MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 1

The top of a water tank is open to air and its water level is maintained. It is giving out

$0.74 m^3$ 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 :

0%
$9.6 m$
0%
$4.8 m$
0%
$2.9 m$
0%
$6.0 m$

Explanation

In flow volume = outflow volume

$\Rightarrow \dfrac{0.74}{60} = (\pi \times 4 \times 10^{-4}) \times \sqrt{2 gh}$

$\Rightarrow \sqrt{2 gh} = \dfrac{74 \times 100}{240 \pi}$

$\Rightarrow \sqrt{2 gh} = \dfrac{740}{24 \pi}$

$\Rightarrow 2gh = \dfrac{740 \times 740}{24 \times 24 \times 10} (\pi^2 = 10)$

$\Rightarrow h = \dfrac{74 \times 74}{2 \times 24 \times 24}$

$\Rightarrow h \approx 4.8 m$

Two liquids of densities

${ \rho }_{ 1 }$ and

${ \rho }_{ 2 }\left( { \rho }_{ 2 }=2{ \rho }_{ 1 } \right)$ are filled up behind a square wall of side

$10m$ as shown in figure. Each liquid has a height of

$5m$ . The ratio of the forces due to these liquids exerted on upper part

$MN$ to that at the lower part

$NO$ is (Assume that the liquids are not mixing)

0%
$1/4$
0%
$2/3$
0%
$1/2$
0%
$1/3$

Explanation

Let

$\rho_1 = \rho$ and

$\rho_2 = 2\rho$

Let

$P_A$ be the pressure at the top

$P_B$ be the pressure at the interface between the liquids and

$P_C$ be the pressure at the bottom of the wall.

Let

$F_1$ and

$F_2$ denote the forces acting on the upper part of the wall and the lower part of the wall respectively.

$P_A = P_0 = 0$

$P_B = \rho_1 g h=\rho g h$

$P_B = \rho g \times 5 = 5 \rho g$

$P_C = \rho_1 g\ 5 + \rho_2 g\ 5=\rho g\ 5 + 2\rho g\ 5$

$P_C=15Pg$

$F_1 = \left(\dfrac{P_A + P_B}{2}\right) = \left( \dfrac{0+5\rho g}{2}\right).A = \dfrac{5\rho gA}{2}$

$F_2 = \left( \dfrac{P_B + P_C}{2}\right) \times A = \left( \dfrac{5\rho g+15\rho g}{2}\right).A = \dfrac{20\rho gA}{2}$

$\dfrac{F_1}{F_2} = \dfrac{\tfrac{5\rho gA}{2}}{ \tfrac{20\rho g A}{2}} = \dfrac{1}{4}$

$\dfrac{F_1}{F_2} = \dfrac{1}{4}$

1477969_1697241_ans_68777e2ca5ad469fafc2ff1665666ad2.PNG

A capillary tube (A) is dipped in water. Another identical tube (B) is dipped in a soap-water solution. Which of the following shows the relative nature of the liquid columns in the two tubes?

Explanation

The weight of the water risen in capillary tube is balanced by the force exerted due to decrease in pressure inside the water column due to the curvature at the water surface.

The decrease in pressure=

$\dfrac{2\sigma}{r}$

Hence from balancing the pressure, we get

$\dfrac{2\sigma}{r}=h\rho g$

Hence

$h\propto \sigma$

Since surface tension(

$\sigma$ ) of water decreases as soap is mixed, the height upto which water rises decreases.

A glass capillary tube is of the shape of truncated cone with an apex angle

$\alpha$ so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height

$h$ , where the radius of its cross section is

$b$ . If the surface tension of water is

$S$ , its density is

$\rho$ , and its contact angle with glass is

$\theta$ , the value of

$h$ will be (

$g$ is the acceleration due to gravity)

0%
$\dfrac{2S}{b \rho g} \cos (\theta - \alpha )$
0%
$\dfrac{2S}{b \rho g} \cos (\theta+ \alpha )$
0%
$\dfrac{2S}{b \rho g} \cos (\theta - \alpha /2)$
0%
$\dfrac{2S}{b \rho g} \cos (\theta + \alpha /2)$

Explanation

Let r = radius of curvature of meniscus

$\displaystyle b = r\, cos \left ( \theta +\frac{\alpha}{2} \right )$
Excess pressure on concave side of meniscus

$\displaystyle \frac {2S}{r}$

$\displaystyle \Rightarrow (P_0 + h\rho g) - P_0 = \frac {2S}{r}$

$\displaystyle h\rho g =\frac {2S cos \left (\theta +\frac{\alpha}{2} \right ) }{b}$

$\displaystyle h =\frac {2S}{b\rho g} cos \left (\theta +\frac{\alpha}{2} \right )$

If the density of air is

$\rho_a$ and that of the liquid

$\rho_l$ , then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to

0%
$\displaystyle \sqrt{\dfrac{\rho_a}{\rho_l}}$
0%
$\sqrt{\rho_a \rho_l}$
0%
$\displaystyle \sqrt{\dfrac{\rho_l}{\rho_a}}$
0%
$\rho_l$

Explanation

From the principle of continuity,

$\dfrac{1}{2} \rho_{1} v_{1} ^{2} = \dfrac{1}{2} \rho_{2} v_{2} ^{2}$

$\therefore \dfrac{v_{1}}{v_{2}} = \sqrt{\dfrac{\rho_{2}}{\rho_{1}} }$

A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance

$d$ of

$1.2 m$ from the person. In the following, state of the lift's motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists.

	List I		List II
P.	Lift is accelerating vertically up.	1.	$d= 1.2 m$
Q.	Lift is accelerating vertically down with an acceleration less than the gravitational acceleration.	2.	$d > 1.2 m$
R.	Lift is moving vertically up with constant speed.	3.	$d < 1.2 m$
S.	Lift is falling freely.	4.	No water leaks out of the jar

0%
P-2, Q-3, R-2, S-4
0%
P-2, Q-3, R-1, S-4
0%
P-1, Q-1, R-1, S-4
0%
P-2, Q-3, R-1, S-1

Explanation

Velocity of efflux

$V=\sqrt{2g_{eff}l}$
where

$l$ is length of jar from top of point.
time taken in covering horizontal distance is

$T= \sqrt{\dfrac{2h}{g_{eff}}}$ , where

$h$ is distance of floor from jar
Horizontal range =

$VT==\sqrt{2g_{eff}l}\sqrt{\dfrac{2h}{g_{eff}}}= \sqrt{4lh}$
which is independent of

$g_{eff}$ .
In free fall velocity of efflux will be zero.

A wind with speed

$40$ m/s blows parallel to the roof of a house. The area of the roof is

$250\ m^2$ . Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be

$(\rho_{air}=1.2 \ kg/m^3)$

0%
$2.4\times 10^5N$ , upward
0%
$2.4\times 10^5N$ , downward
0%
$4.8\times 10^5N$ , downward
0%
$4.8\times 10^5N$ , upward

Explanation

Applying Bernoulli's theorem to two points just inside the house and outside the house,

$P + \dfrac{1}{2}\rho v^2 = constant$

Inside the pressure is

$P_{atm}$

The pressure outside is

$P$

v is zero inside.

$\therefore P - P_{atm} = \frac{1}{2}\rho v^2= \frac{1}{2} \times 1.2 \times 40^2=960$

Force is :

$F = (P - P_{atm}) \times Area= 960 \times 250 = 2.4 \times 10^5\:N$ in upward direction as pressure goes from higher to lower pressure area.

A small hole of an area of cross-section

$2 mm^2$ is present near the bottom of a fully filled open tank of height 2 m. Taking

$g = 10 m/s^2$ , the rate of flow of water through the open hole would be nearly

0%
$12.6 \times 10^{-6} m^3/s$
0%
$8.9 \times 10^{-6} m^3/s$
0%
$2.23 \times 10^{-6} m^3/s$
0%
$6.4 \times 10^{-6} m^3/s$

Explanation

Rate of flow liquid

$Q = au = a \sqrt{2 gh}$

$= 2 \times 10^{-6} m^2 \times \sqrt{2 \times 10 \times 2} m/s$

$= 2 \times 2 \times 3.14 \times 10^{-6} m^3/s$

$= 12.56 \times 10^{-6} m^3/s$

$=12.6 \times 10^{-6} m^3/s$
1259401_1618855_ans_013cef43b63140289f20524598250f44.png

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
0%
Assertion is correct but Reason is incorrect.
0%
Both Assertion and Reason are incorrect.

A spherical body of diameter D is falling in viscous medium. Its terminal velocity is proportional to:

0%
${ V }_{ 1 }\propto { D }^{ 1/2 }$
0%
${ V }_{ 1 }\propto { D }^{ 3/2 }$
0%
${ V }_{ 1 }\propto { D }^{ 2 }$
0%
${ V }_{ 1 }\propto { D }^{ 5/2 }$

Explanation

The formula for terminal velocity is

$v=\dfrac { 2g{ r }^{ 2 }({ \rho }-{ \sigma }) }{ 9{ \eta } }$ , where

${\eta}$ is the viscosity.

So this clearly shows that terminal velocity is directly proportional to square of diameter.

The velocity of efflux of a liquid through an orifice in the bottom of the tank does not depend upon:

0%
size of orifice
0%
height of liquid
0%
acceleration due to gravity
0%
density of liquid

Explanation

$P_o + \dfrac{1}{2}\rho_{bottom} v^2 = P_o +\rho_{liquid} g h$

Velocity of efflux is independent of the size of orifice.

The density of the fluid is actually varying with depth for large heights. In such scenarios, velocity of efflux depends on the density of the liquid.

What is a pressure on the smaller piston if both the pistons are at same horizontal level? (Take

$g=10 ms^{-2})$

0%
3 Pa
0%
$4\times 10^3 Pa$
0%
$6\times 10^5 Pa$
0%
1000 Pa

Explanation

$Pressure=\dfrac {3000\times 10}{5\times 10^{-2}}=6\times 10^5 Pa.$

In turbulent flow, the velocity of the liquid molecules in contact with the walls of the tube.

0%
is zero
0%
is maximum
0%
is equal to critical velocity
0%
may have any value

Which of the following works on Pascal's law?

0%
Sprayer
0%
Venturimeter
0%
Hydraulic lift
0%
Aneroid barometer

The volume of liquid flowing per second out of an orifice at the bottom of the tank does not depend upon:

0%
the density of the liquid
0%
acceleration due to gravity
0%
the height of the liquid above orifice
0%
the area of the orifice

Explanation

$\theta =$ Volume flow per second.

$\theta =$

$v.A$ : where

$v$ is flow velocity

$v =\sqrt{2gh}$

$\theta =A\sqrt{2gh}$
Hence, volume of liquid flowing per second is not depend on the density of the liquid.
So, the correct option is

$(A)$

Stream line motion becomes turbulent motion when the velocity of the liquid is:

0%
beyond critical velocity
0%
critical velocity
0%
below critical velocity
0%
variable velocity

Which of the following is a characteristic of turbulent flow?

0%
velocity more than critical velocity
0%
irregular flow
0%
molecules crossing from one layer to the other
0%
All of the above

In a laminar flow at a given point the magnitude and direction of the velocity of the fluid:

0%
both are constant
0%
magnitude is only constant
0%
direction is only constant
0%
both are not constant

The terminal velocity depends upon:

0%
$\dfrac {1}{r}$
0%
$\dfrac {1}{r^2}$
0%
$\dfrac {1}{r^3}$
0%
$r^2$

Explanation

Terminal velocity,

$v_T = \dfrac {2r^2g}{9\eta}(\rho -\sigma), v_T \propto r^2$

The correct formula of critical velocity

$\left(V_c \right)$ is :

0%
$V_c = \displaystyle \dfrac {k \eta d}{r}$
0%
$V_c = \displaystyle \dfrac {k \eta}{dr}$
0%
$V_c = \displaystyle \dfrac {dr}{k \eta}$
0%
$V_c = \displaystyle \dfrac {r \eta}{dk}$

Explanation

The critical velocity depends on coefficient of viscosity

$\eta$ , density of the liquid

$d$ , and radius of the tube

$r$ as given by

$V_c=\dfrac{k\eta}{d r}$ where

$k$ is the constant of proportionality.

The velocity efflux of a liquid is:

0%
$\sqrt {2 gh}$
0%
$2 gh$
0%
$gh$
0%
$\sqrt {g h}$

Explanation

Torricelli's law states that the speed of efflux

$v$ of a fluid through a sharp-edged hole at the bottom of a tank filled to a depth

$h$ is the same as the speed that a body would acquire in falling freely from a height

$h$ , i.e.

$v=\sqrt{2gh}$

A raindrop falls near the surface of the earth with almost uniform velocity because:

0%
its weight is negligible.
0%
the force of surface tension balances its weight.
0%
the force of viscosity of air balances its weight.
0%
the drops are charged and atmospheric electric field balances its weight.

When a sphere falling in a viscous fluid attains a terminal velocity, then:

0%
the net force acting on the sphere is zero
0%
the drag force balances the buoyant force
0%
the drag force balances the weight of the sphere
0%
the buoyant force balances the weight and drag force

In a stream line (laminar flow) the velocity of flow at any point in the liquid:

0%
does not vary with time
0%
may vary in direction but not in magnitude
0%
may vary in magnitude but not in direction
0%
may vary both in magnitude and direction

The velocity of a liquid coming out of a hole in the tank wall is:

0%
more if the hole is near the upper end.
0%
more if the hole is at the centre of the wall.
0%
more if hole is near the bottom.
0%
velocity of flow does not depend upon the position of the hole.

Explanation

According to Torricelli's Theorem velocity of efflux i.e. the velocity with which the liquid flows out of a hole is equal to

$\sqrt{2gh}$ where

$h$ is the depth of the hole below the liquid surface. Hence more the depth of the hole, more will be the velocity of the liquid coming out of the hole, and will be maximum if the hole is at bottom of the tank.

State whether true or false:

The deeper an underwater swimmer goes, the more strongly the water pushes on his body and the greater is the pressure that he experiences.

0%
True
0%
False

Choose the correct answer from the given options below:

The meniscus formed by water in a test tube is of --------- shape.

0%
horizontal
0%
triangular
0%
concave
0%
convex

Name the kind of meniscus formed in case of water.

0%
Concave
0%
Convex
0%
Both A & B
0%
None

If a capillary tube is tilted to

$45^{\circ}$ and

$60^{\circ}$ from the vertical then the ratio of length

$l_{1}$ and

$l_{2}$ of liquid columns in it will be -

0%
$1: \sqrt{2}$
0%
$\sqrt{2}:1$
0%
1:2
0%
2:1

Explanation

The expression for the capillary rise in a tube is given by,

H = $\dfrac {2T cos \theta}{\rho gr}$

where $\theta$ is the contact angle and not the angle of tilt.

Thus, for all parameters constant,

at $60^o$

$l_1 = H cos 60^0$ = H/2

At $45^o$

$l_2 = H cos 45^0$ = $H/\sqrt 2$

Therefore the ratio of the 2 lengths is given by,

$l_1/l_2 = \dfrac {1}{\sqrt 2}$

If a wax coated capillary tube is dipped in water, then water in it will-

0%
rise up
0%
depress
0%
sometimes rise and sometimes fall
0%
rise up and come out as a fountain

Shape of meniscus for a liquid of zero angle of contact is convex.

0%
True
0%
False

Answer true or false. 'Faster the movement of the air greater is the drop in pressure'.

0%
True
0%
False

A pilot wants to fly his aeroplane in atmosphere free space

0%
He cannot do it
0%
He can do it easily
0%
He can do it but with difficulty
0%
None of these

The height of mercury which exerts the same pressure as 20 cm of a water column, is equal to

0%
$1.48$ cm
0%
$14.8$ cm
0%
$148$ cm
0%
none of these

Explanation

Let

$\rho_w$ be density of water

$\rho_m$ be density of mercury

$h_w$ be height of water

$h_m$ be height of mercury

$g$ be acceleration due to gravity

$P_w$ be pressure exerted by water column

$P_m$ be pressure exerted by mercury column

Now

$P_w=\rho_w g h_w$ (1)

$\rho_w=1g/cc$

$h_w=20cm$

So from (1)

$P_w=1\times 10 \times 20$ (2)

$P_m=\rho_m g h_m$ (3)

$\rho_m=13.6g/cc$

From (3)

$P_m=13.6\times 10 \times h_m$ (4)

$P_w=P_m$

So from (2) and (4)

$1\times 10 \times 20=13.6\times 10 \times h_m$

$h_m=\dfrac{20}{13.6}$

$h_m=1.48cm$

Curve in the upper surface of a liquid to the surface of the container is known as

0%
Surface teusion
0%
Meniscus
0%
Both
0%
Elasticity

If the meniscus of a liquid is concave in shape, then

0%
force of adhesion > force of cohesion
0%
force of adhesion = force of cohesion
0%
force of adhesion < force of cohesion
0%
None of the above

In steady horizontal flow:

0%
the pressure is greatest where the speed is leas
0%
the pressure is independent of speed
0%
the pressure is least where the speed is least
0%
(a) and (c) are correct

In a turbulent flow, the velocity of the liquid molecules in contact with the walls of the tube is :

0%
Zero
0%
Maximum
0%
Equal to critical velocity
0%
May have any value

The horizontal flow of fluid depends upon

0%
Pressure difference
0%
Amount of fluid
0%
Density of fluid
0%
All the above

Pressure (P) exented by liquid on walls of the container when h = height of the liquid column, f = density, g is acceleration due to gravity

0%
$p = h f g$
0%
$p = \dfrac{df}{g}$
0%
$p = \dfrac{hg}{f}$
0%
$p = \dfrac{df}{h}$

The SI unit of hydro-static pressure is :

0%
Pa
0%
$Nm^{-1}$
0%
Nm
0%
$kg$ $wt$ $m^{-2}$

Bernoulli's equation is applicable to points :

0%
In a steadily flowing liquid
0%
In a stream line
0%
In a straight line perpendicular to a stream line
0%
For ideal liquid stream line flow on a stream line

1 mm Hg is equal to

0%
1 atm
0%
1.013 $\times$ 10 $^5$ Pa
0%
133.29 atm
0%
1.316 $\times$ 10 $^{-3}$ atm

Explanation

760 mm Hg

$=1$ atm

$\therefore$ 1 mm Hg

$= \dfrac{1}{760} =1.316\times 10^{-3}$ atm

If a liquid does not wet glass, its angle of contact is:

0%
Zero
0%
Acute
0%
Obtuse
0%
Right angle

If the surface tension of water is

$0.06N/m$ , then the capillary rise in a tube of a diameter

$1mm$ is :(

$\theta={0}^{o}$ )

0%
$1.22cm$
0%
$2.44cm$
0%
$3.12cm$
0%
$3.86cm$

Explanation

$h\cfrac{2T\cos{\theta}}{r\rho g}$

$h=\cfrac{2\times 0.06\times \cos{{0}^{o}}}{(0.5\times {10}^{-3})\times {10}^{3}\times 9.8}$
or

$h=2.44\times {10}^{-2}m$
or

$h=2.44cm$

The ordinary pressure of air surrounding us is

0%
15 pounds
0%
16 pounds
0%
14.2 pounds
0%
14.7 pounds

Explanation

$Answer:-$ D

The ordinary pressure of the air surrounding us is 14.7 pounds per square inch, but the pressure can change when the wind blows or an object, like a car or airplane, accelerates.

One important principle to remember is that wherever the air pressure is higher, there will be a stronger force or push against an object. It’s also helpful to know that when an air particle speeds up, it actually “pushes” less. Imagine that fast-moving air particles are in so much of a hurry that they don’t have time to apply force. This principle is used in airplane wings to make planes fly. When a plane moves along the runway, the air above the wing speeds up more, lowering the pressure, so that the air below the wing can push the plane upward.

When a plane moves along the runway, the air above the wing speeds up more, So the pressure

0%
Lowers
0%
Increases
0%
Raises
0%
No Pressure

Explanation

According to Bernoulli's theorem

$P + \rho gh+ \dfrac{1}{2}\rho v^2= constant$

$\implies$ When

$v$ is increases, then pressure

$(P)$ is lowers.

Choose the correct option. Consider tube to be made of glass.

0%
A: water

B: mercury
0%
A: mercury

B: water
0%
A: hot water

B: cold water
0%
A: hot mercury

B: cold mercury

Choose the correct statement among the following

0%
The upper surface of a stationary liquid is always horizontal.
0%
The pressure of a given liquid is directly proportional to the depth of the liquid.
0%
Pressure at a given depth inside a stationary liquid is different at all points on the horizontal plane.
0%
Both (A) and (B)

Vacuum above the mercury level in a barometer is called:

0%
Torricelli vacuum
0%
Archimedes vacuum
0%
newtons vacuum
0%
Pascal vacuum

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 1 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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