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12th Grade > Physics

MECHANICAL PROPERTIES OF FLUIDS MCQs

Total Questions : 29 | Page 1 of 3 pages
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 1.4×103kg/m3, what is its viscosity?
  1.    1.15 kg/m-s
  2.    15kg/m-s
  3.    2.25 kg/m-s
  4.    19.05 kg/m-s 
 Discuss Question
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=2a2(σρ)ggη=2a2ρg9ηη=2a2ρg9v
where, a = 5mm,ρ=1.4×103kg/m3,v=4mm/s
η=2×25×106×1.4×103×9.89×4×103kg/ms=19.05kg/ms
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 
  1.    2/3
  2.    4/3
  3.    3/2
  4.    3
 Discuss Question
Answer: Option B. -> 4/3
:
B
If two liquids of equal masses and different densities are mixed together then density of mixture
ρ=2ρ1ρ2ρ1+ρ2=2×1×21+2=43
Question 3. Water enters through end A with speed v1 and leaves through end B with speed v2 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 v1 = v2 for
  1.    Case I
  2.    Case II
  3.    Case III
  4.    Each case
 Discuss Question
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
 
  1.    0.1
  2.    0.05
  3.    0.005
  4.    0.0005
 Discuss Question
Answer: Option D. -> 0.0005
:
D
A=(0.1)2=0.01m2, η=0.01 Poise = 0.001 decapoise (M.K.S. unit),
dv=0.1m/s and F = 0.002 N
F=ηAdvdx dx=ηAdvF=0.001×0.01×0.10.002=0.0005m
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
 
  1.    √2
  2.    1√2
  3.    √2−1
  4.    1√2−1
 Discuss Question
Answer: Option C. -> √2−1
:
C
Time taken for the level to fall from H to H' t=AA02g[HH]
According to problem - the time taken for the level to fall from h to h2 t1=AA02g[hh2]
and similarly time taken for the level to fall from h2 to zero t2=AA02g[h20]t1t2=112120=21
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?
  1.    60, yes
  2.    60, no
  3.    120, yes
  4.    120, no
 Discuss Question
Answer: Option C. -> 120, yes
:
C
You know,
Reynolds number, N=ρvDη
Where, ρ = density of fluid
v = velocity of flow of the fluid
D = diameter of the tube
η = coefficient of viscosity of the fluid
N=1000×6100×21000.01
{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
 
  1.    The soap film consists of two surface layers of molecules back to back
  2.    The bubble encloses air inside it
  3.    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
  4.    Because of the elastic property of the film, it will tend 
 Discuss Question
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 × 105 N/m2. When the valve is opened, the reading of manometer falls to 3.0 × 105 N/m2, then velocity of flow of water is
  1.    100 m/s
  2.    10 m/s
  3.    1 m/s
  4.    10 .10m/s
 Discuss Question
Answer: Option B. -> 10 m/s
:
B
Bernoulli's theorem for unit mass of liquid
Pρ+12v2 = constant
As the liquid starts flowing, it pressure energy decreases 12v2=P1p2ρ12v2=3.5×1053×105103
v2=2×0.5×105103v2=100v=10m/s
Question 9. Water is moving with a speed of 5.18 ms1 through a pipe with a cross-sectional area of 4.20 cm2. The water gradually descends 9.66 m as the pipe increase in area to 7.60 cm2. The speed of flow at the lower level is
  1.    3.0 ms−1
  2.    5.7 ms−1
  3.    3.82 ms−1
  4.    2.86 ms−1
 Discuss Question
Answer: Option D. -> 2.86 ms−1
:
D
a1v1=a2v2
4.20×5.18=7.60×v2v2=2.86m/s
Question 10. A cylindrical tank has a hole of 1 cm2 in its bottom. If the water is allowed to flow into the tank from a tube above it at the rate of 70 cm3/sec. then the maximum height up to which water can rise in the tank is
  1.    2.5 cm
  2.    5 cm
  3.    10 cm
  4.    0.25 cm
 Discuss Question
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
=Av=A2gh ......(i)
Volume of water flowing in per second
=70 cm3/sec ......(ii)
From (i) and (ii) we get
A2gh=701×2gh=70
1×2×980×h=70
h=49001960=2.5cm

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