$A$ liquid of density $\rho$ is coming out of a hose pipe of cross-sectional area $A$ with horizontal speed $v$ and hits a mesh. $50\%$ of the liquid passes through the mesh unaffected. $25\%$ loses all of its momentum and $25\%$ comes back with the same speed. The resultant pressure on the mesh will be

  • A
    $\frac{1}{4} \rho v^2$
  • B
    $\frac{3}{4} \rho v^2$
  • C
    $\frac{1}{2} \rho v^2$
  • D
    $\rho v^2$

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$A$ 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 $\beta=(P_1-P_2) / (\rho g d)$,where $\rho$ is the 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 $a=g / 2$

Vapour is injected at a uniform rate into a closed vessel which was initially evacuated. The pressure in the vessel:

Two cylindrical vessels of equal cross-sectional area of $2 \ m^2$ contain water up to heights of $10 \ m$ and $6 \ m$,respectively. If the vessels are connected at their bottom,then the work done by the force of gravity is (Density of water is $10^3 \ kg/m^3$ and $g = 10 \ m/s^2$):

$A$ vertical cylindrical container of base area $A$ and an upper cross-section area $A_1$ making an angle $30^{\circ}$ with the horizontal is placed in an open rainy field as shown,near another cylindrical container having the same base area $A$. The ratio of the rates of collection of water in the two containers will be:

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$A$ spray gun is shown in the figure where a piston pushes air out of a nozzle. $A$ thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle,the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown,the radii of the piston and the nozzle are $20 \ mm$ and $1 \ mm$ respectively. The upper end of the container is open to the atmosphere.
$1.$ If the piston is pushed at a speed of $5 \ mm \ s^{-1}$,the air comes out of the nozzle with a speed of
$(A)$ $0.1 \ m \ s^{-1}$ $(B)$ $1 \ m \ s^{-1}$ $(C)$ $2 \ m \ s^{-1}$ $(D)$ $8 \ m \ s^{-1}$
$2.$ If the density of air is $\rho_{a}$ and that of the liquid is $\rho_{\ell}$,then for a given piston speed,the rate (volume per unit time) at which the liquid is sprayed will be proportional to
$(A)$ $\sqrt{\frac{\rho_{a}}{\rho_{\ell}}}$ $(B)$ $\sqrt{\rho_{a} \rho_{\ell}}$ $(C)$ $\sqrt{\frac{\rho_{\ell}}{\rho_{a}}}$ $(D)$ $\rho_{\ell}$

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