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 \(\mathrm H.\) Water is allowed to come out of a hole \(\mathrm P\) in one of the walls at a depth \(\mathrm D\) below the surface of water. Express the horizontal distance \(\mathrm{x}\) in terms of \(\mathrm H\) and \(\mathrm {D}\text :\)
                    

1. $\mathrm{x}=\sqrt{\mathrm{D}(\mathrm{H}-\mathrm{D})}$

2. $\mathrm{x}=\sqrt{\frac{\mathrm{D}(\mathrm{H}-\mathrm{D})}{2}}$

3. $\mathrm{x}=2\sqrt{\mathrm{D}(\mathrm{H}-\mathrm{D})}$

4. $\mathrm{x}=4\sqrt{\mathrm{D}(\mathrm{H}-\mathrm{D})}$

A streamlined body falls through air from a height h on the surface of a liquid. If d and D(D > d) represents the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, is

1. $\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}$                         

2. $\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}·\frac{D}{d}}$

3. $\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}·\frac{d}{D}}$                   

4. $\sqrt{\frac{2\mathrm{h}}{\mathrm{g}}}\left(\frac{d}{D-d}\right)$

 As the temperature of water increases, its viscosity

1. Remains unchanged

2. Decreases

3. Increases

4. Increases or decreases depending on the external pressure

If a small drop of water falls from rest through a large height h in air, then the final velocity is: