A ball of radius r and density $\mathrm{\rho }$ falls freely under gravity through a distance h before entering water. Velocity of ball does not change even on entering water. If viscosity of water is $\mathrm{\eta }$, the value of h( in CGS units) is given by-

1. $\frac{2}{9}{\mathrm{r}}^2\left(\frac{1-\mathrm{\rho }}{\mathrm{\eta }}\right)\mathrm{g}$

2. $\frac{2}{81}{\mathrm{r}}^2\left(\frac{\mathrm{\rho }-1}{\mathrm{\eta }}\right)\mathrm{g}$

3. $\frac{2}{81}{\mathrm{r}}^4{\left(\frac{\mathrm{\rho }-1}{\mathrm{\eta }}\right)}^2\mathrm{g}$

4. $\frac{2}{9}{\mathrm{r}}^4{\left(\frac{\mathrm{\rho }-1}{\mathrm{\eta }}\right)}^2\mathrm{g}$

A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference P. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is

1. P                             

2. $\frac{3\mathrm{P}}{4}$

3. $\frac{\mathrm{P}}{2}$                           

4. $\frac{\mathrm{P}}{4}$

We have two (narrow) capillary tubes T₁ and T₂. Their lengths are l₁ and l₂ and radii of cross-section are r₁ and r₂ respectively. The rate of flow of water under a pressure difference P through tube T₁ is 8cm³/sec. If l₁ = 2l₂ and r₁ =r₂, what will be the rate of flow when the two tubes are connected in series and pressure difference across the combination is same as before (= P)

1. 4 cm³/sec                           

2. (16/3) cm³/sec

3. (8/17) cm³/sec                   

4. None of these

The Reynolds number of a flow is the ratio of

1. Gravity to viscous force

2. Gravity force to pressure force

3. Inertia forces to viscous force

4. Viscous forces to pressure forces

A liquid flows in a tube from left to right as shown in figure. ${\mathrm{A}}_1$ and ${\mathrm{A}}_2$ are the cross-sections of the portions of the tube as shown. Then the ratio of speeds ${\mathrm{v}}_1/{\mathrm{v}}_2$ will be

1. ${\mathrm{A}}_1/{\mathrm{A}}_2$

2. ${\mathrm{A}}_2/{\mathrm{A}}_1$

3. $\sqrt{{\mathrm{A}}_2}/\sqrt{{\mathrm{A}}_1}$

4. $\sqrt{{\mathrm{A}}_1}/\sqrt{{\mathrm{A}}_2}$

The pans of a physical balance are in equilibrium. If Air is blown under the right-hand pan then the right-hand pan will:

According to Bernoulli's equation

$\frac{\mathrm{P}}{\mathrm{\rho g}}+\mathrm{h}+\frac{1}{2}\frac{{\mathrm{v}}^2}{\mathrm{g}}=\mathrm{constant}$

The terms A, B and C are generally called respectively:

1. Gravitational head, pressure head and velocity head

2. Gravity, gravitational head and velocity head

3. Pressure head, gravitational head and velocity head

4. Gravity, pressure and velocity head

A sniper fires a rifle bullet into a gasoline tank making a hole 53.0 m below the surface of gasoline. The tank was sealed at 3.10 atm. The stored gasoline has a density of 660 ${\mathrm{kgm}}^{-3}$. The velocity with which gasoline begins to shoot out of the hole will be:
 

A tank is filled with water up to a height $H.$ The water is allowed to come out of a hole $P$ in one of the walls at a depth $D$ below the surface of the water. The horizontal distance ${x}$ in terms of $H$ and ${D}$ is:
    
1. $x = \sqrt{D\left(H-D\right)}$
2. $x = \sqrt{\frac{D \left(H - D \right)}{2}}$
3. $x = 2 \sqrt{D \left(H-D\right)}$
4. $x = 4 \sqrt{D \left(H-D\right)}$