IIT-JEE 2008 Linked Comprehension Type Multiple Choice Questions Inviolving Thermal Physics and Properties of Fluids

Today we will discuss three linked comprehension type multiple choice questions which appeared in IIT-JEE 2008 question paper:

Paragraph for Question Nos. 1 to 3

A small spherical monoatomic ideal gas bubble (γ = 5/3) is trapped inside a liquid of density ρ_ℓ (see figure). Assume that the bubble does not exchange any heat with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is T₀, the height of the liquid is H and the atmospheric pressure is P₀ (Neglect surface tension).

1. As the bubble moves upwards, besides the buoyancy force the following forces are acting on it:

(A) Only the force of gravity

(B) The force due to gravity and the force due to the pressure of the liquid

(C) The force due to gravity, the force due to the pressure of the liquid and the force due to viscosity of the liquid

(D) The force due to gravity and the force due to viscosity of the liquid

The force due to the pressure of the liquid is the buoyancy force. Therefore, besides the buoyancy force the forces acting on the bubble are the force due to gravity and the force due to viscosity of the liquid [Option (D)].

2. When the gas bubble is at height y from the bottom, its temperature is

(A) T₀[(P₀ + ρ_ℓ gH) /(P₀ + ρ_ℓ gy)]^2/5

(B) T₀[{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ gH)]^2/5

(C) T₀[(P₀ + ρ_ℓ gH) /(P₀ + ρ_ℓ gy)]^3/5

(D) T₀[{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ gH)]^3/5

Since there is no heat exchange, the process is adiabatic for which

T P^{[(1- γ)/ γ]} = constant where γ is the ratio of specific heats.

[Many of you might be remembering this as T ^γ P^{(1- γ)} = constant].

At the bottom pressure = P₀ + ρ_ℓ gH and temperature = T₀

At height y from the bottom pressure = P₀ + ρ_ℓ g(H –y) and temperature = T (let us say)

Therefore, T₀(P₀ + ρ_ℓ gH)^–2/5 = T[P₀ + ρ_ℓ g(H –y)]^–2/5 so that

T = T₀[(P₀ + ρ_ℓ gH) /{P₀ + ρ_ℓ g(H –y)}]^–2/5

Or, T = T₀ [{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ g(H )]^2/5

So option (B) is correct.

3. The buoyancy force on the gas bubble is (Assume R is the universal gas constant)

(A) ρ_ℓ nRgT₀ [(P₀ + ρ_ℓ gH) ^2/5/(P₀ + ρ_ℓ gy) ^7/5]

(B) ρ_ℓ nRgT₀ /[(P₀ + ρ_ℓ gH) ^2/5{P₀ + ρ_ℓ g(H –y)}^3/5]

(C) ρ_ℓ nRgT₀ [(P₀ + ρ_ℓ gH) ^3/5/(P₀ + ρ_ℓ gy) ^8/5]

(D) ρ_ℓ nRgT₀ /[(P₀ + ρ_ℓ gH) ^3/5{P₀ + ρ_ℓ g(H –y)}^2/5]

Force of buoyancy, F= Vρ_ℓ g

But V = nRT/P so that F = nRTρ_ℓ g /P

Here P = P₀ + ρ_ℓ g(H –y) and T = T₀ [{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ g(H )]^2/5

Substituting, F = nRρ_ℓ g T₀ [{P₀ + ρ_ℓ g(H –y)}/(P₀ + ρ_ℓ g(H )]^2/5/ [P₀ + ρ_ℓ g(H –y)]

Or, F = ρ_ℓ nRgT₀ /[(P₀ + ρ_ℓ gH) ^2/5{P₀ + ρ_ℓ g(H –y)}^3/5]

So option (B) is correct.

Two Questions on Force of Buoyancy

How does a balloon filled with helium rise in air? The weight of the balloon with helium is less than the force of buoyancy on the balloon and hence the balloon rises. Here is a question high lighting this principle:

A balloon of mass 6 kg is to be filled with helium so as to lift an instrument weighing 24 kg. The minimum volume of helium to be filled in the balloon is nearly (Density of air = 1.29 kg m^–3, density of helium = 0.18 kg m^–3 approximately)

(a) ) 22 m³ (b) 27 m³ (c) 30 m³ (d) ) 133 m³ (e) 166 m³

The balloon will start rising when the force of buoyancy on the balloon just exceeds the weight of the balloon with helium and the instrument. The force of buoyancy is equal to the weight of displaced air. Therefore, the minimum volume V of helium to be filled is given by

(6 + 0.18 V + 24)g = 1.29 Vg.

This gives V = 27 m³ approximately.

[The weight of air displaced by the instrument and the empty balloon is negligible and therefore not considered here].

Now consider the following MCQ:

A block of cork floats on the surface of water (in a container) with 30 cm outside water. If the system is placed on the moon where there is no atmosphere and the value of acceleration due to gravity is approximately one sixths that on the earth, the portion outside water will be

(a) 5 cm

(b) slightly less than 5 cm

(c) slightly less than 30 cm

(d) slightly greater than 30 cm

(e) slightly greater than 5 cm

If the acceleration due to gravity has a non-zero value, the exposed portion will remain unchanged since the weight of the floating body and the weight of displaced fluid are directly proportional to the value of the acceleration due to gravity. So, the exposed portion (or, the immersed portion) will remain the same whether you take the system to the moon or mars or inside a coal mine, when you consider the effect of ‘g’ only. But, you have to consider the effect of the atmosphere also. Since there is no atmosphere on the moon, there is no force of buoyancy due to the atmosphere and the net force of buoyancy (which is due to water in the container and due to the air above) is slightly reduced. Therefore, the block of cork gets immersed a little more and the exposed portion is slightly less than 30 cm.

[Note that if ‘g’ is zero, there is no weight and therefore no question of floatation].

You will find similar questions (with solution) at AP Physics Resources: Fluid Mechanics- Questions on Force of Buoyancy