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.