{{1 - cos\alpha }}$
$\;\frac{{1 + sin\alpha }}{{1 - sin\alpha }}$
What is the pressure on a swimmer $20 \,m$ below the surface of water ........ $atm$
Water is filled up to a height $h$ in a beaker of radius $R$ as shown in the figure. The density of water is $\rho$, the surface tension of water is $T$ and the atmospheric pressure is $P_0$. Consider a vertical section $A B C D$ of the water column through a diameter of the beaker. The force on water on one side of this section by water on the other side of this section has magnitude
In the arrangement shown both the vessels $A$ and $B$ are identical but amount of water in $B$ is double of that in $A$. The vessels are closed by identical leak proof pistons at the same height. The pistons are connected to the ends of lever arm. There is no friction between the pistons and the container walls. The system is in equilibrium in the situation shown. Now the valve in the horizontal tube connecting both the vessels is opened. In which direction will the water flow through the tube ?
A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving down with a constant acceleration $a$ along a fixed inclined plane with angle $\theta=45^{\circ} . P_1$ and $P_2$ are pressures at points 1 and 2 , respectively, located at the base of the tube. Let $\beta=\left(P_1-P_2\right) /(\rho g d)$, where $\rho$ is density of water, $d$ is the inner diameter of the tube and $g$ is the acceleration due to gravity. Which of the following statement($s$) is(are) correct?
$(A)$ $\beta=0$ when $a= g / \sqrt{2}$
$(B)$ $\beta>0$ when $a= g / \sqrt{2}$
$(C)$ $\beta=\frac{\sqrt{2}-1}{\sqrt{2}}$ when $a= g / 2$
$(D)$ $\beta=\frac{1}{\sqrt{2}}$ when $a= g / 2$
An air bubble of volume $1\,cm ^3$ rises from the bottom of a lake $40\,m$ deep to the surface at a temperature of $12^{\circ}\,C$. The atmospheric pressure is $1 \times 10^5 Pa$, the density of water is $1000\,kg / m ^3$ and $g =10\,m / s ^2$. There is no difference of the temperature of water at the depth of $40\,m$ and on the surface. The volume of air bubble when it reaches the surface will be $..........\,cm^{3}$