 # JEE Main Fluid Mechanics Previous Year Questions with Solutions

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 Newton’s second law for nonaccelerating bodies, i.e. ∑ F =0.

Fluid dynamics is the study of fluids in motion. The important equation used here is the equation of Newton’s second law for accelerating bodies, i.e.∑ F =ma. Some of the fields where fluid dynamics is applied are meteorology, oceanography, aeronautical engineering, the study of blood flow and many more.

## 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 η. 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)πr3 = (4/3)πr3

Or r = R/3

Terminal velocity, v ∝ r3

Therefore, (v1/v2) = (r12/r22)

v2 = (r12/r22)v1 = [(R/3)/R]2v1 = 1/9

(v1/v2) = 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 = 6πηRv

obviously option (b) holds goods

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 rω. So, the difference in height is given as follows

2gh = ω2r2

h = [(2 x 2π)2(5 x 10-2)2]/(2×10) = 2cm

Q4: Water is flowing continuously from a tap having an internal diameter 8 × 10–3 m. The water velocity as it leaves the tap is 0.4 ms–1. The diameter of the water stream at a distance 2 × 10–1 m below the tap is close to

(a) 5.0 × 10–3 m

(b) 7.5 × 10–3 m

(c) 9.6 × 10–3 m

(d) 3.6 × 10–3 m

Solution

Here, d1 = 8 × 10–3 m

v1 = 0.4 m s–1, h = 0.2 m

According to equation of motion,

$v_{2}=\sqrt{v_{1}^{2}+2gh}=\sqrt{(0.4)^{2}+2+10\times 0.2}$

= 2 m s–1

According to equation of continuity a1v1 = a2v2

$\pi \times \left ( \frac{8\times 10^{-3}}{2} \right )\times 0.4 = \pi \times \left ( \frac{d_{2}}{2} \right )^{2}\times 2$

d2 = 3.6 × 10–3 m

Answer: (d) 3.6 × 10–3 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

Q6: A spherical solid ball of volume V is made of a material of density ρ1. It is falling through a liquid of density ρ22 < 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 = –kv2 (k > 0). The terminal speed of the ball is

(a) Vg(ρ1 – ρ2)

(b) $\sqrt{\frac{Vg(\rho _{1} – \rho _{2})}{k}}$

(c) Vgρ1/k

(d) $\sqrt{\frac{Vg\rho _{1}}{k}}$

Solution The forces acting on the solid ball when it is falling through a liquid is “mg” 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 – Vρ2g – kv2 = ma where V is volume,

v is the terminal velocity

When the ball is moving with terminal velocity, a = 0

Therefore Vρ1g – Vρ2g – kv2 = 0

$\sqrt{\frac{Vg(\rho _{1} – \rho _{2})}{k}}$

Answer: (b) $\sqrt{\frac{Vg(\rho _{1} – \rho _{2})}{k}}$

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 $\sqrt{2gh}$

10-4 = 10-4 x √20 xh

h = (1/20) m = 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 + ρgd1

P2 = P0 + ρgd2

ΔP = P2 – P1 = ρgΔd

3.03 x 106 = 103 x 10 x Δd

Δd = 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 = ρvd/η

Volume flow rate = v x πr2

v = (100 x 10-3/60) x (1/π x 25 x 10-4)

v = (2/3π) m/s

Reynolds number = {(103 x 2 x 10 x 102)/(10-3 x 3π)}

⋍ 2 x 104

Order of 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) = πr2v = (π x 4 x 10-4) x $\sqrt{2gh}$

$\sqrt{2gh}$ =[ (74 x 100)/240π)]

$\sqrt{2gh}$ = 740/24π

2gh = (740/24π)2

h = [(740 x 740)/24 x 24 x 10)] (since π2 =10)

h ≈ 4.8 m

The depth of the centre of the opening from the level of water in the tank is close to 4.8 m