A small sphere of mass m is dropped from a great height. After it has fallen 100 m, it has attained its terminal velocity and continues to fall at that speed. The work done by air friction against the sphere during the first 100 m of fall is

1. Greater than the work done by air friction in the second 100 m

2. Less than the work done by air friction in the second 100 m

3. Equal to 100 mg

4. Greater than 100 mg

Two drops of the same radius are falling through air with a steady velocity of 5 cm per sec. If the two drops coalesce, the terminal velocity would be 

1. 10 cm per sec                               
2. 2.5 cm per sec
3. $5\times (4{)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$ cm per sec                       
4. $5\times \sqrt{2}$ cm per sec

The rate of steady volume flow of water through a capillary tube of length 'l' and radius 'r' under a pressure difference of P is V. This tube is connected with another tube of the same length but half the radius in series. Then the rate of steady volume flow through them is (The pressure difference across the combination is P)

1. $\frac{\mathrm{V}}{16}$                                     

2. $\frac{\mathrm{V}}{17}$

3. $\frac{16\mathrm{V}}{17}$                                   

4. $\frac{17\mathrm{V}}{16}$

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

Water is flowing through a tube of the non-uniform cross-section. The ratio of the radius at the entry and exit end of the pipe is \(3:2\). Then the ratio of velocities at entry and exit of liquid is:
1. \(4:9\)                         
2. \(9:4\) 
3. \(8:27\)                         
4. \(1:1\)

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

An application of Bernoulli's equation for fluid flow is found in 

1. Dynamic lift of an aeroplane

2. Viscosity meter

3. Capillary rise

4. Hydraulic press

The Working of an atomizer depends upon

1. Bernoulli's theorem                 

2. Boyle's law

3. Archimedes principle               

4. Newton's law of motion