Mechanical Properties Of Fluids(12th Grade > Physics ) Questions and answers for exam Preparation

Question 1. An air bubble of radius 5mm rises through a vat of syrup at a steady speed of 4mm/s. If the syrup has a density of

1.4 \times 10^{3} k g / m^{3}

, what is its viscosity?

1.15 kg/m-s
15kg/m-s
2.25 kg/m-s
19.05 kg/m-s

Explanation:-

Answer: Option D. -> 19.05 kg/m-s
:
D
In this case the terminal velocity is upward. Since

σ > ρ

, the density of the air
inside the bubble may be neglected completely.
From equation

v = \frac{2 a^{2} (σ - ρ) g}{g η} = \frac{2 a^{2} ρ g}{9 η} \Rightarrow η = \frac{2 a^{2} ρ g}{9 v}

where, a = 5mm,

ρ = 1.4 \times 10^{3} k g / m^{3}, v = 4 m m / s

∴ η = \frac{2 \times 25 \times 10^{- 6} \times 1.4 \times 10^{3} \times 9.8}{9 \times 4 \times 10^{- 3}} k g / m - s = 19.05 k g / m - s

Hence, (D) is correct.

Question 2. Equal masses of water and a liquid of density 2 are mixed together, then the mixture has a density of

2/3
4/3
3/2
3

Explanation:-

Answer: Option B. -> 4/3
:
B
If two liquids of equal masses and different densities are mixed together then density of mixture

ρ = \frac{2 ρ_{1} ρ_{2}}{ρ_{1} + ρ_{2}} = \frac{2 \times 1 \times 2}{1 + 2} = \frac{4}{3}

Question 3. Water enters through end A with speed

v_{1}

and leaves through end B with speed

v_{2}

of a cylindrical tube AB. The tube is always completely filled with water. In case I tube is horizontal and in case II it is vertical with end A upwards and in case III it is vertical with end B upwards. We have

v_{1}

=

v_{2}

for

Case I
Case II
Case III
Each case

Explanation:-

Answer: Option D. -> Each case
:
D
This happens in accordance with equation of continuity and this equation was derived on the principle of conservation of mass and it is true in every case, either tube remain horizontal or vertical.

Question 4. A square plate of 0.1 m side moves parallel to a second plate with a velocity of 0.1 m/s, both plates being immersed in water. If the viscous force is 0.002 N and the coefficient of viscosity is 0.01 poise, distance between the plates in m is

0.1
0.05
0.005
0.0005

Explanation:-

Answer: Option D. -> 0.0005
:
D

A = (0.1)^{2} = 0.01 m^{2}

,

η = 0.01

Poise = 0.001 decapoise (M.K.S. unit),

d v = 0.1 m / s

and F = 0.002 N

F = η A \frac{d v}{d x}

∴ d x = \frac{η A d v}{F} = \frac{0.001 \times 0.01 \times 0.1}{0.002} = 0.0005 m

Question 5. A large tank filled with water to a height 'h' is to be emptied through a small hole at the bottom. The ratio of time taken for the level of water to fall from h to h/2 and from h/2 to zero is

√2
1√2
√2−1
1√2−1

Explanation:-

Answer: Option C. -> √2−1
:
C
Time taken for the level to fall from H to H'

t = \frac{A}{A_{0}} \sqrt{\frac{2}{g}} [\sqrt{H} - \sqrt{H^{'}}]

According to problem - the time taken for the level to fall from h to

\frac{h}{2}

t_{1} = \frac{A}{A_{0}} \sqrt{\frac{2}{g}} [\sqrt{h} - \sqrt{\frac{h}{2}}]

and similarly time taken for the level to fall from

\frac{h}{2}

to zero

t_{2} = \frac{A}{A_{0}} \sqrt{\frac{2}{g}} [\sqrt{\frac{h}{2}} - 0] ∴ \frac{t_{1}}{t_{2}} = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - 0} = \sqrt{2} - 1

Question 6. Water flows at a speed of 6 cm s^-1 through a tube of radius 1 cm. Coefficient of viscosity of water at room temperature is 0.01 poise. Calculate the Reynolds number. Is it a steady flow?

60, yes
60, no
120, yes
120, no

Explanation:-

Answer: Option C. -> 120, yes
:
C
You know,
Reynolds number,

N = \frac{ρ v D}{η}

Where,

ρ

= density of fluid
v = velocity of flow of the fluid
D = diameter of the tube

η

= coefficient of viscosity of the fluid

\Rightarrow N = \frac{1000 \times \frac{6}{100} \times \frac{2}{100}}{0.01}

{all values in SI}
N = 120 {Dimensionless}, yes the flow is steady

Question 7. A soap bubble assumes a spherical surface. Which of the following statement is wrong

The soap film consists of two surface layers of molecules back to back
The bubble encloses air inside it
The pressure of air inside the bubble is less than the atmospheric pressure; that is why the atmospheric pressure has compressed it equally from all sides to give it a spherical shape
Because of the elastic property of the film, it will tend

Explanation:-

Answer: Option C. -> The pressure of air inside the bubble is less than the atmospheric pressure; that is why the atmospheric pressure has compressed it equally from all sides to give it a spherical shape
:
C
The pressure of air inside the bubble is less than the atmospheric pressure; that is why the atmospheric pressure has compressed it equally from all sides to give it a spherical shape

Question 8. A manometer connected to a closed tap reads 3.5

\times

10^{5}

N/

m^{2}

. When the valve is opened, the reading of manometer falls to 3.0

\times

10^{5}

N/

m^{2}

, then velocity of flow of water is

100 m/s
10 m/s
1 m/s
10 .10m/s

Explanation:-

Answer: Option B. -> 10 m/s
:
B
Bernoulli's theorem for unit mass of liquid

\frac{P}{ρ} + \frac{1}{2} v^{2}

= constant
As the liquid starts flowing, it pressure energy decreases

\frac{1}{2} v^{2} = \frac{P_{1} - p_{2}}{ρ} \Rightarrow \frac{1}{2} v^{2} = \frac{3.5 \times 10^{5} - 3 \times 10^{5}}{10^{3}}

\Rightarrow v^{2} = \frac{2 \times 0.5 \times 10^{5}}{10^{3}} \Rightarrow v^{2} = 100 \Rightarrow v = 10 m / s

Question 9. Water is moving with a speed of 5.18 ms

^{-} 1

through a pipe with a cross-sectional area of 4.20

c m^{2}

. The water gradually descends 9.66 m as the pipe increase in area to 7.60

c m^{2}

. The speed of flow at the lower level is

3.0 ms−1
5.7 ms−1
3.82 ms−1
2.86 ms−1

Explanation:-

Answer: Option D. -> 2.86 ms−1
:
D

a_{1} v_{1} = a_{2} v_{2}

\Rightarrow 4.20 \times 5.18 = 7.60 \times v_{2} \Rightarrow v_{2} = 2.86 m / s

Question 10. A cylindrical tank has a hole of 1

c m^{2}

in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of 70

c m^{3}

/sec. then the maximum height up to which water can rise in the tank is

2.5 cm
5 cm
10 cm
0.25 cm

Explanation:-

Answer: Option A. -> 2.5 cm
:
A
The height of water in the tank becomes maximum when the volume of water flowing into the tank per second becomes equal to the volume flowing out per second. Volume of water flowing out per second

= A v = A \sqrt{2 g h}

......(i)
Volume of water flowing in per second
=70

c m^{3} / s e c

......(ii)
From (i) and (ii) we get

A \sqrt{2 g h} = 70 \Rightarrow 1 \times \sqrt{2 g h} = 70

\Rightarrow 1 \times \sqrt{2 \times 980 \times h} = 70

∴ h = \frac{4900}{1960} = 2.5 c m