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.

MCQ on Viscosity – Poiseuille’s Flow

Poiseuille’s formula for the volume ‘V’ of a liquid of density ‘ρ’ flowing in time ‘t’ through a capillary tube of length ‘L’ and radius ‘r’ under a pressure difference ‘P’ between the ends of the tube is

V = (πPr⁴t)/ (8Lη) where ‘η’ is the coefficient of viscosity of the liquid. [You should note that this formula holds good only if the flow is slow and steady (stream-lined)].

The rate of flow (volume flowing per second) ‘Q’ is given by

Q = (πPr⁴)/ (8Lη)

Here P = hρg where ‘h’ is the height of the liquid column which produces the pressure difference so that

Q = (π hρg r⁴)/ (8Lη)

You will often find questions based on Poiseuille’s formula in entrance tests for admitting students to various courses. Consider the following question:

A capillary tube of length ‘L’ and radius ‘r’ is joined to another capillary tube of length L/4 and radius r/2 A liquid flows through this series combination. If the pressure difference between the ends of the first tube is P, that between the ends of the second tube is

(a) P (b) 4P (c) 8P (d) 16P (e) P/4

Since the tubes are in series, the rates of flow through the tubes are equal so that from Poiseuille’s equation

πPr⁴/8Lη = πP'(r/2)⁴/[8(L/2)η], where P’ is the pressure difference between the ends of the second tube.

From this P' = 16P/2 = 8P.

Now, consider the following MCQ:

Under a constant pressure head, the volume of a liquid flowing per second through a capillary tube of radius 1 mm and length 16 cm is 4 cm³. If another tube of radius 0.5 mm and length 8 cm is connected in series with it and the same pressure head is applied across the combination, the volume of liquid flowing per second will be ( in cm³)

(a) 5/3 (b) 5/6 (c) 4/3 (d) 4/5 (e) 4/9

When tubes are connected in series, the net rate of flow (Q_net) under a given pressure head is given by the reciprocal relation,

1/Q_net = 1/Q₁ + 1/Q₂ + 1/Q₃ +…..etc. where Q₁, Q₂, Q₃.....etc. are the individual rates of flow when the tubes are connected separately to the same pressure heaed.

In the present case, since there are two tubes only, 1/Q_net = 1/Q₁ + 1/Q₂ so that

Q_net = Q₁Q₂/(Q₁+Q₂). Here Q₁ = 4 cm³.

Since the rate of flow through a tube is given by Q = (πPr⁴)/ (8Lη), we have Q α r⁴/L.

Since the radius of the second tube is half that of the first, the rate of flow is reduced to (½)⁴ = 1/16 of that through the first tube on account of this. Since the length of the second tube is half that of the first, the rate of flow is increased to twice that through the first tube on this account. Therefore, rate of flow through the second tube is given by

Q₂ = Q₁×(2/16) = Q₁/8 = 4/8 cm³ = 0.5 cm³.

The net rate of flow is therefore given by Q_net = (4×0.5)/(4+0.5) = 4/9 cm³ [Option (e)].

Properties of Fluids

In any Medical and Engineering entrance test paper you will find a few questions based on the properties of fluids. Let us discuss some typical questions. The following simple question is taken from the A.F.M.C.2004 test paper:
Application of Bernoulli’s theorem can be seen in
(a) dynamic lift of aeroplane (b) hydraulic press (c) helicopter (d) none of these
The correct option is (a). The shape of the aeroplane wings is such that when it moves forward, the air molecules at the top of the wings have a greater velocity (relative to the wings) compared to the air molecules at the bottom. Therefore, in accordance with Bernoulli’s principle, the pressure at the top of the wings is less than that at the bottom. This results in a dynamic lift of the wings which balances the weight of the plane.
Let us now consider another question involving Bernoulli’s principle:
Water is flowing steadily through two horizontal pipes of radii 3cm and 6cm connected in series. The speed of water in the first pipe is 2m/s and the pressure of water in it is 2×10⁴ pascal. The pressure of water in the second pipe will be nearly
(a) 2×10⁴ pascal (b) 2.2×10⁴ pascal (c) 2.4×10⁴ pascal (d) 2.6×10⁴ pascal (e) 3×10⁴ pascal
The speed v₂ of water in the second pipe is given (from the equation of continuity, a₁v₁ = a₂v₂ ) by v₂ = 2×a₁/a₂ = 2×1/4 = 0.5 m/s. (The ratio of cross section areas is ¼ since the radius of the second pipe is twice that of the first).
As per Bernoulli’s principle, we have P₁ + (ρv₁² )/2 + ρgh = P₂ + (ρv₂²)/2 + ρgh with usual notations. Therefore, P₂ = P₁ + ρ(v₁²- v₂²)/2 =2×10⁴ + 1000(2² – 0.5²)/2 = 2.1875×10⁴ pascal. The correct option therefore is (b).
The following question involving surface tension may appear to be difficult for some of you. See how simple it is:
Two soap bubbles of radii ‘R’ and ‘r’ (R > r) are touching each other. The radius of curvature of the common surface at the region of contact is
(a) (R+r)/2 (b) R – r (c) (R – r)/2 (d) Rr/(R+r) (e) Rr/(R-r)
At the contact region, the radius of curvature (R') of the soap film is governed by the net excess of pressure 4T/r – 4T/R where T is the surface tension. Therefore, we have 4T/r – 4T/R = 4T/R' from which R' = Rr/R-r [option (e)].
Consider now the following M.C.Q.:
Two rain drops are falling through air with the same terminal velocity of 8cm/s. If they coalesce, the terminal velocity of the combined single drop will be (in cm/s)
(a) 8/√2 (b) 16 (c) 8×4^1/3 (d) 8×2^1/3 (e) 8√2
On equating the apparent weight of a rain drop to the viscous force (air resistance) opposing the downward motion of the drop, we have, 4/3πr³(ρ-σ)g = 6πrηv, with usual notations. Therefore, the velocity, v α r²…………(1)
Since the drops have the same terminal velocity, it follows that they have the same radius. The radius (r₁ ) of the single large drop (when the two drops coalesce) is given by
2× 4/3πr³ = 4/3 πr₁³. Therefore, r₁=2^1/3 r.
The velocity of the combined single drop, v₁ α [2^1/3 r]² -------------- (2)

From equations (1)&(2), v₁= 4^1/3v = 4^1/3 × 8, since v=8cm/s. The correct option therefore is (c).