A spherical ball of radius \(r\) is falling in a viscous fluid of viscosity \(\eta\) with a velocity \(v.\) The retarding viscous force acting on the spherical ball is:

1.	inversely proportional to \(r\) but directly proportional to velocity \(v.\)
2.	directly proportional to both radius \(r\) and velocity \(v.\)
3.	inversely proportional to both radius \(r\) and velocity \(v.\)
4.	directly proportional to \(r\) but inversely proportional to \(v.\)

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. \(x = 4 \sqrt{D \left(H-D\right)}\)