MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 12

Find the speed with which water comes out of the hole.

0%
$\dfrac { 1 }{ \rho } { \left[ { p }_{ 0 }-\rho g({ h }_{ 1 }-2{ h }_{ 0 }) \right] }^{ 1/2 }$
0%
${ \left[ \dfrac { 2 }{ \rho } { \left[ { p }_{ 0 }+\rho g({ h }_{ 1 }-{ h }_{ 0 }) \right] } \right] }^{ 1/2 }$
0%
${ \left[ \dfrac { 3 }{ \rho } { \left[ { p }_{ 0 }+\rho g({ h }_{ 1 }+{ h }_{ 0 }) \right] } \right] }^{ 1/2 }$
0%
${ \left[ \dfrac { 4 }{ \rho } { \left[ { p }_{ 0 }-\rho g({ h }_{ 1 }-{ h }_{ 0 }) \right] } \right] }^{ 1/2 }$

Explanation

KE of water=pressure energy of the water at the layer

$\cfrac { 1 }{ 2 } m{ v }^{ 2 }=$ Pressure

$\times$ volume

$\cfrac { 1 }{ 2 } m{ v }^{ 2 }=\cfrac { m\times P }{ \rho }$

${ v }^{ 2 }=\cfrac { 2P }{ \rho } =\left[ \cfrac { 2 }{ \rho } \left[ { P }_{ 0 }+\rho g\left( { h }_{ 1 }-{ h }_{ 0 } \right) \right] \right]$

$v=\sqrt { \cfrac { 2 }{ \rho } \left[ { P }_{ 0 }+\rho g\left( { h }_{ 1 }-{ h }_{ 0 } \right) \right] }$

Figure shows a crude type of atomizer. When bulb

$A$ is compressed, air flows swiftly through tube

$BC$ causing a reduced pressure in the particle of the vertical tube. Liquid rises in the tube, enters

$BC$ and is sprayed out. If the pressure in the bulb is

${ P }_{ a }+P,$ where

$P$ is the gauge pressure and

${ P }_{ a }$ is the atmosphere pressure,

$v$ is the speed of air in

$BC$ . Density of air

$=1.3\ kg/{ m }^{ 3 }$ .

0%
$\dfrac { P+\rho gh }{ \sqrt { 0.65 } }$
0%
$\dfrac { Ph\rho g }{ \sqrt { 0.65 } }$
0%
$\dfrac { P\sqrt { h\rho g } }{ 0.65 }$
0%
$\dfrac { P }{ h\rho g\sqrt { 0.65 } }$

What is the reading of the scale when the ball is fully immersed?

0%
$W-\rho Vg$
0%
$W$
0%
$W + mg -\rho Vg$
0%
None

A large open top container of negligible mass and uniform cross-sectional area-A has a small hole of cross-sectional area A/100 in its side wall near the bottom. The container is kept on a smooth horizontal floor and contains a liquid of density

$\rho$ and mass

$m_0$ . Assuming that the liquid starts flowing out horizontally through the hole at t = 0, calculate the acceleration of the container

0%
$\dfrac{g}{50}$
0%
$\dfrac{g}{60}$
0%
$\dfrac{g}{70}$
0%
$\dfrac{g}{80}$

A fixed thermally conducting cylinder has a radius

$R$ and height

$L_{0}$ . The cylinder is open at its bottom and has a small hole at its top. A piston of mass

$M$ is held at a distance

$L$ from the top surface as shown in Fig. The atmospheric pressure is

$p_{0}$ .
The piston is taken completely out of the cylinder. The hole at the top is sealed. A water tank is brought below the cylinder and put in a position so that the water surface in the tank is at the same level as the top of the cylinder as shown in Fig. The density of the water is

$\rho$ . In equilibrium, the height

$H$ of the water column in the cylinder satisfies.

0%
$\rho g(L_{0} - H)^{2} + P_{0}(L_{0} - H) + L_{0}P_{0} = 0$
0%
$\rho g(L_{0} - H)^{2} - P_{0} (L_{0} - H) - L_{0}P_{0} = 0$
0%
$\rho g(L_{0} - H)^{2} + P_{0}(L_{0} - H) - L_{0}P_{0} = 0$
0%
$\rho g(L_{0} - H)^{2} - P_{0}(L_{0} - H) + L_{0}P_{0} = 0$

Explanation

$p_{1} = p_{2}$

$p_{0} + \rho g(L_{0} - H) = p ... (i)$
Now, applying

$p_{1}V_{1} = p_{2}V_{2}$ for the air inside the cylinder, we have

$p_{0}(L_{0}) = p(L_{0} - H)$

$p = \dfrac {p_{0}L_{0}}{L_{0} - H}$
Substituting in Eq. (i), we have

$p_{0} + \rho g (L_{0} - H) = \dfrac {p_{0}L_{0}}{L_{0} - H}$

$\Rightarrow \rho g (L_{0} - H)^{2} + P_{0} (L_{0} - H) - L_{0}P_{0} = 0$
Therefore, option (c) is correct.
1572311_1010878_ans_6e9fa9019a9e423199fd34e8e5e8f0e1.png

The excess pressure inside on soap bubble is three times inside that of the second soap bubble.The ratio of their surface area is:

0%
1:9
0%
1:3
0%
3:1
0%
1:27

A capillary tube

$(A)$ is dropped 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?

Water pressure at the bottom of centre of the box is equal to (atmospheric pressure =

${10}^{5}\ N/{m}^{2}$ , density of water=

$1000\ kg/{m}^{3}, g= 10\ m/{s}^{2}$ )

0%
$1.1\ MPa$
0%
$0.11\ MPa$
0%
$0.101\ MPa$
0%
$0.021\ MPa$

The area for two holes

$A$ and

$B$ are

$2a$ and

$a$ , respectively. The holes are at height

$(H/3)$ and

$(2H/3)$ for the surface of water. Find the correct option(s):

0%
The velocity of efflux at hole $B$ is $2$ times the velocity of efflux at hole $A$
0%
The velocity of efflux at hole $B$ is $\sqrt { 2 }$ times the velocity of efflux at hole $A$
0%
The discharge is same through both the holes.
0%
The discharge through hole $A$ is $\sqrt { 2 }$ time the discharge through hole $B$

A vertical cylinder is filled with liquid. A small hole is made in the wall of the cylinder at a depth

$H$ below the free surface of the liquid. The force exerted on the cylinder by the liquid flowing out of the hole initially will be proportional to :

0%
$\sqrt {H}$
0%
$H$
0%
$H^{3/2}$
0%
$H^2$

Explanation

By using Bernoulli’s principle

The velocity of the liquid coming out is given as,

$v = \sqrt {2gh}$

The mass of the liquid will be

$m = \rho AV$

The change in momentum will be

$P = mv$

The force will be proportional of ${v^2}$ which in turn is proportional to $h$

A small steel ball falls through a syrup at a constant speed of

$1.0 m/s$ . If the steel ball is pulled upwards with a force equal to twice its effective weight, how fast will it move upward?

0%
$3m/s$
0%
$1m/s$
0%
$2m/s$
0%
$5m/s$

Water flows in a stream line manner through a capillary tube of radius a. The pressure difference being P and the rate of flow is Q. If the radius is reduced to

$\cfrac { a }{ 4 }$ and the pressure is increased to 4P, then the rate of flow becomes

0%
$4Q$
0%
$\cfrac { Q }{ 2 }$
0%
$Q$
0%
$\cfrac { Q }{ 64 }$

If the pressure difference between the two points of a horizontal pipe is

$0.1 p_{atm}$ , then the velocity water flow in the pipe is :

0%
$\sqrt{10} ms^{-1}$
0%
$\sqrt{20} ms^{-1}$
0%
$10 ms^{-1}$
0%
$14 ms^{-1}$

Explanation

The difference in pressure is given as,

$\Delta P = \rho gh$

$0.1 \times {10^5} = \rho gh$

The velocity of water flow is given as,

$v = \sqrt {2gh}$

$v = \sqrt {2g \times \dfrac{{0.1 \times {{10}^5}}}{{{{10}^3} \times g}}}$

$v = \sqrt {20} \;{\rm{m/s}}$

The angle of contact at the interface of water glass is

${0^ \circ }$ , Ethyl-alcohol glass is

${0^ \circ }$ Mercury glass is

${140^ \circ }$ and Methyl iodide is

${30^ \circ }$ . A glass capillary is put in a trough containing one of these for liquids it is observed that the meniscus is convex. the liquid in the trough is :

0%
Water
0%
Ethyl alcohol
0%
Mercury
0%
Methyl iodide

Explanation

Given , angle of contact $(θ) = 140°$

Radius of tube $(r) = 1 mm = 10⁻³ m$

Surface tension $(S) = 0.465 N/m$

Density of mercury $(ρ) = 13.6 × 10³ kg/m³$

Height of liquid rise or fall due to surface tension $(h)$

$= 2S cos θ/rρg = 2 × 0.465 × cos 140°/1 × 10⁻³ × 13.6 × 10³ × 9.8$

$= 2 × 0.465 × (- 0.7660) / 10⁻³ × 13.6 × 10³ × 9.8$

$= - 5.34 mm$

Hence, the mercury level will depressed by $5.34 mm.$

A glass capillary is put in a trough containing one of these for liquids it is observed that the meniscus is convex. the liquid in the trough is Ethyl alcohol.

A tank is filled with two immiscible liquids of densities

$2\rho$ and

$\rho$ each of height

$h$ . Two holes are made to the side wall at

$\dfrac{h}{2}$ and

$\dfrac{3h}{2}$ from upper surface of the liquid, then the ratio of velocity of efflux of the liquids through the holes

0%
$\dfrac{\sqrt2}{3}$
0%
$\dfrac{\sqrt3}{1}$
0%
$\dfrac{3}{\sqrt2}$
0%
$\dfrac{1}{\sqrt2}$

Explanation

Let the velocity at hole is

${ v }_{ 1 }$

Height is

$\dfrac { h }{ 2 }$

Therefore velocity is,

${ v }_{ 1 }=\sqrt { 2g\dfrac { h }{ 2 } } \\$

Now pressure at hole 2 is

$h\rho g+\dfrac { h }{ 2 } \times 2\rho g$

Applying Bernoulli's theorem,

${ P }_{ o }+2h\rho g={ P }_{ o }+\dfrac { 1 }{ 2 } (2\rho ){ v }_{ 2 }^{ 2 }$

where

${ P }_{ o }$ is the atmospheric pressure

Velocity at hole 2 is

${ v }_{ 2 }=\sqrt { 2g\dfrac { 3h }{ 2 } } =\sqrt { 3gh }$

Now,

$\dfrac { { v }_{ 2 } }{ { v }_{ 1 } } =\dfrac { \sqrt { 3gh } }{ \sqrt { gh } }$

$\dfrac { { v }_{ 2 } }{ { v }_{ 1 } }$ =

$\sqrt { 3 } :1$

A cylindrical vessel has some water in it. A small hole at the bottom is now opened. The ratio of times it takes to become

$75\%$ empty to completely empty is

0%
$3:4$
0%
$1:2$
0%
$1:3$
0%
$2:3$

Water is filled upto same height in two identical closed containers

$A$ and

$B$ . Container

$A$ has vacuum over the water while container

$B$ has air over the water. At the same depth of both the containers there is an opening on which identical balloons

$A$ and

$B$ are attached as shown in the figure given below. Then

0%
Balloon - $A$ will bulge more than balloon - $B$
0%
Balloon - $B$ will bulge more than balloon - $A$
0%
Both the balloons will bulge equally
0%
None of the balloons will bulge

Explanation

Solution:- (A) Balloon

$A$ will bulge more than balloon

$B$ .

Since the container

$A$ has vacuum over the water, thus water will cover the vacuum area first and hence the balloon

$A$ will bulge more than balloon

$B$ .

In a vessel of water a hole is made at a depth of

$3.5\ m$ from the free surface. The velocity of efflux will be:

0%
$8.4\ m/s$
0%
$84\ m/s$
0%
$8.4\ cm/s$
0%
$84\ cm/s$

$A$ denotes the area of free surface of a liquid and

$h$ the depth of an orifice of area of cross-section

$a$ , below the liquid surface, then the velocity

$v$ of flow through the orifice is given by:

0%
$v = \sqrt {\left( {2gh} \right)}$
0%
$v = \sqrt {\left( {2gh} \right)} \sqrt {\left( {\frac{{{A^2}}}{{{A^2} - {a^2}}}} \right)}$
0%
$v = \sqrt {\left( {2gh} \right)} \sqrt {\left( {\frac{A}{{A - a}}} \right)}$
0%
$v = \sqrt {\left( {2gh} \right)} \sqrt {\left( {\frac{{{A^2} - {a^2}}}{{{A^2}}}} \right)}$

Explanation

$v_{1}=$ velocity of surface of liquid

$v_{2}=$ velocity of liquid far om orifice

Acc. to continuity theorem.

$A v_{1}=a v_{2} \Rightarrow \dfrac{A v_{1}}{a}=v_{2}$

$\Rightarrow \dfrac{a v_{2}}{A}=v_{1}$

Acc. to Bernoulli's Theorem

$P_{0}+\rho g h+\dfrac{1}{2} \rho v_{1}^{2}=P_{0}+\rho g(0)+\dfrac{1}{2} \rho v_{L}^{2}$

$\rho g h+\dfrac{1}{2} \rho\left[\dfrac{a^{2} v_{1}^{2}}{A^{2}}-v_{1}^{2}\right]=0$

$\dfrac{v_{1}^{2}}{2}\left[\dfrac{A^{2}-a^{2}}{A^{2}}\right]=g h$

$v_{1}=\sqrt{2 g h} \sqrt{\dfrac{A^{2}}{A^{2}-a^{2}}}$

Answer: $\sqrt{2 g h} \sqrt{ \dfrac{A^{2}}{A^{2}-a^{2}}}$

The angle of contact at the interface of water-glass is

$0^\circ$ . Ethyl alcohol-glass is

$0^\circ$ . Mercury-glass is

$140^\circ$ and Methyl iodide-glass is

$30^\circ$ . A glass capillary is put in trough containing one of these four liquids. It is observed that the meniscus is convex. The liquid in the trough is:

0%
water
0%
ethylalcohol
0%
mercury
0%
methyliodide

Explanation

Given ,

Angle of contact

$(\theta)=140^o$

Radius of tube

$(r)= 1 mm =10^{-3}m$

Surface tension

$(S) = 0.465N/m$

Density of mercury

$(ρ) = 13.6 \times 10^{3} kg/m^3$

Height of liquid rise or fall due to surface tension (h)

$=\dfrac{2Scos{\theta}}{rρg}$

$=\dfrac{2 \times 0.465 \times cos140^o }{1×10^{-3} \times 13.6 \times 10^{3} \times 9.8}$

$=2 \times 0.465 \times (−0.7660)/10^{-3} \times 13.6 \times 10^{3} \times 9.8$

$=−5.34\,mm$

Hence, the mercury level will depressed by

$5.34$ mm.

The meniscus of liquid in a capillary tube will be convex upward if angle of contact is obtuse. It is so when one end of glass capillary tube is immersed in a though of mercury.

A balloon of volume

$1500{m}^{3}$ and weighing

$1650 kg$ with all its equipment is filled with helium (density

$0.2 kg/{m}^{3}$ ). if the density of air is

$1.3 kg/{m}^{3}$ , the pull on the rope tied to the ballon will be

0%
$300 kg$
0%
$16500 kg$
0%
$1950 kg$
0%
$zero$

Calculate the mass of lead which will be required to just submerge in water a block of wood weighing

$50$ g (a) when lead is attached underneath, (b) when lead is placed on the top of the block. Take sp. gravity of lead =

$11.3$ and of wood =

$0.75$ .

0%
(a) $28.28$ g
(b) $16.67$ g
0%
(a) $18.28$ g
(b) $26.67$ g
0%
(a) $18.28$ g
(b) $16.67$ g
0%
(a) $18.28$ g
(b) $50.67$ g

Water flows through a tube as shows below. Then

0%
h is measure of pressure difference at M and N
0%
Pressure at M> pressure at N
0%
Velocity at N> velocity at M
0%
h is direct measure of difference in velocity

An ice cube of volume

$1\ m^{3}$ is kept in a gravity free room. when ice has melted, the surface area of ice is:

0%
$(9\ \pi)^{1/3}$
0%
$(18\ \pi)^{1/3}$
0%
$(36\ \pi)^{1/3}$
0%
$(27\ \pi)^{1/3}$

A stream of water escapes from a hole

$2\ m$ above the base of a tank and strikes the ground at a horizontal distance of

$6\ m$ . The depth of the hole from the free surface of water is

0%
$4\ m$
0%
$4.5\ m$
0%
$9\ m$
0%
$6\ m$

Explanation

$H=\text { height of Beaker } \\$

$h=\text { height of hole from Bottom } \\$

$R=6 \mathrm{~m} \\$

$\text { We know that }$

$R=\sqrt{4 h(H-h)} \\$

$6=\sqrt{4 \times 2(H-2)} \\$

$36=8(H-2) \\$

$(H-2)=4.5$

height of hole from

Surface

$=4.5 \mathrm{~m}$

Answer:-(B)

$\begin{array} { l } { \text { A soap film is formed on a triangular wire frame having } } \\ { \text { two fixed and one movable side. The free side is } } \\ { \text { moved with a constant acceleration 'a'. Take side is } } \\ { \text { length to be zero at } t = 0 \text { as shown. If the surface } } \\ { \text { tension is constant and } U \text { is the surface potential } } \\ { \text { energy of the soap film then } } \end{array}$

0%
$\frac { d U } { d t } \propto t$
0%
$\frac { d U } { d t } \propto t ^ { 2 }$
0%
$\frac { d U } { d t } \propto t ^ { 3 }$
0%
$\frac { d U } { d t } \propto t ^ { 0 }$

if the blood vessels in a human being acted as simple pipes what would be the difference in blood pressure between the blood in a

$1.8\ m$ tall man's feet and in his head when he is standing? Assume the spacific gravity of blood to be

$1.06$ .

0%
$16.239kPa$
0%
$18.7kPa$
0%
$17.3kPa$
0%
$15.5kPa$

Two water drops of radii

$R_{1}$ and

$R_{2}$ are falling with their terminal velocity. If some how gravity is switched off, what is the ratio of their initial accelerations (Just after gravity is switched off).

0%
$R_{1}^{3} : R_{2}^{3}$
0%
$R_{1}^{2} : R_{2}^{2}$
0%
$R_{1} : R_{2}$
0%
$1 : 1$

The value of

$l$ is

0%
$2\ cm$
0%
$4\ cm$
0%
$8\ cm$
0%
$6\ cm$

An air bubble rises from the bottom of a lake to the surface if its radius increases by 200% and atmospheric pressure are equal to a water column of height H, then depth of the lake is (assume Temperature to be constant at top and bottom of the lake)

0%
21 H
0%
8 H
0%
9 H
0%
26 H

A small steel ball of mass m and radius r is falling under gravity through a viscous liquid of coefficient of viscosity

$\eta$ . If g is the value of acceleration due to gravity, then the terminal velocity of the ball is proportional to (ignore buoyancy)

0%
$\dfrac{mg\eta}{r}$
0%
$mg\eta r$
0%
$\dfrac{mg}{r}$
0%
$\dfrac{mg}{r\eta}$

The given table shown properties of four cells system

$A, B, C$ and

$D$ . The maxium rate of inward diffusion of water will be observed in which of these system?

System	Intracellular concentration of water	Extracellular concentration of water
A	0.09M	0.11M
B	0.2 M	0.5 M
C	0.05M	0.7M
D	0.03M	0.6M

0%
System A
0%
System B
0%
System C
0%
System D

A rain drop of radius

$1.5$ mm, experience a drag force

$F=(12\times10^{-5}v)N$ , while falling through air from a height

$2$ km, with a velocity v. The terminal velocity of the rain drop will be nearly

$(use {\,}g=10m/s^2)$

0%
$200$ m/s
0%
$80$ m/s
0%
$7$ m/s
0%
$3$ m/s

A tube has two area of cross-section as shown in figure. The diameters of the tube are 8 mm and 2 mm. Find range of water falling on horizontal surface, if piston is moving with a constant velocity of 0.25 m/s, h = 1.25 m (g = 10

$m/s^2$ ).

0%
$2 m$
0%
$4 m$
0%
$6 m$
0%
$8 m$

A water tank is filled with water up to a height

$8 \:m$ a hole is made in the tank wall at a depth

$3\: m$ from the surface of water. Then the time taken to reach the ground

0%
$1 sec$
0%
$2 sec$
0%
$3 sec$
0%
$4 sec$

There is a hole at the bottom of a large open vessel. If water is filled upto a height

$h$ , it flows out in time

$t$ . If water is filled to a height

$4h$ , it will flow out in time.

0%
$\dfrac {t}{3}$
0%
$\dfrac {t}{4}$
0%
$3t$
0%
$2t$

Explanation

$\text { We know that } \\$

$V_{i}=\sqrt{2 g h} \\$

$\text { and } \\$

$V_{f}=0$

$\text { so } \\$

$V_{f}=V_{i}-g t \\$

$V_{i}=g t$

$t=\sqrt{\frac{2 h}{g}}$

$t_{2}=\sqrt{\frac{2 \times 4 h}{g}}$

$t_{2}=2 t$

Answer:-

$(D)$

Given figure shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height

$h_0$ and pressure

$2P_0$ where

$P_0$ is the atmospheric pressure. There is a hole in the wall of the tank at a depth

$h_1$ below the top from which water comes out. A long vertical tube is connected as shown
Find the speed with which water comes out of the hole

0%
$\frac{1}{\rho} [ P_0 -\rho g (h_1 - 2h_0)]^{1/2}$
0%
$[\frac{2}{\rho} [ P_0 +\rho g (h_1 - h_0)]]^{1/2}$
0%
$[\frac{3}{\rho} [ P_0 + \rho g (h_1 + h_0)]]^{1/2}$
0%
$[\frac{4}{\rho} [ P_0 -\rho g (h_1 - h_0)]]^{1/2}$

A water tank is filled with water up to a height H hole is made in the tank wall at a depth h from the aurface of water. Then the volcity of efflux

0%
$\sqrt{gh}$
0%
$\sqrt{2gh}$
0%
$2 gh$
0%
$pgh$

Explanation

Acc. to bernoulli's Theorem.

$P_{0}+\rho g H+0=P_{0}+0+\frac{1}{2} \rho v^{2}$

So,

$v=\sqrt{2 g H}$

Answer:

$(B) \sqrt{2 g h}$

Given figure shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height

$h_0$ and pressure

$2P_0$ where

$P_0$ is the atmospheric pressure. There is a hole in the wall of the tank at a depth

$h_1$ below the top from which water comes out. A long vertical tube is connected as shown
Find the height

$h_2$ of the water in the long tube above the top initially

0%
$\dfrac{3P_0}{\rho g} - \dfrac{h_0}{3}$
0%
$\dfrac{2P_0}{\rho g} - \dfrac{h_0}{2}$
0%
$\dfrac{P_0}{\rho g} - h_0$
0%
$\dfrac{P_0}{2\rho g} - 2h_0$

Two balloons are blown into spherical shape of unequal sizes and are connected through a narrow tube as shown in the figure. Then

0%
the smaller balloon becomes bigger
0%
the bigger balloons becomes smaller
0%
there is no change in their sizes
0%
the smaller becomes smaller and the bigger becomes higher

In a cylindrical vessel containing liquid of density

$p$ , there are two holes in the side wall at heights

$h_{1}$ and

$h_{2}$ respectively such that the range of efflux at the bottom of the vessel is same. The height of a hole, for which the range of efflux would be maximum, will be:

0%
$(h_{2}-h_{1})$
0%
$\dfrac{h_{2}-h_{1}}{2}$
0%
$h_{1}+h_{2}$
0%
$\dfrac{h_{1}+h_{2}}{2}$

Explanation

let height of vessel = H.

from, Bernoulli's theorem.

velocity of liquid at

$h_{1} ,V_{1}=\sqrt{2 g\left(H-h_{1}\right)}$

Range

$R_{1}=v_{1} t$

for

$t$ ,

$h_{1}=\frac{1}{2} \times g \times t^{2}$

$t=\sqrt{\frac{2 h_{1}}{g}}$

$R_{1}=V_{1} t=2 \sqrt{(H-h)} h_{1}$

For

$h_{2}$ height, Range

$R_{2}=2 \sqrt{\left(H-h_{2}\right) h_{2}}$

$\because R_{1}=R_{2}$

$2 \sqrt{\left(H-h_{1}\right) h_{1}} =2 \sqrt{\left(H-h_{2}\right) h_{2}}$

$H\left(h_{1}-h_{2}\right)=\left(h_{1}+h_{2}\right)\left(h_{1}-h_{2}\right)$

$H=h_{1}+h_{2}$

now, let

$h_{1}$ height is

$'y'$ height below than

$'H'$

$R=2 \sqrt{(H-y) y}$

$\dfrac{d R^{2}}{d y}=\dfrac{d}{d y}\left(H y-y^{2}\right)$

$\quad$

$\quad H-2 y=0$

$y=H / 2$

for

$\max$ value

$y=\dfrac{h_{1}+h_{2}}{2}$

Correct option is (D).

In a horizontal pipe line of uniform cross section as pressure falls between two points separated by certain distance, the change in the kinetic energy of the oil flowing between these two points is

$0.01 J \mathrm { kg }$ and density of the oil is

$800\mathrm { kg } \mathrm { m } ^ { - 3 }.$ Then the fall in pressure between those two points is

0%
$2 N m ^ { - 2 }$
0%
$4 N m ^ { - 2 }$
0%
$6 N m ^ { - 2 }$
0%
$8 N m ^ { - 2 }$

A vision liquid flows through a horizontal pipe of varying cross-sectional area. Identify the option which correctly represents the variation of height of rise of liquid in each vertical tube

0%
0%
0%
0%
None of these

Water flows through a horizotal pipe of varying cross-section. The pressure of water equals to 0.1 m of mercury at a place where the velocity of flow is 0.4

$ms^{-1}$ . What will be the pressure at another place where the velocity of flow is 0.5

$ms^{-1}$ ?

0%
0.08966 m of Hg column.
0%
0.09966 m of Hg column.
0%
0.09977 m of Hg column.
0%
0.09999 m of Hg column.

A vessel with water is placed on a weighing pan and reads

$600$ g.Now a ball of

$40$ g and density 0.80g/cc is sunk into the water with a pin as shown in fig., keeping it sunk. The weighing pan will show a reading .

0%
$600$ g
0%
$550$ g
0%
$650$ g
0%
$632$ g

The reading of pressure-meter attached with a closed pipe is

$3.5 \times 10 ^ { 5 } \mathrm { nm } ^ { - 2 }$ . On opening the valve of the pipe the reading of the pressure-meter is reduced to

$3 \times 10 ^ { 5 } \mathrm { nm } ^ { - 2 }$ . What is the speed of water flowing in the pipe?

0%
$20 \mathrm { ms } ^ { - 1 }$
0%
$10 \mathrm { ms } ^ { - 1 }$
0%
$15 \mathrm { cms } ^ { - 1 }$
0%
$10 \mathrm { cms } ^ { - 1 }$

Eight drops of water of

$0.6\mathrm { mm }$ radius each merge to form one big drop. If the surface tension of water is

$0.072\mathrm { N } / \mathrm { m } ,$ the energy dissipated in the process is

0%
$16 \pi \times 10 ^ { - 7 } \mathrm { J }$
0%
$4.15 \times 10 ^ { - 7 } \mathrm { J }$
0%
$2.075 \pi \times 10 ^ { - 7 } J$
0%
$8 \pi \times 10 ^ { - 7 } \mathrm { J }$

A tank containing water has an orifice in one vertical side. It is 10 m below the level of water in tank. Then velocity of efflux from that small orifice is

0%
${ 7ms }^{ -1 }$
0%
${ 3.5ms }^{ -1 }$
0%
${ 9.8ms }^{ -1 }$
0%
${ 14ms }^{ -1 }$

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

0%
obtuse
0%
parabolic
0%
hemi-spherical
0%
cylindrical

An ideal liquid of density

$p$ is following through a uniform horizontal tube

$OPQ$ of cross sectional area

$A$ with uniform speed

$V$ as shown in figure. The magnitude of force exerted by liquid on the tube is

0%
${ pAV }^{ 2 }$
0%
$2{ pAV }^{ 2 }$
0%
$\sqrt { 3p } { AV }^{ 2 }$
0%
$2\sqrt { 3p } { AV }^{ 2 }$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 12 - 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 12 - MCQExams.com

0:0:2

Practice Class 11 Engineering Physics Quiz Questions and Answers

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