There are two identical small holes of area of cross-section a on opposite sides of a tank containin

English

Gujarati

Hindi

9-1.Fluid Mechanics

normal

There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density $\rho $. The difference in height between the holes is $h$. Tank is resting on a smooth horizontal surface. Horizontal force which will have to be applied on the tank to keep it in equilibrium is

$g\,h\,\rho a$

$\frac{{2gh}}{{\rho a}}$

$2g\,h\,\rho a$

$\frac{{\rho gh}}{a}$

Solution

$F=u \frac{\mathrm{d} m}{\mathrm{dt}}=v \frac{\mathrm{d}}{\mathrm{dt}} \mathrm{a} \rho \ell=\operatorname{apv} \frac{\mathrm{d} \ell}{\mathrm{dt}}=\mathrm{a} \rho \mathrm{v}^{2}$

$F=2 \mathrm{a} \rho \mathrm{gh}$

Standard 11

Physics

A spherical solid ball of volume $V$ is made of a material of density $\rho _1$ . It is falling through a liquid of density $\rho _2(\rho _2 < \rho _1)$ . Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed $v$ , i.e., $F_{viscous} =\, -kv^2 (k > 0)$ . Then terminal speed of the ball is

normal

The work done in splitting a drop of water of $1\, mm$ radius into $10^6$ droplets is (surface tension of water $72\times10^{-3}\, N/m$) :

normal

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6\, cm -s^{-1}$ . If they coalesce to form one big drop, what will be the terminal speed of the bigger drop ? (Neglect the buoyancy of the air) ……. $cm -s^{-1}$

normal

The velocity of small ball of mass $m$ and density $\rho $ when dropped in a container filled with glycerine of density $\sigma $ becomes constant after sometime. The viscous force acting on the ball in the final stage is

normal

A wooden block, with a coin placed on its top, floats in water as shown in fig. the distance $l$ and $h$ are shown there. After some time the coin falls into the water. Then

normal

Solution

Similar Questions

The work done in splitting a drop of water of $1\, mm$ radius into $10^6$ droplets is (surface tension of water $72\times10^{-3}\, N/m$) :

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of $6\, cm -s^{-1}$ . If they coalesce to form one big drop, what will be the terminal speed of the bigger drop ? (Neglect the buoyancy of the air) ……. $cm -s^{-1}$

The velocity of small ball of mass $m$ and density $\rho $ when dropped in a container filled with glycerine of density $\sigma $ becomes constant after sometime. The viscous force acting on the ball in the final stage is

A wooden block, with a coin placed on its top, floats in water as shown in fig. the distance $l$ and $h$ are shown there. After some time the coin falls into the water. Then

Start a Free Trial Now