MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 2

The correct assumption(s) made to perform a venturi-meter experiment is(are):

0%
that Bernoulli's theorem holds good
0%
that equation of continuity is maintained
0%
that the fluid used is highly incompressible
0%
that all of the above are true

The height up to which water will rise in a capillary tube will be

0%
Minimum when water temperature is $4^{\circ}C$
0%
Maximum when water temperature is $4^{\circ}C$
0%
Maximum when water temperature is $0^{\circ}C$
0%
Minimum when water temperature is $0^{\circ}C$

Explanation

The height of rise in capillary,

$h = \dfrac {2T\cos \theta}{r\rho g}$
For water, the density is maximum at

$4^{\circ}C$ , so height is minimum at

$4^{\circ}C$ .

For a liquid which is rising in a capillary, the angle of contact is

0%
Obtuse
0%
Acute
0%
$180^o$
0%
$90^o$

A liquid is allowed to flow in a tube of truncated cone shape. Identify correct statement from the following.

0%
The speed is high at the wider end and low at the narrow end
0%
The speed is low at the wider end and high at the narrow end
0%
The speed is same at both ends in a stream line flow
0%
The liquid flows with uniform velocity in the tube

Explanation

For an incompressible liquid equation of continuity

$Av = constant$

or,

$A \propto \cfrac{1}{v}$

Therefore at the wider end speed will be low and at the narrow end speed will be hgih.

The gauge pressure of

$3 \times 10^5$ N/m

$^2$ must be maintained in the main water pipes of a city. How much work must be done to pump

$50,000$ m

$^3$ of water at a pressure of

$1.0 \times 10^5$ N/m

$^2$ -

0%
$10^{11}$ J
0%
$10^{10}$ J
0%
$10^{9}$ J
0%
$10^{8}$ J

When two capillary tubes of different diameters are dipped vertically, the rise of the liquid is :

0%
same in both the tubes
0%
more in tube of larger diameter
0%
less in the tube of smaller diameter
0%
more in the tube of smaller diameter

Explanation

The rise of liquid columns in cappileries

$15$ related to its radius as

$h=\dfrac { 2Tcos\theta }{ PgR }$

So rise is more for the tube of smaller diameter.

Two drops of the same radius are falling through air with a steady velocity of 5 cm/s. If the two drops coalesce, the terminal velocity would be

0%
$10 cm/s$
0%
$2.5 cm/s$
0%
$5(4)^{1/3} cm/s$
0%
$5(3)^{1/3} cm/s$

Explanation

terminal velocity

$(V) \alpha \space r^2$

$= Kr^2$

Now the volume of each drop =

$(\frac{4}{3} \pi r^3)$

The volume of coalesced =

$2 \times \dfrac{4}{3} \pi R^3$

Radius of coalesce drop =

$(2)^{\frac{1}{3} } r$

Hence, ter,inal velocity of coalesced drop

$v^1 = k[(2)^{\frac{1}{3}}r]$

$= (2)^{\frac{2}{3} } \times 5$

$V' = (4)^{\frac{1}{3}} \times 5$

Hence option (C) is correct

A drop of mercury of radius

$2 mm$ is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is

$0.465{ J/m }^{ 2 }$ )

0%
$23.4\mu J$
0%
$18.5\mu J$
0%
$26.8\mu J$
0%
$16.8\mu J$

Explanation

As we know,

Increase in surface energy or work is done in splitting a big drop

$=4\pi R^{2}T(n^{1/3}-1)$

$\Rightarrow W=4\pi *(2*10^{-3})^{2}*0.465(8^{1/3}-1)=23.4\mu J$

Water does not wet an oily glass, because

0%
cohesive force of oil $>>$ adhesive force between oil and glass
0%
cohesive force of oil $>$ cohesive force of water
0%
oil repels water
0%
cohesive force of water $>$ adhesive force between water and oil molecules

Explanation

Water does not wet an oily glass, because of, Cohesive force for water

$>$ adhesive force between water and oil molecules.

No capillarity will take place if:-
(i) The liquid is at its boiling point
(ii) The liquid is at its freezing point
(iii) The angle of contact is

$0^{\circ}$
(iv) The angle of contact is

$90^{\circ}$

0%
(i,ii)
0%
(i,iv)
0%
(i,ii,iii)
0%
(ii,iv)

Explanation

, the liquid is at its freezing point and the angle of contact is $0^0$

When the temperature is increased the angle of contact of a liquid

0%
increases
0%
decreases
0%
remains the same
0%
first increases and then decreases

Nature of meniscus for liquid of

$0^{0}$ angle of contact

0%
plane
0%
parabolic
0%
semi-spherical
0%
cylindrical

Explanation

Nature of meniscus for liquid of

$0^{0}$ angle of contact is Semi-spherical.

A student uses a measuring cylinder to measure the volume of some water. The diagram shows
part of the measuring cylinder. The top and bottom of the meniscus are labelled.
What is the volume of the water?

0%
$47.0\ cm^{3}$
0%
$47.5\ cm^{3}$
0%
$49.0\ cm^{3}$
0%
$49.5\ cm^{3}$

In order to float a ring of area

$0.04 m^{2}$ in a liquid of surface tension

$75 N/m$ , the required surface energy will be

0%
$3J$
0%
$6.5J$
0%
$1.5J$
0%
$4J$

Explanation

Given,

area=

$0.04 m^{2}$

surface tension =

$75 N/m$ ,

$E=T*\Delta A=75*0.04=3J$

The pressure at the bottom of a tank containing a liquid does not depend on

0%
acceleration due to gravity
0%
height of the liquid column
0%
area of the bottom surface
0%
nature of the liquid

Explanation

As we know,

$P=h\rho g$

hence,

The pressure at the bottom of a tank containing a liquid does not depend on the area of the bottom surface.

A capillary tube, made of glass is dipped into mercury. Then:

0%
mercury rises in the capillary tube.
0%
mercury descends in capillary tube.
0%
mercury rises and flows out of capillary tube.
0%
mercury neither rises nor descends in the capillary tube.

Explanation

It descends because the contact angle of mercury is generally

$135^{o}$ .
Therefore, surface tension applies a downward force on mercury pushing it down.

A capillary tube is dipped in water vertically.Water rises to a height of 10mm. The tube is now tilted and makes an angle 60

$^{o}$ with vertical.Now water rises to a height of:

0%
10 mm
0%
5 mm
0%
20 mm
0%
40 mm

The liquid meniscus in a capillary tube will be convex, if the angle of contact is :

0%
greater than $90^{o}$
0%
less than $90^{o}$
0%
equal to $90^{o}$
0%
equal to zero

Explanation

A convex meniscus occurs when the particles in the liquid have a stronger attraction to each other(cohesion) than to the material of the container.
It occurs when mercury is placed in a glass capillary.

The types of meniscus are shown in the figure.

In a convex meniscus, the angle of contact is greater than

$90^{\circ}$ , which is angle between tangent to the liquid surface and the capillary surface, inside the liquid.

A capillary tube when immersed vertically in a liquid rises to 3 cm. If the tube is held immersed in the liquid at an angle of 60

$^{o}$ with the vertical,the length of the liquid column along the tube will be:

0%
2 cm
0%
4.5 cm
0%
6 cm
0%
7.5 cm

Explanation

$h = \dfrac {2T cos \theta}{r \rho g}$
$h \alpha cos \theta$
for $\theta = 60^0 cos \theta = \dfrac {1}{2}$
$\therefore$ h is double $\Rightarrow h = 6 cm$ .

A vessel whose bottom has round holes with diameter of

$1mm$ is filled with water. Assuming that surface tension acts only at holes, then the maximum height to which the water can be filled in vessel without leakage is

(Given surface tension of water is

$75 \times 10 ^{-3} N/m$ and

$g =10m / s^{2}$ )

0%
$3 cm$
0%
$0.3 cm$
0%
$3 mm$
0%
$3 m$

Explanation

Using formula

$h = \dfrac{2T \cos\theta}{\rho rg}$
=

$\dfrac{2 \times 75 \times 10^{-3} \times 1}{1000 \times \dfrac{1}{2} \times 10^{-3} \times 10} \quad(\cos 0^o = 1)$

$= 0.03m = 3 cm$

Three tubes A, B, C are connected to a horizontal pipe in which liquid is flowing. The radii of the pipes at the joints of A, B and C are 2 cm, 1 cm and 2 cm respectively. The height of the liquid:

0%
in A is maximum
0%
in A and C is equal
0%
is same in all the three
0%
in A and B is same

Explanation

According to the Bernoulli's theorem, for the flow of liquid through a horizontal pipe

$P+\frac{1}{2}\rho V^2$ = constant
where variables have their usual meanings.
Hence, when the tubes A, B and C are connected to the pipe with different radii, the height of liquid in tubes is different and depends on radii of tubes in inverse proportion. Hence, height of the liquid in the tubes A and C is the same.

A water barrel having water up to depth '

$d$ ' is placed on a table of height '

$h$ '. A small hole is made on the wall of the barrel at its bottom. If the stream of water coming out of the hole falls on the ground at a horizontal distance '

$R$ ' from the barrel, then the value of '

$d$ ' is:

0%
$\dfrac{4h}{R^{2}}$
0%
$4hR^{2}$
0%
$\dfrac{R^{2}}{4h}$
0%
$\dfrac{h}{4R^{2}}$

Explanation

By Bernoulli's equation:

mgd =

$\dfrac{1}{2}mv^2$

v =

$\sqrt{2gd}$ ....... (1)

By Projectile motion equation:

$\dfrac{1}{2}gt^2 = h$

t =

$\sqrt{\dfrac{2h}{g}}$

Therefore,

$R = vt$

$\sqrt{2gd} \sqrt{\dfrac{2h}{g}}$

= 2

$\sqrt{dh}$

Hence d =

$\dfrac{R^2}{4h}$

Match List I with List - II

List - I	List - II
a) Meniscus of water in a glass Capillary tube	e) convex
b) Meniscus of water in silver capillary tube	f) do not over flow
c) Meniscus of mercury in glass capillary tube	g) flat
d) Water in glass capillary tube of insufficient length	h) concave

The correct match is :

0%
a -- h; b-- g; c-- e; d-- f
0%
a -- e; b-- f;c-- g; d -- h
0%
a -- f; b-- e; c--g; d-- h
0%
a-- h; b-- g; c-- f; d-- e

Four identical capillary tubes

$a, b, c$ and

$d$ are dipped in four beakers containing water with tube ‘

$a$ ’ vertically, tube ‘

$b$ ’ at

$30^{o}$ , tube ‘

$c$ ’ at

$45^{o}$ and tube ‘

$d$ ’ at

$60^{o}$ inclination with the vertical. Arrange the lengths of water column in the tubes in descending order.

0%
$d, c, b, a$
0%
$d, a, b, c$
0%
$a, c, d, b$
0%
$a, b, c, d$

Explanation

In capillary tube fluid always rise to the same vertical height as when the tube is perfectly vertical. So, the tube which is making greater angle with vertical will get more water in it.
So, order of lengths of water column will be

$d > c > b > a$ .

Rain drops fall with terminal velocity due to:

0%
Buoyancy
0%
Viscosity
0%
Low weight
0%
surface tension

The velocity of the wind over the surface of the wing of an aeroplane is 80 ms

$^{-1}$ and under the wing 60 ms

$^{-1}$ . If the area of the wing is 4m

$^{2}$ , the dynamic lift experienced by the wing is [ density of air

$=$ 1.3 kg. m

$^{-3}$ ]:

0%
3640 N
0%
7280 N
0%
14560 N
0%
72800 N

Explanation

Dynamic lift

$= \dfrac{1}{2}A\rho \left ( v^{2}_{2}- v^{2}_{1}\right )$

$= \dfrac{1}{2}\times 1.3\times \left ( 2800 \right )\times 4$

$= 7280$ N

A solid rubber ball of density 'd' and radius 'R' falls vertically through air. Assume that the air resistance acting on the ball is F

$=$ KRV where K is constant and V is its velocity. Because of this air resistance the ball attains a constant velocity called terminal velocity

$V_{T}$ after some time.Then

$V_{T}$ is:

0%
$\dfrac{4\pi R^{2}dg}{3K}$
0%
$\dfrac{3K}{4\pi R^{2}dg}$
0%
$\dfrac{4}{3}\dfrac{\pi r^{3}dg}{K}$
0%
$\pi rdgk$

Explanation

Let any time t, the velocity be V.
Acceleration is then

$\dfrac {mg - KRV}{m} = \dfrac {dV}{dt}$
but once it attains terminal velocity

$V_T$ ;

$\dfrac {dV}{dt} = 0$

$\Rightarrow mg = KRV_T$

$(\dfrac {4}{3} \pi R^3)dg = KRV_T$

$\Rightarrow V_T = \dfrac {4 \pi R^2dg}{3K}$

Assertion: A raindrop after falling through some height attains a constant velocity

Reason: At constant velocity, the viscous drag is equal to its weight.

0%
Both assertion and reason are true and reason is correct explanation of assertion.
0%
Both assertion and reason are true but reason is not the correct explanation of assertion.
0%
Assertion is true but reason is false.
0%
Both assertion and reason are false.

A drop of water of radius r is falling through the air of coefficient of viscosity

$\eta$ with a constant velocity of v. The resultant force on the drop is:

0%
$\dfrac{1}{6\pi \eta rv }$
0%
$6\pi \eta rv$
0%
$\sqrt{6\pi \eta rv }$
0%
zero

Read the following statements and pick the correct choice
Statement A: With increase in temperature, viscosity of a gas increases and that of a liquid decreases.
Statement B: If the density of a small sphere is equal to the density of the liquid in which it is dropped, then the terminal velocity of the sphere will be zero.

0%
Both A and B are true
0%
Both A and B are false
0%
A is true and B is false
0%
B is true but A is false.

A small ball is dropped in a viscous liquid. Its fall in the liquid is best described by the figure:

0%
Curve A
0%
Curve B
0%
Curve C
0%
Curve D

After terminal velocity is reached the acceleration of a body falling through a viscous fluid is :

0%
zero
0%
g
0%
less than g
0%
greater than g

Section-A	Section-B
a) Incompressible liquid	e) Density constant
b) Turbulent flow	f) Stream lines
c) Tube of flow	g) Constant
d) Fluid flux rate in laminar flow	h) Reynold's no $> 2000$

0%
a-f; b-e; c-g; d-h
0%
a-e; b-h; c-f; d-g
0%
a-g; b-f; c-e; d-h
0%
a-h; b-g; c-e; d-f

Explanation

Incompressibility means that the density of fluid is same at all the points and remains same as time passes.
The motion of water in high fall of at high speed is called turbulent flow. Reynold's no >

$2000$ for turbulent flow.
Streamline flow is flow of liquid in a line of flow.
In laminar flow, fluid flux rate is constant.

When a metallic sphere is dropped in a long column of a liquid, the motion of the sphere is opposed by the viscous force of the liquid. If the apparent weight of the sphere equals to the retarding forces on it, the sphere moves down with a velocity called:

0%
critical velocity
0%
terminal velocity
0%
velocity gradient
0%
constant velocity

A ball is dropped into coaltar. Its velocity-time curve will be

The reading of a pressure meter attached with a closed water pipe is 3.5 x 10

$^{5}$ N m

$^{-2}$ . On opening the valve of the pipe, the reading of pressure meter is reduced to 3 x 10

$^{5}$ N m

$^{-2}$ . Calculate the speed of water flowing in the pipe.

0%
$10 cm/s$
0%
$10 m/s$
0%
$0.1 m/s$
0%
$0.1 cm/s$

Explanation

Applying Bernoulli's principle when the pipe was closed and after it was opened,
$\Rightarrow P_{1}=P_{2}+\dfrac{1}{2}\rho V^{2}$
$\Rightarrow 3.5 \times 10^{5}= 3 \times 10^{5} + \dfrac{1}{2}\rho V^{2}$
$\Rightarrow \dfrac{0.5 \times 10^{5} \times 2}{10^{3}}=V^{2}$
$\Rightarrow 10^{2}= V^{2}\Rightarrow V=10 m/s$

The initial speed with which water strikes the ground in

$ms^{-1}$ is:

0%
$10$
0%
$5$
0%
$5\sqrt{2}$
0%
$10\sqrt{2}$

Explanation

Velocity of efflux for a given body =

$( \dfrac{ 2gh }{ 1 - \dfrac {a ^{2}} {A^{2}} } )^{0.5}$ where "h" is the height from the ground , 'A' is the area of the tank and "a" is the area of the orifice .

now here "h" = 10m , a =

$10^{-4} m^{2}$ , A =

$\pi r^{2}$ =

$\pi m^{2}$

$( 1 - \dfrac{ a^{2} }{ A ^{2} } )$ is approximately = 1 in this case

so velocity of efflux in this case =

$( 2(10)(10) )^ {\dfrac{1}{2} }$ =

$(200)^{0.5}$

so option (D) is correct

There is a hole at the side-bottom of a big water tank. The area of the hole is

$2\ mm^{2}$ . Through it, a pipe is connected. The upper surface of the water is

$5\ m$ above the hole. The rate of flow of water through the pipe is ( in

$m^{3}s^{-1}$ ) (

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

0%
$4 \times 10^{-5}$
0%
$4 \times 10^{5}$
0%
$4 \times 10^{-6}$
0%
$28 \times 10^{-6}$

Explanation

Velocity of water at the side bottom hole

$=\sqrt{2gh}$

$=\sqrt{2 \times 10 \times 5}=10 \ m/s$
Area of hole =

$4 mm^{2}= 4 \times 10^{-6}\ m^{2}$
So, rate of flow of water through pipe =

$4 \times 10^{-6} \times 10 \ m^{3}/s$

$=4 \times 10^{-5} \ m^{-3}/s$

At what speed will the velocity head of stream of water be equal to 40cm?

0%
280 ms $^{-1}$
0%
2.80 cm/s
0%
280 cms $^{-1}$
0%
0.280 m/s

Explanation

Velocity head

$=\dfrac{V^{2}}{2g}=40 cm$

$\Rightarrow V^{2}=40 \times 980 \times 2$

$\Rightarrow V=280 cm/s$

The level of water in a tank is 5m high. A hole of area of cross section 1 cm

$^{2}$ is made at the bottom of the tank. The rate of leakage of water from the hole in

$m^{3}s^{-1}$ is ( g

$=$ 10ms

$^{-2}$ ):

0%
$10^{-3}$
0%
$10^{-4}$
0%
$10$
0%
$10^{-2}$

Explanation

Velocity of water from the hole=

$\sqrt{2 \times 5 \times 10}=10 m/s$
So,

$Q=AV=1 \times 10^{-4} m^{2} \times 10m/s$

$= 10^{-3} m^{3}/s$

Calculate the velocity of efflux of kerosene oil from an orifice of a tank in which pressure is

$4 atm$ . The density of kerosene oil

$= 720 kg m^{-3}$ and

$1$ atmospheric pressure

$= 1.013 \times 10^{5}N m^{-2}$ .

0%
$3.355 m/s$
0%
$33.55 m/s$
0%
$335.5 m/s$
0%
$33.55 cm/s$

Explanation

From Bernoulli's Theorem,

$P_{1}=\dfrac{1}{2}\rho v^{2}$

$\Rightarrow 4 atm = \dfrac{1}{2}\times \rho V^{2}$

$\Rightarrow \dfrac{4 \times 1.013 \times 10^{5} \times 2}{720}=V^{2}$

$\Rightarrow V=33.549 m/s$

$\Rightarrow V \approx 33.55 m/s$

In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are

$70m s^{-1}$ and

$63 ms^{-1}$ respectively. What is the lift on the wing, if its area is

$2.5 m^{2}$ ? Take the density of air to be

$1.3kg m^{-3}$ .

0%
$1513 N$
0%
$1513\ dynes$
0%
$151.3 N$
0%
$151.3\ dynes$

Explanation

Applying Bernoulli's principle to above and below the using,

$P_{1}+\dfrac{1}{2}PV_{1}^{2}=P_{2}+\dfrac{1}{2}PV_{2}^{2}$

$\Rightarrow P_{1}-P_{2}=\dfrac{1}{2}\times 1.3 \times (70^{2}-63^{2})$

$\Rightarrow$ Pressure differnece across sing =

$605.15N/m^{2}$
Lift on the using = pressure difference across surface of using

$\times$ Area.

$= 605.15 N/m^{2} \times 2.5 m^{2}$

$=1512.875 N$

$\approx 1513 N$

The initial speed with which water flows out from the orifice in ms

$^{-1}$ is (g

$=$ 10ms

$^{-2}$ ):

0%
10
0%
5
0%
5. $\sqrt{2}$
0%
10. $\sqrt{2}$

Explanation

Speed at which water flows out from the orifice=

$\sqrt{2 \times g \times h}=\sqrt{5 \times 2 \times 10}$ =10 m/s

where h is the height of water in the tank.

In a horizontal pipe line of uniform cross-section,pressure falls by

$5\ Pa$ between two points separated by

$1\ km$ . The change in the kinetic energy per kg of the oil flowing at these points is :(Density of oil

$= 800\ kgm^{-3}$ )

0%
$6.25 \times 10^{-3}\ Jkg^{-1}$
0%
$5.25 \times 10^{-4}\ Jkg^{-1}$
0%
$3.25 \times 10^{-5}\ Jkg^{-1}$
0%
$4.25 \times 10^{-2}\ Jkg^{-1}$

Explanation

from Bernoulli's therom,

$P_{1}+\dfrac{1}{2}\rho V_{1}^{2}=P_{2}+\dfrac{1}{2}\rho V_{2}^{2}$

$\Rightarrow 5 Pa- \dfrac{1}{2} \times 800 \times (V_{2}^{2}-V_{1}^{2})$

$\Rightarrow \dfrac{100}{800}= V_{2}^{2}-V_{1}^{2}\Rightarrow 0.0125=V_{2}^{2}-V_{1}^{2}$

$\Rightarrow \dfrac{1}{2}m(V_{2}^{2}-V_{1}^{2})=\dfrac{1}{2} \times 1 \times0.0125$ [where m=1kgs]

$\Rightarrow$ Change in K.E per kg

$= 6.24 \times 10^{-3}/kh^{-1}$

A horizontal pipe of non uniform cross section has water flow through it such that the velocity is

$2ms^{-1}$ at a point where the pressure is

$40 kPa$ . The pressure at a point where the velocity of water flow is

$3 ms^{-1}$ is : ( in

$kPa$ )

0%
$27$
0%
$60$
0%
$37.5$
0%
$40$

Explanation

Using the bernoulli's therom,

$P_{1}+\dfrac{1}{2}pV_{1}^{2}=P_{2}+\dfrac{1}{2}pV_{2}^{2}$

$\Rightarrow 40 \times 10^{3}+\dfrac{1}{2} \times 500 \times 4=P_{2}+\dfrac{1}{2}\times{1000}\times 9$

$\Rightarrow 40 \times 10^{3}+2000 - 45000 =P_{2}$

$\Rightarrow 37.5 \times 10^{3} Pa= P_{2}$

$\Rightarrow P_{2}=37.5 kPa$

In a horizontal oil pipe line of constant cross sectional area the decrease of pressure between two points 100 km apart is 1500 pa. The loss of energy per unit volume per unit distance is ----Joule.

0%
15
0%
0.015
0%
0.03
0%
zero

Explanation

Decreases in pressure $10^{5}m$ \ apart $=1500pa$
energy cost per unit volume per unit distance $=\dfrac{1500\times 10^{5}A}{10^{5}\times A\times 10^{5}}=\dfrac{work done}{lenght\times total volume}$
$=0.015\ joules$

A room has a window of area

$A$ . Out side of the room wind is blowing parallel to the window with a velocity

$v$ . If the density of air is

$\rho$ , then the force acting on the window is:

0%
$\dfrac{1}{2}\dfrac{\rho v^{2}}{A}$
0%
$\dfrac{1}{2}\rho v^{2}A$
0%
$\rho v^{2}A$
0%
$2\rho v^{2}A$

Explanation

As the wind is blowing with velocity

$v$ outside the window so, pressure will decrease from Bernoulli's equation as following :
Applying bernouillis theorem across the window ,

$p_{1}+\cfrac{1}{2}\rho v^{2}=p_{2}$

$\Rightarrow p_{2}-p_{1}=\cfrac{1}{2}\rho v^{2}$
So, force on the window =pressure difference x Area

$= \Delta p\times A$

$=\cfrac{1}{2}\rho Av^{2}$

Water is maintained at a height of 10m in a tank.The diameter of a circular aperture needed at the base of the tank to discharge water at the rate of 26.4m

$^{3}$ min

$^{-1}$ is(Given that g

$=$ 9.8 m s

$^{-2})$ :

0%
0.2 cm
0%
2 m
0%
0.2 m
0%
2 cm

Explanation

Velocity of water from the bottom operture $=\sqrt{2gh}$
$=\sqrt{2 \times 9.8 \times 10}m/s$
So,

$Q=26.4 m^{3}=\pi r^{2}V$
$\Rightarrow 26.4 m^{3} = \pi \times r^{2} v$
$\Rightarrow \dfrac{26.4}{60}m^{3}/s = \pi r^{2} \times \sqrt{2 \times 9.8 \times 10}$
$\Rightarrow r^{2}=0.010004024$
$\Rightarrow r =0.1 m$

So, Diameter of aperture $=0.2\ m$

An aeroplane of mass

$6000 \ kg$ is flying at an altitude of

$3 \ km$ . If the area of the wings is

$30 \ m^{2}$ and the pressure at the lower surface of wings is

$0.6\times10^{5}\ pa$ , the pressure on the upper surface of wings is : ( in pascal) (

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

0%
$5.8 \times 10^{4}$
0%
$6.8 \times 10^{4}$
0%
$7.8 \times 10^{4}$
0%
$8.8 \times 10^{4}$

Explanation

Pressure at lower surface of wing

$=0.6\times 10^{5}Pa$
Let pressure at upper surface of wing

$=p$
Weight of the airplane

$=5\times 10^{4}N$
For equilibrium, Weight

$=$ upthrust due to air

$\Rightarrow 5\times 10^{4}=\left ( 0.6\times 10^{5}-p \right )\times 50$

$\Rightarrow 10^{3}=0.6\times 10^{5}-p$

$\Rightarrow p=0.6\times 10^{5}-10^{3}$

$\Rightarrow p=59\times 10^{3}Pa$

The pressure at the top of a building of height

$200m$ is

$500kPa$ . The minimum pressure required at the base of the building to pump water to a closed tank on the top of the building is(

$g=10m/s^{2}$ )

0%
$2.5 \times 10^{4}Pa$
0%
$2.5 \times 10^{5}Pa$
0%
$2.5 \times 10^{6} Pa$
0%
$2.5 \times 10^{3} Pa$

Explanation

So,
Minimum pressure required at the base of building

$=500\times 10^{3}+1000\times 10\times 200$

$=500\times 10^{5}+20\times 10^{5}=25\times 10^{5}Pa$

$=2.5\times 10^{6}Pa$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 2 - 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 2 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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