AP Physics B - Multiple Choice Questions on Fluid Mechanics

Essential things you need to remember in fluid mechanics were discussed in the post dated 3^rd December 2007. A few multiple choice questions (with solution) in this section were discussed in the posts dated 4^th and 5^th December 2007. You can access all posts related to fluid mechanics on this site by clicking on the label ‘fluid mechanics’ below this post.

Today we will discuss a few more multiple choice questions on fluid mechanics:

(1) A water storage tank has a square hole of side 2 cm at its bottom. A plumber, unaware of the hole, admits water into the tank at a constant rate of √2 litre per second. Up to what height can water remain in the tank? (g = 10 ms^–2)

(a) 12.25 m

(b) 8.25 m

(c) 2.45 m

(d) 1.25 m

(e) 0.625 m

If h is the height up to which water can remain in the tank, the velocity v of water flowing out through the hole in the steady state is given by

v = √(2gh)

The volume of water flowing out in the steady state is av where a is the area of the hole.

This must be equal to the volume of incoming water which is given as √2 litre = √2 ×10^–3

m³. Thus we have

a√(2gh) = √2 ×10^–3

Here a = 4 cm² = 4×10^–4 m².

Substituting, 4×10^–4×√(2×10×h) = √2 ×10^–3

Or, √(10h) = 2.5 from which h = 6.25/10 = 0.625 m.

(2) A piece of granite floats at the interface of mercury and water contained in a beaker (Fig.). If the densities of granite, water and mercury are ρ, ρ₁ and ρ₂ respectively, the ratio of the volume of granite in water to the volume in mercury is

(a) (ρ₂ – ρ) /(ρ – ρ₁)

(b) (ρ₂ + ρ) /(ρ+ ρ₁)

(c) ρ₁ ρ₂ /ρ

(d) ρ₁ /ρ₂

(e) ρ₂ /ρ₁

The weight of a floating body is equal to the weight of the displaced fluid. If V and v represent the total volume of the piece of granite and volume of granite in water respectively, we have

V ρg = v ρ₁g + (V – v) ρ₂g

Or, v(ρ₁ – ρ₂) = V(ρ – ρ₂)

Therefore, v/V = (ρ – ρ₂) /(ρ₁ – ρ₂)

The ratio required in the question is v/ (V–v) and is given by

v/ (V–v) = (ρ – ρ₂) /[(ρ₁ – ρ₂) –(ρ – ρ₂)]

Or, v/ (V–v) = (ρ – ρ₂) /(ρ₁ – ρ) = (ρ₂ – ρ) /(ρ – ρ₁)

(3) The pressure of water at the bottom of a water tank exceeds the atmospheric pressure by 10⁴ pascal. The velocity of efflux of water through an orifice at the bottom of the water tank will be (g = 10 ms^–2)

(a) √5 ms^–1

(b) √10 ms^–1

(c) √15 ms^–1

(d) √20 ms^–1

(e) √30 ms^–1

If h is the height of water column in the tank, we have

hρg = 10⁴

Since the density (ρ) of water is 1000 kg m^–3,

h×1000×10 = 10⁴ from which h = 1m.

The velocity of efflux (Torricelli’s theorem) is √(2gh) = √(2×10×1) = √20 ms^–1

(4) In a wind tunnel the flow speeds (of air) on the upper and lower surfaces of the wing of a model airplane are v₁ and v₂ respectively (v₁> v₂). If the wing area is A and the density of air is ρ, the lift on the wing is

(a) ρ (v₁² – v₂²)A

(b) ½ ρ (v₁ – v₂)A

(c) ½ ρ (v₁² – v₂²)A

(d) ½ (v₁² – v₂²)A /ρ

(e) ρ (v₁ – v₂)A

If P₁ and P₂ are the pressures on the upper and lower surfaces of the wing we have, by Bernoulli’s theorem,

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

[We have ignored the small height difference between the top and bottom of the wing so that the gravitational potential energy is treated as constant].

The pressure at the top is less than the pressure at the bottom since v₁> v₂.

The lift on the wing is (P₂ – P₁)A = ½ ρ (v₁² – v₂²)A

You will find similar useful multiple choice questions with solution here as well as here at Physicsplus.

Fluid Mechanics- Questions on Bernoulli’s Theorem

As promised in the post dated 4^th December 2007, I give below more questions from Fluid Mechanics. Consider the following question:

In the figure, a tall cylindrical empty tank with a small side hole of square shape near its bottom is shown. The area of the hole is A. Water flowing with a velocity ‘v’ through a pipe of circular cross section having the same area A (as that of the hole) starts falling vertically down in to the tank. Which one of the following statements is correct?

(a) No water can remain in the tank

(b) The height of water column in the tank will go on increasing

(c) The height of water column in the tank will increase initially and will remain steady at v/2g

(d) The height of water column in the tank will increase initially and will remain steady at √(v/2g)

(e) The height of water column in the tank will increase initially and will remain steady at v²/2g.

The velocity with which water will flow out through the side hole in the tank will be small initially and hence the height of water column will go on increasing until the volume of water flowing into the tank becomes exactly equal to the volume flowing out through the side hole. The height of water column remains fixed at the value ‘h’ for attaining this condition. Since the area of the hole is equal to the area of cross section of the inlet pipe, the steady height condition is attained when the velocity of efflux (of water through the hole), as given by Torricelli’s theorem, is equal to the velocity of water through the inlet pipe.

Therefore we have, v = √(2gh) from which h = v²/2g [ Option (e)].

Now, consider the following MCQ:

Water is flowing steadily out through the end of a vertical pipe (fig) with a velocity of 4 ms^–1. At what distance from the end of the pipe will the area of cross section of the stream of water be (2/3)A where A is the area of cross section of the pipe?

(a) 0.5 m (b) 1 m (c) 1.5 m (d) 2.2 m (e) 2.5 m

From the equation of continuity, we have A₁v₁ = A₂v₂.

Here A₁ = A, A₂ = 2A/3 and v₁ = 4 ms^–1 so that v₂ = 6 ms^–1. This is the velocity of the stream at distance ‘x’ (let us say) below the end of the pipe, where the area of cross section of the stream reduces to (2/3)A

Since the stream is open to the atmosphere, the pressures are the same at all regions of the stream. The kinetic energy of the water particles will be increased because of an equal decrease in the gravitational potential energy.

Considering a mass m of water, we can write ½ mv₂² = ½ mv₁² + mgx. [Note that we have taken the reference (zero) level for the gravitational potential energy at distance x below the end of the pipe].

This gives

v₂ = √(v₁² + 2gx). Or, 6 = √(4² + 2gx).

Taking the acceleration due to gravity to be 10 ms^–2, x = 1m.

[If you blindly write Bernoulli’s equation, you will have P + ρgx + (½)ρv₁ ² = P + (ρg×0) + (½)ρv₂² from which you will get v₂ = √(v₁² + 2gx). You will then proceed to find ‘x’].

Fluid Mechanics – Equations to be remembered

“There is no substitute for hard work.”

– Thomas A. Edison

You won’t be allowed to use equation tables and calculators to work out multiple choice questions (MCQ). (These are allowed in the case of free-response questions). You will have to remember certain basic formulas for solving multiple choice questions within the permitted time. The essential things you must remember in the section, ‘Fluid Mechanics’ are given below:

1. Weight of floating body = weight of displaced liquid

This can be modified as mass of floating body = mass of displaced fluid

2. Force of buoyancy (F_buoy)= weight of displaced fluid

or, F_buoy = Vρg where V is the volume of fluid displaced, ρ is the density of the fluid and g is the acceleration due to gravity.

3. Equation of continuity in fluid dynamics, as applied to liquids which are supposed to be incompressible, is A₁v₁= A₂v₂ where A₁ and A₂ are the cross section areas of the tube and v₁ and v₂ are the velocities of the liquid at these sections respectively.

4. Bernoulli’s equation (Bernoulli’s theorem) is

P + ρgh + (½)ρv² = constant where P is the pressure, ρ is the density, ‘h’ is the height (with respect to the reference level for estimating the gravitational potential energy) and ‘v’ is the velocity of the fluid. The symbol ‘y’ also may be used in place of ‘h’.

Note that Bernoulli’s theorem follows from the law of conservation of energy in the case of a small mass ‘m’ of a fluid throughout its flow and can be written as

PV+ mgh + (½ )mv² = constant

Another useful form of Bernoulli’s equation is

P/ρ + gh + (½)v² = constant

In the case of a liquid flowing through a horizontal pipe (constant gravitational potential energy), the pressure of liquid will be smaller at points where the velocity is greater and vice versa.

5.Torricelli’s Theorem:

Velocity of efflux (of liquid flowing out through a hole) =√(2gh)

where ‘h’ is the depth of the hole (fig).

(It is interesting to note that a particle dropped from a height ‘h’ will strike the ground with the above velocity)

The expression for the efflux velocity follows from Bernoulli’s theorem by considering the cases at the free surface of the liquid in the tank (fig) and at the hole:

P + ρgH + 0 = P + ρg(H–h) + (½)ρv² so that v =√(2gh),

where P is the atmospheric pressure. Note that the hole is open to the atmosphere, as is the free liquid surface in the tank.

6. The horizontal range of liquid jet (fig), R= 2√[h(H–h)]

Note that the above expression is obtained by considering the horizontal range (on the ground) of a particle projected horizontally from a height (H–h). Do it as an exercise.

The range R will be maximum if h = H/2. [You may show this by putting dR/dh = 0 when R is maximum]. The maximum range is H.

In the next post we will consider some typical questions in this section. Of course, their solution also will be discussed.