Fluid Mechanics is a study of fluids (liquids, gases and plasmas) and the forces acting on it. The fluid is a substance that flows under the action of an applied force and does not have a shape of its own. Liquids and gases are classified together as fluids because they exhibit the same flow phenomena and have an identical equation of motion. There are two branches of fluid mechanics, namely fluid statics (hydrostatics) and fluid dynamics.
Fluid statics is the study of fluids at rest. The main equation used in fluid statics is the equation of Newtonβs second law for nonaccelerating bodies, i.e. β F =0.
Fluid dynamics is the study of fluids in motion. The important equation used here is the equation of Newtonβs second law for accelerating bodies, i.e.β F =ma. Some of the fields where fluid dynamics is applied are meteorology, oceanography, aeronautical engineering, the study of blood flow and many more.
Download Fluid Mechanics JEE Advanced Previous Year Solved Questions PDF
JEE Main Previous Year Solved Questions on Fluid Mechanics
Q1: A solid sphere of radius R acquires a terminal velocity v1 when falling (due to gravity) through a viscous fluid having a coefficient of viscosity Ξ·. The sphere is broken into 27 identical spheres. If each of these acquires a terminal velocity v2, when falling through the same fluid, the ratio (v1/v2) equals
(a) 9
(b) 1/27
(c) 1/9
(d) 27
Solution
27 x (4/3)Οr3 = (4/3)ΟR3
Or r = R/3
Terminal velocity, v β r3
Therefore, (v1/v2) = (R2/r2)
v1/v2= [R/(R/3)]2= 9
(v1/v2) = 9
Answer: (a) 9
Q2: Spherical balls of radius R are falling in a viscous fluid of viscosity with a velocity v. The retarding viscous force acting on the spherical ball is
(a) directly proportional to R but inversely proportional to v
(b) directly proportional to both radius R and velocity v
(c) inversely proportional to both radius R and velocity v
(d) inversely proportional to R but directly proportional to velocity v
Solution
Retarding viscous force = 6ΟΞ·Rv
obviously option (b) holds goods
Answer: (b) directly proportional to both radius R and velocity v
Q3: A long cylindrical vessel is half-filled with a liquid. When the vessel is rotated about its own vertical axis, the liquid rises up near the wall. If the radius of the vessel is 5 cm and its rotational speed is 2 rotations per second, then the difference in the heights between the centre and the sides, in cm, will be
(a) 0.4
(b) 2.0
(c) 0.1
(d) 1.2
Solution
The linear speed of the liquid at the sides is rΟ. So, the difference in height is given as follows
2gh = Ο2r2
h = Ο2r2/2g
here Ο = 2Οf
Therefore, h = [(2 x 2Ο)2(5 x 10-2)2]/(2×10) = 2cm
Answer: (b) 2.0
Q4: Water is flowing continuously from a tap having an internal diameter 8 Γ 10β3 m. The water velocity as it leaves the tap is 0.4 msβ1. The diameter of the water stream at a distance 2 Γ 10β1 m below the tap is close to
(a) 5.0 Γ 10β3 m
(b) 7.5 Γ 10β3 m
(c) 9.6 Γ 10β3 m
(d) 3.6 Γ 10β3 m
Solution
Here, d1 = 8 Γ 10β3 m
v1 = 0.4 m sβ1, h = 0.2 m
According to equation of motion,
= 2 m sβ1
According to equation of continuity a1v1 = a2v2
According to equation of continuity a1v1 = a2v2
(ΟD12/4) v1 = (ΟD22/4) v2
D22 = (v1/v2)D12
D2 = [β(v1/v2)]D1
= [β(0.4/2)]x 8 Γ 10β3 m
D2 = 3.6 Γ 10β3 m
Answer: (d) 3.6 Γ 10β3 m
Q5: A 20 cm long capillary tube is dipped in water. The water rises up to 8 cm. If the entire arrangement is put in a freely falling elevator the length of the water column in the capillary tube will be
(a) 4 cm
(b) 20 cm
(c) 8 cm
(d) 10 cm
Solution
In a freely falling elevator, g = 0 Water will rise to the full length i.e., 20 cm to tube
Answer: (b) 20 cm
Q6: A spherical solid ball of volume V is made of a material of density Ο1. It is falling through a liquid of density Ο2 (Ο2 < 1). Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed v, i.e., Fviscous = βkv2 (k > 0). The terminal speed of the ball is
(a) Vg(Ο1 – Ο2)
(c) VgΟ1/k
Solution
The forces acting on the solid ball when it is falling through a liquid is βmgβ downwards, thrust by Archimedes principle upwards and the force due to the force of friction also acting upwards. The viscous force rapidly increases with velocity, attaining a maximum when the ball reaches the terminal velocity.
Then the acceleration is zero
mg – VΟ2g – kvt2 = ma where V is volume,
vt is the terminal velocity
When the ball is moving with terminal velocity, a = 0
Therefore VΟ1g β VΟ2g β kvt2 = 0
Answer: (b)
Q7: Water flows into a large tank with a flat bottom at the rate of 10-4 m3s-1. Water is also leaking out of a hole of area 1 cm2 at its button. If the height of the water in the tank remains steady, then this height is
(a) 5 cm
(b) 7 cm
(c) 4 cm
(d) 9 cm
Solution
Since the height of the water column is constant
Water inflow rate (Qin) = Water outflow rate (Qout)
Qin = 10-4 m3s-1
Qout = 10-4 x β(2gh)
10-4 = 10-4 x β20 xh
h = (1/20) m = 5 cm
Answer: (a) 5 cm
Q8: A submarine experiences a pressure of 5.05 x106 Pa at depth of d1 in a sea. When it goes further to a depth of d2, it experiences a pressure of 8.08 x 106 Pa. Then d1 – d2 is approximately (density of water = 103 ms-2 and acceleration due to gravity = 10 ms-2)
(a) 300 m
(b) 400 m
(c) 600 m
(d) 500 m
Solution
P1 = P0 + Οgd1
P2 = P0 + Οgd2
ΞP = P2 – P1 = ΟgΞd
(8.08 x 106 – 5.05 x106) = 103 x 10 x Ξd
3.03 x 106 = 103 x 10 x Ξd
Ξd = 303 m β 300 m
Answer: (a) 300 m
Q9: Water from a pipe is coming at a rate of 100 litres per minute. If the radius of the pipe is 5 cm, the Reynolds number for the flow is of the order (density of water = 1000 kg/m3, coefficient of viscosity of water = 1 mPa s)
(a)103
(b) 104
(c)102
(d) 106
Solution
Reynolds number = Οvd/Ξ·
Volume flow rate = v x Οr2
v = (100 x 10-3/60) x (1/Ο x 25 x 10-4)
v = (2/3Ο) m/s
Reynolds number = {(103 x 2 x 10 x 10-2)/(10-3 x 3Ο)}
β 2 x 104
Order of 104
Answer : (b) 104
Q10: The top of a water tank is open to the air and its water level is maintained. It is giving out 0.74 m3 water per minute through a circular opening of 2 cm radius in its wall. The depth of the centre of the opening from the level of water in the tank is close to
(a) 6.0 m
(b) 4.8 m
(c)9.6 m
(d) 2.9 m
Solution
Here, volumetric flow rate = (0.74/60) = Οr2v = (Ο x 4 x 10-4) x β2gh
β β2gh =[ (74 x 100)/240Ο)]
β β2gh= 740/24Ο
2gh = (740/24Ο)2
h = [(740 x 740)/24 x 24 x 10)] (since Ο2 =10)
h β 4.8 m
Answer: (b) 4.8 m
Video Lessons
Equation of Continuity
Surface Tension
Also Read:
Fluid Mechanics JEE Advanced Previous Year Questions With Solutions
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