Fluid Mechanics MCQs with Answers and Detailed Explanations

Pressure is force per unit area.

\[
P=\frac{F}{A}
\]

Its SI unit is

\[
\text{N/m}^2
\]

This unit is called the Pascal, written as Pa.

Hydrostatic pressure is

\[
P=P_0+\rho gh
\]

It depends on liquid density \(\rho\), gravitational acceleration \(g\), and depth \(h\).

Container shape and total liquid volume do not affect pressure at the same depth.

Pressure due to a liquid column is

\[
P=\rho gh
\]

At 5 m depth, pressure is proportional to 5.

At 15 m depth, pressure is proportional to 15.

Therefore,

\[
\frac{P_{15}}{P_5}=\frac{15}{5}=3
\]

So the pressure due to water becomes three times.

According to Pascal’s law, pressure is transmitted equally.

\[
\frac{F_1}{A_1}=\frac{F_2}{A_2}
\]

\[
F_2=F_1\frac{A_2}{A_1}
\]

\[
F_2=100\cdot\frac{0.50}{0.02}
\]

\[
F_2=2500\text{ N}
\]

The hydraulic lift increases force, although the large piston moves a smaller distance.

Archimedes’ principle states that an immersed object experiences an upward buoyant force equal to the weight of the fluid displaced.

\[
F_B=\rho_{\text{fluid}}V_{\text{displaced}}g
\]

The object may float, sink, or remain suspended depending on how this buoyant force compares with its weight.

A floating object is in vertical equilibrium.

Therefore,

\[
F_B=W
\]

The block displaces exactly enough water so that the displaced water weighs the same as the block.

For a floating object,

\[
\frac{V_{\text{submerged}}}{V_{\text{total}}}
=
\frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}
\]

Therefore,

\[
\frac{V_{\text{submerged}}}{V_{\text{total}}}
=
\frac{800}{1000}
\]

\[
=\frac{4}{5}
\]

So 80 percent of the block remains below the water surface.

Apparent weight is reduced because of buoyancy.

\[
F_B=W_{\text{air}}-W_{\text{apparent}}
\]

\[
F_B=50-35
\]

\[
F_B=15\text{ N}
\]

The buoyant force acts upward.

At the same horizontal level in a connected liquid, pressure must be equal.

Since both surfaces experience the same atmospheric pressure, the liquid settles until both free surfaces are at the same height.

The diameter of the arms does not change this final result.

Pressure is

\[
P=\frac{F}{A}
\]

For the same force, a smaller contact area produces greater pressure.

Therefore, a sharp needle can pierce a surface more easily than a blunt needle.

Gauge pressure is the extra pressure created by the liquid column above a point.

\[
P_{\text{gauge}}=\rho gh
\]

Here,

\[
\rho=800\text{ kg/m}^3,\qquad g=10\text{ m/s}^2,\qquad h=5\text{ m}
\]

Therefore,

\[
P_{\text{gauge}}=800\times10\times5
\]

\[
P_{\text{gauge}}=40000\text{ Pa}
\]

\[
P_{\text{gauge}}=4.0\times10^4\text{ Pa}
\]

Atmospheric pressure is not included because the question asks for gauge pressure.

For an incompressible liquid, displaced volume is conserved.

\[
A_1d_1=A_2d_2
\]

Therefore,

\[
d_2=\frac{A_1d_1}{A_2}
\]

\[
d_2=\frac{5\times20}{500}
\]

\[
d_2=0.2\text{ cm}
\]

A hydraulic press multiplies force, but the large piston moves a much smaller distance. This is consistent with conservation of work in an ideal system.

The buoyant force equals the weight of displaced water.

\[
F_B=\rho_{\text{water}}Vg
\]

Since the cube is fully submerged, it displaces its full volume.

\[
F_B=1000\times0.020\times10
\]

\[
F_B=200\text{ N}
\]

The cube’s own density does not change the buoyant force while its full volume remains submerged.

The object experiences weight downward and buoyant force upward.

Weight:

\[
W=mg=3\times10=30\text{ N}
\]

Buoyant force:

\[
F_B=\rho Vg
\]

\[
F_B=1000\times2.0\times10^{-4}\times10
\]

\[
F_B=2\text{ N}
\]

Net downward force is

\[
F_{\text{net}}=30-2=28\text{ N}
\]

Therefore,

\[
a=\frac{F_{\text{net}}}{m}
\]

\[
a=\frac{28}{3}
\]

\[
a=9.33\text{ m/s}^2
\]

The acceleration is downward because the object is much denser than water.

For steady incompressible flow, the continuity equation is

\[
A_1v_1=A_2v_2
\]

Therefore,

\[
v_2=\frac{A_1v_1}{A_2}
\]

\[
v_2=\frac{0.040\times2}{0.010}
\]

\[
v_2=8\text{ m/s}
\]

The fluid moves faster in the narrow section because the same volume of water must pass through every cross-section each second.

Volume flow rate is

\[
Q=Av
\]

Therefore,

\[
v=\frac{Q}{A}
\]

\[
v=\frac{0.012}{0.003}
\]

\[
v=4\text{ m/s}
\]

The unit check is also useful:

\[
\frac{\text{m}^3/\text{s}}{\text{m}^2}=\text{m/s}
\]

For a horizontal pipe, Bernoulli’s equation becomes

\[
P_1+\frac{1}{2}\rho v_1^2
=
P_2+\frac{1}{2}\rho v_2^2
\]

Thus,

\[
P_2=P_1+\frac{1}{2}\rho(v_1^2-v_2^2)
\]

\[
P_2=1.20\times10^5+\frac{1}{2}(1000)(2^2-6^2)
\]

\[
P_2=1.20\times10^5+500(4-36)
\]

\[
P_2=1.20\times10^5-16000
\]

\[
P_2=1.04\times10^5\text{ Pa}
\]

The pressure decreases because the water speed increases.

For a large open tank, the water surface speed is approximately zero.

Torricelli’s theorem gives

\[
v=\sqrt{2gh}
\]

\[
v=\sqrt{2\times10\times1.8}
\]

\[
v=\sqrt{36}
\]

\[
v=6\text{ m/s}
\]

This result comes directly from Bernoulli’s principle or conservation of mechanical energy.

The gas pushes the mercury lower on its side, so gas pressure is greater than atmospheric pressure.

Pressure difference is

\[
\Delta P=\rho gh
\]

Here,

\[
h=5\text{ cm}=0.05\text{ m}
\]

Therefore,

\[
\Delta P=13600\times10\times0.05
\]

\[
\Delta P=6800\text{ Pa}
\]

This is the gauge pressure of the gas chamber.

For a floating object,

\[
\frac{V_{\text{submerged}}}{V}
=
\frac{\rho_{\text{object}}}{\rho_{\text{fluid}}}
\]

In water,

\[
\rho_{\text{object}}=\frac{3}{4}\rho_{\text{water}}
\]

In the new liquid,

\[
\rho_{\text{new}}=1.5\rho_{\text{water}}
\]

Thus,

\[
\frac{V_{\text{submerged}}}{V}
=
\frac{\frac{3}{4}\rho_{\text{water}}}{1.5\rho_{\text{water}}}
\]

\[
=
\frac{3}{4}\times\frac{2}{3}
\]

\[
=
\frac{1}{2}
\]

The denser liquid provides the required buoyant force with less displaced volume.

The speed of water leaving the orifice follows Torricelli’s theorem:

\[
v=\sqrt{2gh}
\]

The hole is at height

\[
H-h
\]

above the ground. Therefore, the time taken by water to reach the ground is

\[
t=\sqrt{\frac{2(H-h)}{g}}
\]

The horizontal range is

\[
R=vt
\]

\[
R=
\sqrt{2gh}
\sqrt{\frac{2(H-h)}{g}}
\]

\[
R=2\sqrt{h(H-h)}
\]

For maximum range, the product

\[
h(H-h)
\]

must be maximum. This occurs when

\[
h=H-h
\]

Therefore,

\[
h=\frac{H}{2}
\]

The hole should be placed halfway below the water surface.

Water is expelled backward, so the tank moves forward.

The thrust produced by the jet is

\[
F=\dot{m}v
\]

Here, \(\dot{m}\) is the mass leaving per second and \(v\) is the water speed relative to the tank.

Using Newton’s second law for the tank and remaining water,

\[
a=\frac{F}{M_{\text{total}}}
\]

Therefore,

\[
a=\frac{\dot{m}v}{M_{\text{total}}}
\]

As water leaves, \(M_{\text{total}}\) decreases. Therefore, the tank acceleration increases if \(\dot{m}\) and \(v\) remain constant.

First use continuity equation:

\[
A_1v_1=A_2v_2
\]

Area is proportional to diameter squared.

\[
\frac{A_1}{A_2}
=
\left(\frac{4}{2}\right)^2
=
4
\]

Therefore,

\[
v_2=4v_1
\]

\[
v_2=4(2)=8\text{ m/s}
\]

For a horizontal pipe, Bernoulli’s equation gives

\[
P_1+\frac{1}{2}\rho v_1^2
=
P_2+\frac{1}{2}\rho v_2^2
\]

Thus,

\[
P_1-P_2
=
\frac{1}{2}\rho(v_2^2-v_1^2)
\]

\[
P_1-P_2
=
\frac{1}{2}(1000)(8^2-2^2)
\]

\[
P_1-P_2
=
500(64-4)
\]

\[
P_1-P_2=30000\text{ Pa}
\]

The correct answer is

\[
30000\text{ Pa}
\]

The marked option should be option 4 before publishing.

At equilibrium, buoyant force equals the block weight.

When the block is pushed downward by a small distance \(x\), the extra displaced liquid volume is

\[
Ax
\]

Therefore, the extra upward buoyant force is

\[
F=\sigma gAx
\]

This force acts opposite to displacement, so

\[
F=-\sigma gAx
\]

The mass of the block is

\[
m=\rho AL
\]

Using

\[
m\omega^2x=\sigma gAx
\]

we get

\[
\rho AL\omega^2x=\sigma gAx
\]

Cancel \(A\) and \(x\):

\[
\omega^2=\frac{\sigma g}{\rho L}
\]

Therefore,

\[
\omega=\sqrt{\frac{\sigma g}{\rho L}}
\]

Pressure produced by each liquid layer is added separately.

The upper liquid produces pressure

\[
P_1=\rho gh
\]

The lower liquid produces pressure

\[
P_2=(2\rho)gh
\]

Therefore, total gauge pressure is

\[
P=P_1+P_2
\]

\[
P=\rho gh+2\rho gh
\]

\[
P=3\rho gh
\]

Pressure depends on the density and vertical height of each layer, not on the total container shape.

In the accelerating tank frame, water experiences effective gravity.

There is normal gravity downward:

\[
g
\]

There is pseudo acceleration opposite to tank acceleration:

\[
a
\]

The free surface must remain perpendicular to the effective gravity direction.

Therefore,

\[
\tan\theta=\frac{a}{g}
\]

The water level rises on the side opposite to the direction of tank acceleration.

Capillary rise is given by

\[
h=\frac{2T\cos\theta}{\rho gr}
\]

Therefore,

\[
h\propto\frac{1}{r}
\]

If radius becomes

\[
\frac{r}{2}
\]

then the new height becomes

\[
h’=\frac{2T\cos\theta}{\rho g(r/2)}
\]

\[
h’=2h
\]

A narrower capillary tube produces a larger capillary rise.

Under Stokes’ law,

\[
F_{\text{viscous}}=6\pi\eta rv
\]

At terminal speed, viscous force balances effective weight.

Effective weight is proportional to sphere volume:

\[
V\propto r^3
\]

Thus,

\[
6\pi\eta rv_t\propto r^3
\]

Therefore,

\[
v_t\propto r^2
\]

If radius changes from \(r\) to \(2r\),

\[
v_t’\propto(2r)^2
\]

\[
v_t’=4v_t
\]

This result is valid only when the flow remains laminar and Stokes’ law applies.

The sphere weight is

\[
W=\rho_sVg
\]

The buoyant force is

\[
F_B=\rho_lVg
\]

Since \(\rho_s>\rho_l\), the sphere moves downward.

Net downward force is

\[
F_{\text{net}}=(\rho_s-\rho_l)Vg
\]

Mass of the sphere is

\[
m=\rho_sV
\]

Therefore,

\[
a=\frac{F_{\text{net}}}{m}
\]

\[
a=
\frac{(\rho_s-\rho_l)Vg}{\rho_sV}
\]

\[
a=g\left(1-\frac{\rho_l}{\rho_s}\right)
\]

The acceleration is downward.

For a hole at depth \(h\), the jet speed is

\[
v=\sqrt{2gh}
\]

The height of the hole above the ground is

\[
H-h
\]

The fall time is

\[
t=\sqrt{\frac{2(H-h)}{g}}
\]

Thus, horizontal range is

\[
R=vt
\]

\[
R=2\sqrt{h(H-h)}
\]

Two holes give the same range when

\[
h_1(H-h_1)=h_2(H-h_2)
\]

For different holes, this happens when

\[
h_1=H-h_2
\]

Therefore,

\[
h_1+h_2=H
\]