MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 4

The area of cross-section of the wider tube shown in figure is

$800{ cm }^{ 2 }$ . if mass of

$12 kg$ is placed on the massless piston, the difference in heights h in the level of water in the two tubes is:

0%
$10 cm$
0%
$6 cm$
0%
$15 cm$
0%
$2 cm$

Explanation

Given that

$A = 800 cm^3$

$m = 12 Kg =12000 g$
Here, the force on piston is

$F = mg$
Hence, Increase in pressure on the liquid in the wider tube is

$P = \dfrac{F}{A} = \dfrac{mg}{A}$
since, h is the difference in level of water in the two tubes, thus

$P = h\rho g$

$h = \dfrac{P}{\rho g}$

$h = \dfrac{mg}{A\rho g} = \dfrac{m}{A\rho}$

$h = \dfrac{12000}{800}$ .......(

$\because \rho = 1$ )

$\therefore h = 15 cm$

The diagram (fig.) shows a venturimeter, through which water is flowing. The speed of water at X is 2 cm/sec. The speed of water at Y (taking

$g=1000 cm/sec^2)$ is:

0%
23 cm/sec
0%
32 cm/sec
0%
101 cm/sec
0%
1024 cm/sec

Explanation

$p_x - p_y = \dfrac{\rho}{2}(v_y^2 - v_x^2) \\\Rightarrow \Delta h \rho g=\dfrac{\rho}{2}(v_y^2 - v_x^2) \\\Rightarrow \Delta h g=\dfrac{1}{2}(v_y^2 - v_x^2) \\\Rightarrow 0.51 \times 1000=\dfrac{1}{2}(v_y^2 - 2^2) \\\Rightarrow 1020= v_y^2 - 4 \\\Rightarrow 1024= v_y^2 \\\Rightarrow v_y=32 cm/s$

Which of the following is the incorrect graph for a sphere falling in a viscous liquid? (Given at

$t=0$ , velocity

$v=0$ and displacement

$x=0$ )

Explanation

A sphere falling in a viscous liquid increases its speed with time and finally achieves a terminal velocity at infinite time.

Velocity varies with time as shown.

Net acceleration of the sphere is (

$mg-$ Buoyant force-Viscous Force) downwards

$a=d\dfrac{4}{3}\pi R^3 g-\rho\dfrac{4}{3}\pi R^3 g-6\pi\eta Rv$

Hence a varies linearly with velocity. Hence option C is the correct answer.

A long capillary tube of radius '

$r$ ' initially just vertically completely imerged inside a liquid of angle of contact

${ 0 }^{ \circ }$ . If the tube is slowly raised then relation between radius of curvature of miniscus inside the capillary tube and displacement

$(h)$ of tube can be represented by

Explanation

The curvature equation relating the radius of meniscus inside the capillary tube and the displacement of the tube is given by,

$H = \frac {1}{2} \times (\frac {1}{{R}_{1}} + \frac {1}{{R}_{2}})$

For a hemispherical, meniscus (which is the case for

${0}^{o}$ contact angle liquid,

${R}_{1} = {R}_{2} = r$

Substituting the values in the equation,

H =

$\frac {1}{r}$

Thus, the graph should resemble a hyperbolic parabola with a value r at a displacement h.

Hence, option B

A solid metallic sphere of radius

$r$ is allowed to fall freely through air. If the frictional resistance due to air is proportional to the cross-sectional area and to the square of the velocity, then the terminal velocity of the sphere is proportional to which of the following?

0%
${ r }^{ 2 }$
0%
$r$
0%
${ r }^{ { 3 }/{ 2 } }$
0%
${ r }^{ { 1 }/{ 2 } }$

Explanation

The frictional resistance

$f_r$ is given as:

$f_r\propto A$ and

$f_r\propto v^2$

$f_r=kAv^2$ (here k is a constant)

The downwards force is given as:

$mg=\displaystyle\frac{4}{3}\pi r^3\rho g$

Balancing the above we get the terminal velocity as:

$\displaystyle\frac{4}{3}\pi r^3\rho g=k(\pi r^2)v^2$

$r\propto v^2$

$v\propto r^{1/2}$

A tube is attached as shown in closed vessel containing water. The velocity of water coming out from a small hole is

0%
$\sqrt { 2 } m/s$
0%
$2 m/s$
0%
Depends on pressure of air inside vessel
0%
None of these

Explanation

Hint: Use Torricelli's equation for velocity

$\textbf{Step1: Torricelli's equation for velocity}$

Torricelli's theorem states that speed v by which water comes out from the tank is proportional to the square toot of twice the acceleration created by gravity and vertical height between surface and center of the opening.

$V = \sqrt{2\ *\ g\ *\ h}$

$\textbf{Step2: Calculation of velocity of water coming out}$

On putting the values, g = 10 $m/s^{2}$ and h = 20 cm = 0.2 m.

$V = \sqrt{2\ *\ 10\ *\ 0.2}$

= 2 m/s

Answer:

Hence, option B (2 m/s) is the correct answer.

The displacement of a ball falling from rest in a viscous medium is plotted against time. Choose a possible option:

A water barrel stands on a table of height

$h$ . If a small hole is punched in the side of the barrel at its base, it is found that the resultant stream of water strikes the ground at a horizontal distance

$R$ from the barrel. The depth of water in the barrel is

0%
$\cfrac { R }{ 2 }$
0%
$\cfrac { { R }^{ 2 } }{ 4h }$
0%
$\cfrac { { R }^{ 2 } }{ h }$
0%
$\cfrac { h }{ 2 }$

Explanation

Hint:

Use Torricelli’s equation for velocity and projectile motion’s horizontal range formula

$\textbf{Step1: Torricelli's equation for velocity}$

Torricelli's theorem stats that speed v by which water come out from tank is proportional to square toot of twice the acceleration created by gravity and vertical height between surface and center of opening.

V =

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

$\textbf{Step2:Expression of horizontal range}$

The horizontal range where the water strikes the ground R is given by,

R =

$V * \sqrt{\frac{2\ h}{g}}$

$\textbf{Step3:Calculation of the depth of water in the barrel}$

On putting the values and solving the equation,

R =

$\sqrt{2\ \ast\ g\ \ast\ h}\ * \sqrt{\frac{2\ h}{g}}$

So,

$R^2 = 4hH$

$H = \frac{R^2}{4h}$

Answer:

Hence, option B is the correct answer.

A large tank is filled with water (density

$= { 10 }^{ 3 } kg/{ m }^{ 3 }$ ). A small hole is made at a depth

$10 m$ below water surface. the range of water issuing out of the hole is R on ground. What extra pressure must be applied on the water surface so that the range becomes

$2R$ ( take

$1atm={ 10 }^{ 5 } Pa$ and

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

0%
$9 atm$
0%
$4 atm$
0%
$5 atm$
0%
$3 atm$

Explanation

Range will become twice, if the velocity of efflux becomes twice. Since velocity of efflux is

$\sqrt{2gh}$ , it will become twice, if

$h$ becomes

$4$ times, or

$40m.$
Thus, an extra pressure equivalent to

$30m$ of water should be applied.

$1 atm =0.76 \times 13.6$ m of water

$=10.336 m$ of water

$30m$ of water

$\approx 3 atm$

Water is flowing steadily through a horizontal tube of non-uniform cross-section.If the pressure of water is

$4\times10^4N/mm^2$ at a point when cross section is

$0.02m^2$ velocity of flow is 2m/s what is pressure at a point where cross section reduces to

$0.01m^2$ .

0%
$1.4\times { 10 }^{ 4 } N{ m }^{ 2 }$
0%
$3.4\times { 10 }^{ 4 } N{ m }^{ 2 }$
0%
$2.4\times { 10 }^{ -4 } N{ m }^{ 2 }$
0%
none of these

Explanation

As we know

$A_1V_1=A_2V_2$

Where

$A_1=0.02\ m^2$

$V_1=2\ m/s$

$A_2=0.01\ m^2$

$\implies 0.02\times 2=0.01\times V_2$

$\implies V_2=4\ m/s$

Now using Bernoulli's theorem

$\cfrac{P_1}{\rho}+\cfrac{1}{2}V_1^2=\cfrac{P_2}{\rho}+\cfrac{1}{2}v_2^2$

$\implies P_2=P_1+\cfrac{\rho_1}{2}[V_1^2-V_1^2]$

$\implies P_2=4\times 10^{10}+\cfrac{1000}{2}[2^2-4^2]-4\times 10^{10}=3.4\times 10^{4}\ N/m^2$

Equal volumes of two immiscible liquids of densities

$\rho$ and

$2\rho$ are filled in a vessel as shown figure. Two small holes are punches at depth

$h/2$ and

$3h/2$ from the surface of lighter liquid. If

${ v }_{ 1 }$ and

${ v }_{ 2 }$ are the velocities of a flux at these two holes then

${ v }_{ 1 }/{ v }_{ 2 }$ is :

0%
$\dfrac { 1 }{ 2\sqrt { 2 } }$
0%
$\dfrac { 1 }{ 2 }$
0%
$\dfrac { 1 }{ 4 }$
0%
$\dfrac { 1 }{ \sqrt { 2 } }$

Explanation

Here, two small holes are punches at depth

$\dfrac{h}{2}$ and

$\dfrac{3h}{2}$ from the surface of lighter liquid. Hence, the height of first hole from the surface of lighter liquid is

$h_1 = \dfrac{h}{2}$

and the height of second hole from the first hole is(i.e. from surface of remaining liquid)

$h_2 = \dfrac{3h}{2} - \dfrac{h}{2} = h$

Now, we have the velocity of flux from first hole is

$v_1 = \sqrt {2h_1g}$

$v_1 = \sqrt {2(\dfrac{h}{2})g}$

$v_1 = \sqrt {hg}$

Applying Bernoulli's equation to point just inside and outside the second hole,

$P_{a} + \rho gh + 2\rho g \displaystyle \dfrac{h}{2} = P_{a} + \displaystyle \dfrac{1}{2} 2\rho v^2$

$v = \sqrt {2gh}$

So, the velocity of flux from second hole is

$v_2 = \sqrt {2hg}$

Therefore,

$\dfrac{v_1}{v_2} = \dfrac{\sqrt {hg}}{\sqrt {2hg}}$

$\dfrac{v_1}{v_2} = \dfrac{1}{\sqrt 2}$

A rectangular tank is placed on a horizontal ground and is filled with water to a height

$H$ above the base. A small hole is made on one vertical side at a depth

$D$ below the level of water in the tank. The distance

$x$ from the bottom of the tank at which the water jet from the tank will hit the ground is

0%
$2\sqrt { D(H-D) }$
0%
$2\sqrt { DH }$
0%
$2\sqrt { D(H+D) }$
0%
$\cfrac { 1 }{ 2 } \sqrt { DH }$

Explanation

This setup can be visualised as a horizontal projectile from height h .

here

$h=$ height of the hole above the bottom of the tank.

$h=H-D$

let Horizontal velocity of water be

$u_x$ .

so velocity of liquid falling at depth d is given by

$\sqrt{2gd}$

here

$d=D$ so

$u_x=\sqrt{2gD}$

Time taken to fall at surface can be found using

$s=ut+\dfrac{1}{2}at^2$

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

here

$u_y=0$ , no component of velocity in vertical direction.

and

$h=H-D$

$H-D=\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2(H-D)}{g}}$

since there is no acceleration in horizontal direction

Range in Horizontal direction=

$u_xt$

putting values from above

$r=\sqrt{2gD}\times$

$\sqrt{\dfrac{2(H-D)}{g}}$

So, range of the water stream

$x=2\sqrt{D(H-D)}$

1656069_125832_ans_2fbfc2834a304a348eef18ecd92cd378.png

A water tank is filled with water upto height H. A hole is made in a tank wall at a depth D from the surface of water. The distance X from the lower end of wall where the water stream from tank strikes the ground is:

0%
$2 \sqrt {gD}$
0%
$2 \sqrt {D \left ( H + D \right)}$
0%
$2 \sqrt {D \left ( H - D \right)}$
0%
$\sqrt D$

Explanation

Velocity of efflux at point D =

$\sqrt {2gD}$
Time taken to reach the ground =

$\sqrt{ \displaystyle\frac{2(H-D)}{g}}$
So horizontal range = velocity

$\times$ time =

$\sqrt {2gD} \times \sqrt{ \displaystyle\frac{2(H-D)}{g}}$
So range of the water stream

$X=2\sqrt{D(H-D)}$

To measure the radius of the drop Millikan used _____ law of freely falling drops.

0%
Poiseuille's
0%
Ostwald's
0%
Brewester's
0%
Stoke's

A spherical ball is dropped in a long column of a viscous liquid. The speed

$v$ of ball as a function of time may be best represented by graph

Explanation

$\textbf{Explanation:}$

As soon as a spherical ball is started to fall in a viscous liquid, its speed will increase due to gravity, and at that time dragging force will be low as it is not that deep in liquid.
When the ball falls in a long column, its speed will be zero at t = 0. So, options C and D are incorrect. Now, with acceleration, velocity will increase, but after enough depth is reached, dragging force will start to work and when its value becomes equal to gravity force mg, net acceleration will be zero. And velocity will remain constant. So, in the graph, A velocity keeps on increasing with time so it is wrong.

Answer:

Hence, option B is the correct answer.

The difference of two liquid levels in a manometer is

$10 cm$ and its density is

$0.8 gm/cm^3$ . If the density of air is

$1.3 \times 10^3 gm/cm^3$ then the velocity of air will be (in

$cm/s$ )

0%
$347$
0%
$34.7$
0%
$3470$
0%
$0.347$

Explanation

The manometer here gives the stagnation pressure i.e. the velocity of liquid is zero at point B. $\\$ Hence the pressure drop is given by $\\ \Delta P=\rho gh={ P }_{ B }-{ P }_{ A }\\ \rho =0.8{ gm }/{ { cm }^{ 3 } }=800kg/{ m }^{ 3 } , h=0.1m , { \rho }_{ air }=1.3\times { 10 }^{ 3 }{ gm }/{ { cm }^{ 3 } }=1.3{ kg }/{ { m }^{ 3 } }\\ \Delta P=800\times 9.81\times 0.1\\ \Delta P=784.8 N/{ m }^{ 2 }\\$

From Bernoulli's equation for stagnation pressure

$\\ { P }_{ A }+\dfrac { { \rho }_{ air }{ { V } }^{ 2 } }{ 2 } ={ P }_{ B }\\ V=\sqrt { \dfrac { 2\Delta P }{ { \rho }_{ air } } } \\ V=\sqrt { \dfrac { 2\times 784.8 }{ 1.3 } } \\ V=34.74{ m }/{ s }\\ V\cong 3470{ cm }/{ s }$

Water flows through two identical tubes

$A$ and

$B$ . A volume

$V_{0}$ of water passes through the tube

$A$ and

$2 V_{0}$ through

$B$ in a given time. Which of the following may be correct?

0%
Flow in both the tubes are steady
0%
Flow in both the tubes are turbulent
0%
Flow is steady in $A$ but turbulent in $B$
0%
Flow is steady in $B$ but turbulent in $A$

Explanation

Reynolds number, R =

$\dfrac{\rho ud}{\mu}$
where

$\rho$ = density, u = mean velocity, d = diameter and

$\mu$ = viscosity
We know flow is Turbulent if R > 4000
In identical tubes, if volume V flows through A and 2V flows through B in a given time, then mean velocity,u in tube B is greater than that in A

$u^{}_{B}$ >

$u^{}_{A}$

$\implies$

$R^{}_{B} > R^{}_{A}$
option(D)

$\implies$

$R^{}_{A}$ > 4000

$\implies$

$R^{}_{B}$ >

$R^{}_{A}$ > 4000
So

$R^{}_{B}$ > 4000

$\implies$ Flow is turbulent in B
option (C)

$\implies$

$R^{}_{B}$ > 4000 , 2000 >

$R^{}_{A}$ is possible
(A)(B) options are also possible

A barometer tube, containing mercury, is lowered in a vessel containing mercury until only

$50 cm$ of the tube is above the level of mercury in the vessel. If the atmospheric pressure is

$75 cm$ of mercury, what is the pressure at the top of the tube

0%
$33.3 kPa$
0%
$66.7kPa$
0%
$3.33 MPa$
0%
$6.67 MPa$

Explanation

Since the mercury rises only

$50cm$ it implies the rest of the excess atmospheric pressure is being countered by the pressure at the top of the tube.
So pressure at top=atmospheric pressure -

$50cm$ of Hg=

$(75-50)cm=25cm$ of Hg=

$1/3\times 100kPa= 33.33kPa$

Water rises in a vertical capillary tube upto a length of

$10cm.$ If the tube is inclined at

$45^o$ , the length of water risen in the tube will be,

0%
$10 cm$
0%
$10 \sqrt2 cm$
0%
$\displaystyle \dfrac {10}{\sqrt2}$
0%
none of these

Explanation

The vertical rise in the level of liquid is constant.
Hence for an inclined tube, the effective length is

$\displaystyle\dfrac {h}{\cos\theta}$
So, the length of water risen in the tube will be

$10 \sqrt 2 cm$

The height of water level in a tank is

$H$ . The range of water stream coming out of a hole at depth

$\dfrac{H}{4}$ from upper water level will be.

0%
$\displaystyle \frac {\sqrt 3 H}{2}$
0%
$\displaystyle \frac {2H}{\sqrt 3}$
0%
$\displaystyle \frac {H}{\sqrt 3}$
0%
$\sqrt 3 H$

Explanation

The horizontal range of the liquid coming out a hole is equal to

$2\sqrt{hh'}$ , where

$h=$ depth of the hole below the free surface of the liquid

$h'=$ height of the hole above the bottom of the tank

Here,

$h=\dfrac{H}{4}$ and

$h'=H-\dfrac{H}{4} =\dfrac{3H}{4}$

So range of the water stream

$=2\sqrt{\dfrac{H}{4} \times \dfrac{3H}{4}}=\dfrac{\sqrt{3}H}{2}$

Two equal drops are falling through air with a steady velocity of

$5 cms^{-1}$ . If the drop coalesces, the new terminal velocity will become:

0%
$5 \times 2 cms^{-1}$
0%
$5 \times \sqrt 2 cms^{-1}$
0%
$5 \times (4)^{\frac {1}{3}} cms^{-1}$
0%
$\displaystyle \frac {5}{\sqrt 2} cms^{-1}$

Explanation

when the two equal drops coalesces, volume of water remains same and a bigger drop is formed. Lets say

$r$ is radius of drop before these combine and

$R$ is radius after these combine. we have

$2 \times \frac{4}{3} \pi r^3 =\frac{4}{3} \pi R^3 \\\Rightarrow R=2^{1/3}r$
and we know that terminal velocity

$V_T \propto$ radius

$^2$ , so we have

$\dfrac{V_T'}{V_T}=\dfrac{R^2}{r^2} \\\Rightarrow V_T' =\left (\frac{R}{r} \right)^2 V_T = \left (2^{1/3} \right)^2 \times 5 cms^{-1}=5 \times 4^{\frac{1}{3}} cms^{-1}$

The diagram shows a venturimeter through which water is flowing. The speed of water

$X$ is

$2 \ \text{cms}^{-1}$ . The speed of water at

$Y$

$(\ taking\ g = 10 \text{ms}^{-2})$ is

0%
$23\ \text{cms}^{-1}$
0%
$32\ \text{cms}^{-1}$
0%
$101\ \text{cms}^{-1}$
0%
$1024\ \text{cms}^{-1}$

Explanation

There is a small hole of diameter 2 mm in the wall of a water tank at a depth of 10 m below the free water surface. The velocity of efflux of water from the hole will be:

0%
0.14 m/s
0%
1.4 m/s
0%
0.014 m/s
0%
14 m/s

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.
Here

$h=10m$

$\Rightarrow v=\sqrt{2 \times 9.8 \times 10}=\sqrt{196}=14m/s$

Water stored in a tank flows out through a hole of radius 1mm at a depth 10 m below the surface of water.The rate of flow of water in

$m^3 /s$ will be

0%
$4.4 \times 10^{-5}$
0%
$4.4 \times 10^{-4}$
0%
$4.4 \times 10^{-3}$
0%
$4.4 \times 10^{-2}$

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. here

$h=10m$

$\Rightarrow v=\sqrt{2 \times 9.8 \times 10}=\sqrt{196}=14m/s$
and rate of flow

$=av=\pi (0.001)^2 \times 14=4.4 \times 10^{-5} m^3/s$

One end of a horizontal pipe is closed with the help of a valve and the reading of a barometer attached to the pipe is

$3 \times 10^5$ pascal. When the valve in the pipe is opened then the reading of the barometer falls to

$10^5$ pascal. The velocity of water flowing through the pipe will be in m/s.

0%
$0.2$
0%
$2$
0%
$20$
0%
$200$

Explanation

Applying Bernoulli's principle on a horizontal pipe, we have

$P_1+\frac{1}{2} \rho {v_1}^2=P_2+\frac{1}{2} \rho {v_2}^2$
here

$P_1= 3 \times 10^5$ pascal ,

$v_1=0$ ,

$P_2=10^5$ Pascal,

$\rho=1000 kg/m^3$

$P_1+\frac{1}{2} \rho {v_1}^2=P_2+\frac{1}{2} \rho {v_2}^2 \\\Rightarrow 3 \times 10^5=10^5+\frac{1}{2} \times 1000 \times {v_2}^2$
Solving the above equation we have,

$v_2=20m/s$

Bernoulli's equation includes as a special case:

0%
Archimede's principle
0%
Pascal's law
0%
Toricelli's law
0%
Hooke's law

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 (in this case a drop of water) would acquire in falling freely from a height

$h$ .

This problem is solved using Bernoulli's equation,

$gh+\dfrac{P_{atm}}{\rho}=\dfrac{v^2}{2}+\dfrac{P_{atm}}{\rho}$

$\implies v=\sqrt{2gh}$

Water flows steadily through a horizontal tube of variable cross-section. If the pressure of water is p at a point where the velocity of flow is v, what is the pressure at another point where the velocity of flow is 2v;

$\rho$ being the density of water?

0%
$p - \displaystyle \frac {3}{2} \rho v^2$
0%
$p + \displaystyle \frac {3}{2} \rho v^2$
0%
$p - 2 \rho v^2$
0%
$p + 2 \rho v^2$

Explanation

Applying Bernoullis's Equation at two points:

${ p }_{ 1 }+\displaystyle\frac { \rho { { v }_{ 1 } }^{ 2 } }{ 2 } ={ p }_{ 2 }+\displaystyle\frac { \rho { { v }_{ 2 } }^{ 2 } }{ 2 }$

Given,

${ p }_{ 1 }=p,{ \quad v }_{ 1 }=v\quad { v }_{ 2 }={ 2v }\\ { p }_{ 2 }={ p }+\displaystyle\frac { \rho { { v } }^{ 2 } }{ 2 } -\displaystyle\frac { 4\rho { { v } }^{ 2 } }{ 2 } \\ { p }_{ 2 }={ p }-\displaystyle\frac { 3\rho { { v } }^{ 2 } }{ 2 }$

A spherical ball falls through viscous medium with terminal velocity

$v$ . If this ball is replaced by another ball of the same mass but half the radius, then the terminal velocity will be (neglect the effect of buoyancy)

0%
$v$
0%
$2v$
0%
$4v$
0%
$8v$

Explanation

In case of spherical ball

$m=\dfrac{4}{3}\pi r^3 \rho$

keeping

$m$ constant, if

$r$ is halved,

$\rho$ will increased by a factor of

$8$ ;

If buoyancy is neglected,

${ V }_{ T }\propto { r }^{ 2 }\rho; \text{ otherwise } { V }_{ T }\propto { r }^{ 2 }\left ( \rho-\rho_o \right ) \\ \Rightarrow \dfrac { { V }_{ T }' }{ { V }_{ T } } =\dfrac { { r' }^{ 2 }\rho ' }{ { r }^{ 2 }\rho } \\ \Rightarrow \dfrac { { V }_{ T }' }{ { V }_{ T } } =\dfrac { { \left( \dfrac { r }{ 2 } \right) }^{ 2 }8\rho }{ { r }^{ 2 }\rho } \\ \Rightarrow \dfrac { { V }_{ T }' }{ { V }_{ T } } =\dfrac { 2{ r }^{ 2 }\rho }{ { r }^{ 2 }\rho } \\ \Rightarrow \dfrac { { V }_{ T }' }{ { V }_{ T } } =2\\ \Rightarrow { V }_{ T }'=2{ V }_{ T }\\ \Rightarrow \boxed{{ V }_{ T }'=2v}$

Water is floating smoothly through a closed-pipe system. At one point A, the speed of the water is 3.0 m/s while at another point B, 1.0 m higher, the speed is 4.0 m/s. The pressure at A is 20 kPa when the water is flowing and 18 kPa when the water flow stops. Then

0%
the pressure at B when water is flowing is 6.5 kPa
0%
the pressure at B when water is flowing is 8.0 kPa
0%
the pressure at B when water stops flowing is 10 kPa
0%
the pressure at B when water stops flowing is 8.0 kPa

Explanation

Apply Bernoulli's principle,
When water is flowing,

${P}^{}_{1} + \dfrac{1}{2}{\rho}^{}_{1} {V}^{2}_{1} = \rho gh + {P}^{}_{2} + \dfrac{1}{2}{\rho}^{}_{1} {V}^{2}_{2}$

$20\times {10}^3 + \dfrac{1}{2} {10}^3 \times {3}^2$ =

${10}^3\times (10) (1) + \dfrac{1}{2} {10}^3 \times {4}^2 + {P}^{}_{2}$

${P}^{}_{2} = 6.5kPa$
When water doesnot flow,

${P}^{}_{2} = {P}^{}_{1} - \rho gh$
=

$18\times {10}^3 - {10}^3\times (10) (1)$
=

$8.0 kPa$

Two liquid jets coming out of the small holes at P and Q intersect at the point R. Find the position of R if we maintain the liquid level constant

0%
$\sqrt 2h$
0%
$2\sqrt 2h$
0%
$3\sqrt 2h$
0%
$5\sqrt 2h$

Explanation

Velocity of liquid coming out of point P

$V_p = \sqrt{2gh}$

Time taken by liquid stream to reach R

$T_1 = \sqrt{\dfrac{2 (h+y)}{g}}$

$\therefore$

$x = V_p T_1 = \sqrt{4h (h+y)}$ .........(1)

Velocity of liquid coming out of point Q

$V_Q = \sqrt{2g(2h)}$

Time taken by liquid stream to reach R

$T_2 = \sqrt{\dfrac{2 y}{g}}$

$\therefore$

$x = V_Q T_2 = \sqrt{4h (2h)}$ .........(2)

From (1) and (2) we get

$4h (h+y) = 4h (2h)$

$\implies y = h$

$\therefore$ Position of R

$x = \sqrt{4h (h+y)} = \sqrt{4h (h+ h)} =2\sqrt{2} h$

A tank of large base area is filled with water up to a height of

$5\ m$ . A hole of

$2\ cm^{2}$ cross section section in the bottom allows the water to drain out in continuous streams. For this situation, mark out the correct statement(s) (take

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

0%
The cross sectional area of the emerging stream of water decreases as it falls down
0%
The cross sectional area of the emerging stream of water increases as it falls down
0%
At a distance of $5m$ below the bottom of the tank, the cross-sectional area of the stream is $1.414 cm^{2}$
0%
At a distance of $5m$ below the bottom of the tank, the cross-sectional area of the stream is $2.86 cm^{2}$

Explanation

As the water falls down in the stream, the speed of the stream will increase due to gravity. So according to equation of continuity

$(av = \text{a constant})$ , the area of the stream will decrease to keep the product of area and speed constant.

$\Rightarrow$ option A is correct.
Velocity of water just coming out of the hole is

$v_1=\sqrt{2gh}=\sqrt{2 \times 9.8 \times 5}=7\sqrt{2} \text{ m/s}$
Velocity of water as it falls down 5m more

$v_2=\sqrt{(7\sqrt{2})^2+2 \times 9.8 \times 5}=14\text{ m/s}$

$\ \because v_2=\sqrt{{v_1}^2+2gh}$
Now using the equation of continuity, we have

$a_{ 1 }v_{ 1 }=a_{ 2 }v_{ 2 }\\ \Rightarrow 2\times 7\sqrt { 2 } =a_{ 2 }\times 14$

$\Rightarrow a_{ 2 }=\sqrt { 2 } \text{ cm}^2=1.414 \text{ cm}^2$

$\Rightarrow$ option C is correct.

A pressure meter attached to a closed water tap reads

$1.5\times 10^5\ \text{Pa}$ . When the tap is opened, the velocity of flow of water is

$10\ \text{ms}^{-1}$ and the reading of the pressure meter is

0%
$1.5\times 10^5\ \text{Pa}$
0%
$3\times 10^5\ \text{Pa}$
0%
$0.5\times 10^5\ \text{Pa}$
0%
$10^5\ \text{Pa}$

Explanation

Applying Bernoulli's principle on a horizontal pipe, we have

$P_1+\frac{1}{2} \rho {v_1}^2=P_2+\frac{1}{2} \rho {v_2}^2$
here

$P_1= 1.5 \times 10^5\ \text{pascal},\ v_1=0,\ v_2=10 \text{m/s}$ ,

$\rho=1000\ \text{kg/m}^3$

$P_1+\dfrac{1}{2} \rho {v_1}^2=P_2+\dfrac{1}{2} \rho {v_2}^2 \\\Rightarrow 1.5 \times 10^5=P_2+\dfrac{1}{2} \times 1000 \times {(10)}^2$
solving the above equation we have

$P_2= 10^5 \text{pascal}$

In a horizontal pipeline of uniform cross section the pressure falls by

$8N/m^2$ between two points separated by 1 km. If oil of density

$800 kg/m^3$ flows through the pipe, find the change in KE per kg of oil at these points.

0%
$10^{-2}J/kg$
0%
$10^{-3}J/kg$
0%
$10^{-4}J/kg$
0%
$10^{-1}J/kg$

Explanation

Applying Bernoulli's principle on a horizontal pipe, we have
$P_{ 1 }+\dfrac { 1 }{ 2 } \rho { v_{ 1 } }^{ 2 }=P_{ 2 }+\dfrac { 1 }{ 2 } \rho { v_{ 2 } }^{ 2 }\\ \Rightarrow \dfrac { 1 }{ 2 } \rho { v_{ 2 } }^{ 2 }-\dfrac { 1 }{ 2 } \rho { v_{ 1 } }^{ 2 }=P_{ 1 }-P_{ 2 }\\ \Rightarrow \dfrac { 1 }{ 2 } { v_{ 2 } }^{ 2 }-\dfrac { 1 }{ 2 } { v_{ 1 } }^{ 2 }=\dfrac { P_{ 1 }-P_{ 2 } }{ \rho } \\ \Rightarrow \text{Change in kinetic energy per unit mass } (\Delta K.E.)=\dfrac { P_{ 1 }-P_{ 2 } }{ \rho } \\\Rightarrow \Delta K.E.=\dfrac{8}{800}=10^{-2}\mathrm{J/Kg}$

The difference of pressure between two points along a horizontal pipe through which water is flowing is

$1.4 cm$ of Hg. If due to non-uniform cross section the speed of flow at a point of greater cross section is

$60 cm/s$ , the speed of flow at the other point is

0%
$2 m/s$
0%
less than $60 cm/s$
0%
not affected by non-uniform cross section
0%
greater than $60 cm/s$

The meniscus formed by mercury in a test tube is

0%
convex
0%
concave
0%
no meniscus is formed
0%
none of these

The figure shows capillary tube of radius

$r$ dipped into water. The atmospheric pressure is

$P_{0}$ and the capillary rise of water is

$h. S$ is the surface tension for water-glass.
Initially,

$h = 10\ cm$ . If the capillary tube is now inclined at

$45^{\circ}$ , the length of water rising in the tube will be

0%
$10 cm$
0%
$10\sqrt {2} cm$
0%
$\dfrac {10}{\sqrt {2}}cm$
0%
None of these

Explanation

When the capillary tube is tilted by an angle

$\theta$ with the vertical, the capillary rise to distance

$l$ in the tube will be such that the meniscus will remain at the same vertical height above the level of water in the container.
Hence,

$h = l\cos \theta$

$l = \dfrac {h}{\cos \theta} = \dfrac {10}{\cos 45^{\circ}} = 10\sqrt {2} cm$ .
1556956_158118_ans_637c7add7a4247e49a583e370fe4d5f7.jpg

The figure shows capillary tube of radius

$r$ dipped into water. The atmospheric pressure is

$P_{0}$ and the capillary rise of water is

$h. s$ is the surface tension for water-glass.
The pressure inside water at the point

$A$ (lowest point of the meniscus) is

0%
$P_{0}$
0%
$P_{0} + \dfrac {2s}{r}$
0%
$P_{0} - \dfrac {2s}{r}$
0%
$P_{0} - \dfrac {4s}{r}$

Explanation

Since the pressure above the meniscus is

$P_0$ .
Difference between pressures outside and inside of the meniscus is 2s/r
Since Meniscus is concave upwards, this means pressure inside meniscus is less as compared to outside
So Pressure below the meniscus of radius

$r$ is

$P_0-2s/r$
( assuming meniscus to be spherical)

The incident intensity on a horizontal surface at sea level from the sun is about

$1 kW m^{-2}$ . Find the ratio of this pressure to atmospheric pressure

$p_0$ (about

$1\times 10^5 Pa)$ at sea level.
[Assuming that

$50 %$ % of this intensity is reflected and

$50 %$ % is absorbed]

0%
$5\times 10^{-11}$
0%
$4\times 10^{-8}$
0%
$6\times 10^{-12}$
0%
$8\times 10^{-11}$

Explanation

Given,

$I=1k W/m^2$

$c=3\times 10^8 m/s$

$P_0=1\times 10^5 Pa$
Pressure exerted due to absorbed light,

$P_a= \dfrac{I}{2c}$

Pressure exerted due to reflected light,

$P_r=\dfrac{2I}{2c}$
Hence, the total radiation pressure on the surface can be given as:

$P_t=P_a+P_r=\dfrac{3I}{2c}$

$P_t=\dfrac{3\times 10^3}{3\times2\times 10^8}=\dfrac{ 10^3}{2\times 10^8}$

$\Rightarrow P_t=5\times 10^{-6}Pa$

So, the ratio of the radiation and atmospheric pressure will be:

$\dfrac{P}{P_0}=\dfrac{5\times 10^{-6}}{1\times 10^5}$

$\Rightarrow \dfrac{P}{P_0}=5\times 10^{-11}$
Hence, the correct option is

$(A)$

Mercury in a test tube forms ______ meniscus

0%
concave
0%
convex
0%
no
0%
cant say

Explanation

Answer:

$convex$

Mercury in a test tube forms

$convex$ meniscus.

The

$cohesive \space force$ between the molecules of the material i.e

$mercury$ is higher than the

$adhesive \space force.$
The molecules tend to bind each other tightly hence the force in the test tube is inwards due to the surface tension.

The direction of the excess pressure in the meniscus of a liquid of angle of contact

$2\pi/3$ is:

0%
Upward
0%
Downward
0%
Horizontal
0%
Cannot determined

Explanation

Excess pressure is always on the concave side of the surface. For angle of contact of

$2\pi/3$ , the liquid should have a convex surface. So, the excess pressure should be in the upward direction.

A small ball (menu) falling under gravity in a viscous medium experiences a drag force proportional to the instantaneous speed u such that

$F_{drag}$ = Ku. Then the terminal speed of the ball within the viscous medium is:

0%
$\dfrac {K}{mg}$
0%
$\dfrac {mg}{K}$
0%
$\sqrt{\dfrac {mg}{K}}$
0%
$\left ( \dfrac{mg}{K} \right)^2$

Explanation

For terminal velocity , weight should be balanced by upward force.

Here we have excluded buoyancy force.

Now,

$Ku=mg$

$u=\dfrac{mg}{K}$

There is a hole in the bottom of tank having water. If total pressure at bottom is 3 atm (1 atm =

$10^5 N/m^2$ ) then the velocity of water flowing from hole is:

0%
$\sqrt{400}m/s$
0%
$\sqrt{600}m/s$
0%
$\sqrt{60}m/s$
0%
None of these

Explanation

Pressure at the bottom of tank

$P = h\rho g = 3 x 10^5 \dfrac{N}{m^2}$
Pressure due to liquid column

$P_1 = 3 x 10^5 - 1 x 10^5 = 2 x 10^5$
and velocity of water

$v = \sqrt{2gh}$

$\therefore v = \sqrt{\dfrac{2P_1}{\rho}}=\sqrt{\dfrac{2x 2 x 10^5}{10^3}}= \sqrt{400}m/s$

A spherical ball of iron of radius

$2\ \text{mm}$ is falling through a column of glycerine. If densities of glycerine and iron are respectively

$1.3\times 10^3\ \text{kg/m}^3$ and

$8\times 10^3\ \text{kg/m}^3$ .

$\eta\ {for \ glycerine} = 0.83\ \text{Nm}^{-2}\ \text{sec},$ then the terminal velocity is:

0%
$0.7\ \text{m/s}$
0%
$0.07\ \text{m/s}$
0%
$0.007\ \text{m/s}$
0%
$0.0007\ \text{m/s}$

Explanation

Terminal velocity

$, v_0 = \dfrac{2r^2 (\rho -\rho_0)g}{9\eta }$

$v_0= \dfrac{2\times (2 \times 10^{-3})^2\times (8 -1.3) \times 10^3 \times 9.8} {9 \times 0.83} = 0.07\ \text{ms}^{-1}$

A cylinder is filled with non-viscous liquid of density d to a height

$h_0$ and a hole is made at a height

$h_1$ from the bottom of the cylinder. The velocity of the liquid issuing out of the hole is:

0%
$\sqrt{2gh_0}$
0%
$\sqrt{2g(h_0-h_1)}$
0%
$\sqrt{dgh_1}$
0%
$\sqrt{dgh_0}$

Explanation

$\text{Velocity of liquid flowing out of hole} = \sqrt{2gh}$ ;
Here

$h= (h_0 - h_1)$

The rain drops falling from the sky neither injure us nor make holes on the ground because they move with:

0%
constant acceleration
0%
variable acceleration
0%
variable speed
0%
constant terminal velocity

Explanation

Gravity does cause things to fall with increasing speed, but as they speed up air resistance increases. Eventually the force of the air resistance is enough to balance the force of gravity, so the acceleration stops and the raindrop reaches a constant terminal velocity. Apparently the terminal velocity of rain isn't high enough to cause damage. Since

$F=Ma$ so the force required to stop the raindrop wouldn't be that much because

a raindrop doesn't have a lot of mass
it's speed (terminal velocity) isn't that high.

Two drops of the same radius are falling through air with a steady velocity of

$5\ \text{cm per sec}.$ If the two drops coalesce, the terminal velocity would be

0%
$10\ \text{cm per sec}$
0%
$2.5\ \text{cm per sec}$
0%
$5\times (4)^{1/3}\ \text{cm per sec}$
0%
$5\sqrt{3}\ \text{cm per sec}$

Explanation

$R$ is radius of bigger drop formed, then

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

$R = 2^{1/3}r$
As

$v_0 \propto r^2$

$\therefore \dfrac{v_{01}}{v_0} = \dfrac{R^2}{r^2}= \dfrac{(2^{1/3})^2}{r^2}= 2^{2/3}$

$v_{01} = v_0 2^{2/3} = 5 (4)^{1/3}$

In a container having water filled upto a height h, a hole is made in the bottom. The velocity of the water flowing out of the hole is:

0%
independent of h
0%
proportional to $h^{1/2}$
0%
proportional to h
0%
proportional to $h^2$

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.

$v\propto \sqrt{h}$ .

Which molecule of a liquid has higher potential energy?

0%
One at the centre of gravity of the liquid
0%
One at maximum distance from centre of gravity of liquid
0%
One in the surface film
0%
One at the bottom of the vessel

A cylinder of height

$20$ m is completely filled with water. The velocity of efflux of water (in ms

$^{-1}$ ) through a small hole on the side wall or the cylinder near its bottom is:

0%
$10$ m/s
0%
$20$ m/s
0%
$25.5$ m/s
0%
$5$ m/s

Explanation

P.E. = K.E.
mgh =

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

$\sqrt{2gh}= \sqrt{2 x 10 x 20}(Here g = 10 m/s^2)$

$= 20 m/s$

For flow of a fluid to be turbulent:

0%
fluid should have high density
0%
velocity should be large
0%
reynold number should be less than 2000
0%
both (a) and (b)

Explanation

1) For turbulent flow both velocity and density of fluid should be high.

2) Value of reynold's number

${Re}>{4000}$

note

${Re}<{2000}$ is for laminar flow and

${2000}<{Re}<{4000}$ transition flow.

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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