MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 7

Water containing air bubbles flows without turbulence through a horizontal pipe which has a region of narrow cross-section. In this region the bubbles.

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Move with greater speed and are smaller than in the rest of the pipe
0%
Move with greater speed and are larger in size than in the rest of the pipe
0%
Move with lesser speed and are smaller than in the rest of the pipe
0%
Move with lesser speed and are of the same size as in the rest of the pipe

Explanation

Since the water flow is assumed to be laminar, we can apply Bernoulli's equation between two points in the flow.

Bernoulli's equation is given by

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

The continuity equation is given by

$Av = \textrm{constant}$

At the narrow section, we have

$A$ to be minimum. Thus,

$v$ is maximum.

So, the bubbles travel faster at the narrow section.

Since,

$v$ is maximum, by Bernoulli's equation, we have

$\frac{1}{2} \rho v^2$ also maximum. Thus,

$P$ is minimum.

$P$ is minimum, we have the size of the bubbles as maximum.

Hence, at the minimum cross-section, the bubbles are larger in size and move with greater speed than rest of the pipe.

When one end of the capillary is dipped in water, the height of water column is

$'h'$ . The upward force of

$105$ dyne due to surface tension balanced by the force due to the weight of water column. The inner circumference of the capillary is
(Surface tension of water

$= 7\times 10^{-2}N/m)$

0%
$1.5\ cm$
0%
$2\ cm$
0%
$2.5\ cm$
0%
$3\ cm$

Explanation

Upward force acting

$F = 105$ dyne

$= 105\times 10^{-5}$

$N$

Surface tension of water

$T =7\times 10^{-2} N/m$

Surface tension acts at the circumference

$C$ of the capillary tube.

$\therefore$

$F = C T$

$105\times 10^{-5} = C\times 7\times 10^{-2}$

$C = 15\times 10^{-3} m$

$\implies$

$C = 15\times 10^{-1} cm = 1.5 cm$

A particle released from rest is falling through a thick fluid under gravity. The fluid exerts a resistive force on the particle proportional to the square of its speed. Which one of the following graphs best depicts the variation of its speed

$v$ with time

$t$ ?

Explanation

Let the effective

$g=g'$

Now we know

$mg'-\alpha v^2=m\dfrac{dv}{dt}$

Initially velocity will rise and later

$mg'=\alpha v^2$ then velocity will become constant .

An open tank filled with water density

$\rho$ has a narrow hole at a depth of

$h$ below, then the velocity of water flowing out is:

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

Explanation

From Bernoulli's equation, we can write:

$\rho g h=\dfrac{1}{2}\rho v^2$

So,

$v=\sqrt{2gh}$

Equal volume of two immiscible liquids of densities

$\rho$ and

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

$\dfrac {h}{2}$ and

$\dfrac {3h}{2}$ from the surface of lighter liquid. If

$v_{1}$ and

$v_{2}$ are the velocities of efflux at these two holes, then

$\dfrac{v_{1}}{v_{2}}$ will be

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

Explanation

We have

$v_{1} = \sqrt {2gh(h/2)} = \sqrt {gh} .... (i)$
and by using Bernoulli's theorem

$\rho gh + 2\rho g\left (\dfrac {h}{2}\right ) = \dfrac {1}{2}(2\rho) v_{2}^{2}$

$\Rightarrow v_{2} = \sqrt {2}gh .... (ii)$
From Eqs. (i) and (ii)

$\dfrac {v_{1}}{v_{2}} = \dfrac {1}{\sqrt {2}}$ .

A cylindrical tank has a hole of

$1$

$cm^2$ m bottom. If the water is allowed to flow into the tank from a tube above it at the rate of

$70cm^2/s$ , then the maximum height up to which water can rise in the tank is?

0%
$2.5$ cm
0%
$5$ cm
0%
$10$ cm
0%
$0.25$ cm

Explanation

The height of water in the tank becomes maximum when the volume of water flows into the tank per second becomes equal to the volume flowing out per second.
Volume of water flowing out per second

$=A\sqrt{2gh}$
Volume of water flowing in per second

$=70cm^3/s$

$\therefore A\sqrt{2gh}=70$

$1\sqrt{2\times 980\times h}=70$

$h=\displaystyle\frac{4900}{1960}=2.5$ cm.

A rectangular vessel when full of water, takes

$10\ min$ to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water?

0%
$9\ min$
0%
$7\ min$
0%
$5\ min$
0%
$3\ min$

Explanation

$A_{0}$ is the area of the orifice at the bottom below the free surface and

$A$ that of vessel, time

$t$ taken to be empited the tank,

$t = \dfrac {A}{A_{0}}\sqrt {\dfrac {2H}{g}}$

$\therefore \dfrac {t_{1}}{t_{2}} = \sqrt {\dfrac {H_{1}}{H_{2}}}$

$\Rightarrow \dfrac {t}{t_{2}} = \sqrt {\dfrac {H_{1}}{H_{1}/2}}$

$\Rightarrow \dfrac {t}{t_{2}} = \sqrt {2}$

$\therefore t_{2} = \dfrac {t}{\sqrt {2}}$

$= \dfrac {10}{\sqrt {2}} = 5\sqrt {2}$

$\approx 7\ min$

The pressure at depth

$h$ below the surface of a liquid of density

$\rho$ open to the atmosphere is

0%
Greater than the atmospheric pressure by $\rho gh$
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Less than the atmospheric pressure by $\rho gh$
0%
Equal to the atmospheric pressure
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Decreases exponentially with depth
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Increases exponentially with depth

Explanation

Pressure at depth

$h$ to the open surface

$p={ p }_{ 0 }+h\rho g$ .
i.e., the pressure at depth

$h$ below the surface of a liquid of density

$\rho$ open to the atmosphere is greater than the atmosphere pressure by

$\rho gh$ .

The material of a wire has a density of 1.4 g/cm

$^3$ . If it is not wetted by a liquid of surface tension 44 dyne/cm, then the maximum radius of the wire which can float on the surface of liquid is :

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$\dfrac{10}{28} cm$
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$\dfrac{10}{14} cm$
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$\dfrac{10}{7} mm$
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0.7 cm

Explanation

$2 Tl = \pi r^2 Id \times g$

$r = \sqrt{\left(\dfrac{2T}{\pi dg} \right )} = \sqrt{\dfrac{2 \times 44 \times 7}{22 \times 1.4 \times 980}}$

$= \dfrac{1}{7} cm = \dfrac{10}{7} mm$

A tank is filled with water of density 1 g

$cm^{-3}$ and oil of density 0.9 g

$cm^{-3}$ . The height of water layer is 100 cm and of the oil layer is 400 cm. If g = 980 cm

$s^{-2}$ , then the velocity of efflux from an opening in the bottom of the tank is :

0%
$\sqrt{920\times 980}cms^{-1}$
0%
$\sqrt{900\times 980}cms^{-1}$
0%
$\sqrt{1000\times 980}cms^{-1}$
0%
$\sqrt{92\times 980}cms^{-1}$

Explanation

The pressure at the bottom of the tank must be equal to the pressure due to water of height h.

$d_w$ and

$d_o$ be the densities of water and oil respectively, then the pressure at the bottom of the tank

$= h_wd_wg + h_od_og$

If this pressure is equivalent to pressure due to water of height h.

Then,

$hd_wg=h_wd_wg+h_od_og$

$h=h_w+\dfrac{h_od_o}{d_w}=100+\dfrac{400\times 0.9}{1}$

$=100+360=460$

According to Torricelli's theorem

$v=\sqrt{2gh}=\sqrt{2\times 980\times 460}$

$=\sqrt{920\times 980}cms^{-1}$

The potential energy of a molecules on the surface of a liquid compared to the one inside the liquid is :

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Zero
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Lesser
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Equal
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Greater

A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference p. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled, is

0%
$\rho$
0%
$\displaystyle\frac{3\rho}{4}$
0%
$\displaystyle\frac{\rho}{2}$
0%
$\displaystyle\frac{\rho}{4}$

Explanation

$\because$ Rate of flow,

$Q=\displaystyle\frac{v}{t}=\frac{\pi pr^4}{8\eta l}$
and rate of flow,

$2Q=\displaystyle\frac{\pi p'(2r)^4}{8\eta (2l)}=\frac{8\pi p'\pi^4}{8\eta l}$

$\displaystyle\frac{Q}{2Q}=\frac{p}{8p'}$

$\Rightarrow p'=\frac{p}{4}$ .

Water flows along a horizontal pipe whose cross.section is not constant. The pressure is 1 cm of Hg, where the velocity is

$35\, cms^{-1}$ . At a point where the velocity is

$65\, cms^{-1}$ , the pressure will be

0%
0.89 cm Hg
0%
8.9 cm of Hg
0%
0.5 cm of Hg
0%
1 cm of Kg

Explanation

Using Bernoulli's theorem in horizontal pipe,

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

Here,

$P_1 =\rho_mgh_1= 13600 \times 9.8 \times 10^{-2}$

$P_2 = 13600 \times 9.80 \, h$

$\rho_w= 1000\, kg/m^3$

$v_1 = 35\times 10^{-2}\, m/s$

$v_2 = 65\times 10^{-2}\, m/s$

So, we have

$13600 \times 9.8 \times 10^{-2}+\dfrac{1}{2}\times 1000 \times (0.35)^2$

$= 13600 \times 9.8 \times h+\dfrac{1}{2}\times 1000 \times (0.65)^2$
After solving, we get

$h=0.98 \ cm$ of Hg.

A liquid is filled upto a height of

$20\ m$ in a cylindrical vessel. The speed of liquid coming out of a small hole at the bottom of the vessel is (take,

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

0%
$1.2\ ms^{-1}$
0%
$1\ ms^{-1}$
0%
$2\ ms^{-1}$
0%
$3.2\ ms^{-1}$
0%
$1.4\ ms^{-1}$

Explanation

Given,

$h = 20\ cm, = 0.2\ m$
and

$g = 10\ ms^{-2}$
We know that,

$v = \sqrt {2gh}$
Speed of liquid coming out of small hole,

$\Rightarrow v = \sqrt {2\times 10\times 0.2} = 2\ ms^{-1}$ .

A cylindrical vessel of base radius R and height H has a narrow neck of height h and radius r at one end(see figure). The vessel is filled with water(density

$\rho_w$ ) and its neck is filled with immiscible oil (density

$\rho_o$ ). Then the pressure at :

0%
M is $g(h\rho_o+H\rho_w)$
0%
N is $g(h\rho_o+H\rho_w)\displaystyle\frac{r^2}{R^2}$
0%
M is $gH\rho_w$
0%
N is $g\displaystyle\frac{\rho_wHR^2+\rho_ohr^2}{R^2+r^2}$

Explanation

Pressure at M equals pressure at N (same depth) which is given by:

$P_M = P_N = \rho_w g H + \rho_o g h$

Equal volumes of two immisible liquids of densities

$\rho$ and

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

$h/2$ and

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

$v_1$ and

$v_2$ are the velocities of efflux at these two holes, then

$v_1/v_2$ is?

0%
$\displaystyle\frac{1}{2\sqrt{2}}$
0%
$1/2$
0%
$1/4$
0%
$\displaystyle\frac{1}{\sqrt{2}}$

Explanation

$v_1=\sqrt{2g\left(\displaystyle\frac{h}{2}\right)}=\sqrt{gh}$ .......(i)
From Bernoulli's theorem

$\rho gh+2\rho g\left(\displaystyle\frac{h}{2}\right)=\displaystyle \frac{1}{2}(2\rho)v^2_2$

$\therefore \displaystyle v_2=\sqrt{2gh}$ .........(ii)

$\therefore \displaystyle\frac{v_1}{v_2}=\frac{1}{\sqrt{2}}$ .

A tank of height

$5\ m$ is full of water. There is a hole of cross sectional area

$1\ cm^2$ in its bottom. The initial volume of water that will come out from this hole per second is

0%
$10^{-3}m^3/s$
0%
$10^{-4}m^3/s$
0%
$10m^3/s$
0%
$10^{-2}m^3/s$

Explanation

Since velocity of water coming out =

$\sqrt { 2gh }$

$\sqrt { 2\times 10\times 5 }$

= 10 m/s

So volume of water = velocity

$\times$ area

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

volume flow rate =

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

A large tank is filled with water to a height H. A small hole is made at the base of the tank. It takes 71 time to decrease the height of water to

$\frac {H}{\eta }(\eta > 1)$ and

$T_2$ time to take out the rest of 11 water. If

$T_1 = T_2$ then the value of

$\eta$ is

0%
2
0%
4
0%
6
0%
8

Explanation

$t=\dfrac{A}{a}\sqrt{\dfrac{2}{g}}[\sqrt{H_1}-\sqrt{H_2}]$

$T_1 =\dfrac{A}{a}\sqrt{\dfrac{2}{g}} \left [ \sqrt{H_1} \sqrt{\dfrac{H}{\eta }}\right ]$

$T_2 =\dfrac{A}{a}\sqrt{\dfrac{2}{g}} \left [ \sqrt{\dfrac{H}{\eta }-0}\right ]$
Given,

$T_1=T_2$

$\sqrt{H}-\sqrt{\dfrac{H}{\eta }}=\sqrt{\dfrac{H}{\eta }}-0$

$\Rightarrow \sqrt{H}=2 \sqrt{\dfrac{H}{\eta }}$

$\Rightarrow \eta =4$

A capillary tube of radius

$r$ is immersed in water and water rises in it to a height

$h$ . Mass of water in the capillary tube is

$m$ . If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now be

0%
$m$
0%
$2m$
0%
$\cfrac { m }{ 2 }$
0%
$4m$

Explanation

$\textbf{Step 1 - Calculation of new height}$

Let initial and final mass of water be

$m_{1}$ and

$m_{2}$

Let initial and final height be

$h_{1}$ and

$h_{2}$

Capillary height is given by :

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

Where

$T\rightarrow \text {Surface Tension}$

$\theta \rightarrow \text {contact angle}$

$\rho \rightarrow \text{density}$

$r \rightarrow \text{radius of tube}$

Here,

$T , \theta, \rho, g$ are constant

$\therefore r_{1}h_{1} = r_{2}h_{2}$

$\Rightarrow h_{2} = \dfrac {r_{1}\cdot h_{2}}{r_{2}}$

$h_{2} = \dfrac {r\cdot h}{2r} = \dfrac {h}{2}$

$\textbf{Step 2 - Calculation of new mass of water}$

Now, mass of water in capillary tube

$\therefore \dfrac {m_{2}}{m_{1}} = \dfrac {V_{2}}{V_{1}}$

$\Rightarrow m_{2} = \dfrac {V_{2}}{V_{1}}\times m_{1}$

$\Rightarrow m_{2} = \dfrac {\pi \cdot (r_{2})^{2} \cdot h_{2}}{\pi \cdot (r_{1})^{2} \cdot h_{1}} \times m$

$\Rightarrow m_{2} = \dfrac {\pi \cdot 4r^{2} \cdot (h/2)}{\pi\cdot r^{2} \cdot h} \times m$

$\Rightarrow m_{2} = 2m$

Therefore, mass of water will be

$2m$ .

Two mercury drops(each of radius r) merge to form a bigger drop. The surface energy of the bigger drop if T is the surface tensions

0%
$2^{5/3}\pi r^2T$
0%
$4\pi r^2T$
0%
$2\pi r^2T$
0%
$2^{8/3}\pi r^2T$

Explanation

Since final volume = 2 initial volume

${ R }^{ 3 }=2{ r }^{ 3 }$

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

So final surface area =

$4\pi { R }^{ 2 }$

$4\pi { 2 }^{ 2/3 }{ r }^{ 2 }$

Area =

${ 2 }^{ 8/3 }\pi { r }^{ 2 }$

So surface energy = T

$\times$ area

${ 2 }^{ 8/3 }\pi { r }^{ 2 }T$

$20$ cm long capillary tube is dipped in water. The water rises up to

$8$ cm. If the entire arrangement is put in a freely falling elevator, the length of water column in the capillary tube will be:

0%
$8$ cm
0%
$6$ cm
0%
$10$ cm
0%
$20$ cm

Water rises in a vertical capillary tube up to a length of

$10$ cm. If the tube is inclined at

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

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

The heart of a man pumps

$5\ litres$ of blood through the arteries per minute at a pressure of

$150\ mm$ of mercury. If the density of mercury be

$13.6\times { 10 }^{ 3 }\ kg/{ m }^{ 3 }$ and

$g=10\ m/{ s }^{ 2 }$ then the power of heart in watt is

0%
$1.50$
0%
$1.70$
0%
$2.35$
0%
$3.0$

Explanation

Given,

$\dfrac{dV}{dt}=5L=\dfrac{5\times 10^{-3}}{60}m^3/s$

Power of heat,

$=P\cdot \dfrac{dV}{dt}$

$=\rho gh\times \dfrac{dV}{dt}$

$=(13.6 \times 10^3)(10)(0.15)\times \dfrac{5\times 10^{-3}}{60}$

$=1.70W$

A spherical solid ball of volume

$V$ is made of a material of density

$\rho_{1}$ . It is falling through a liquid of density

$\rho_{1}(\rho_{2} < \rho_{1})$ . Assume that the liquid applies a viscous force on the ball that is proportional to the square of its speed

$v$ , i.e.

$F_{viscous} = kv^{2}(k > 0)$ . The terminal speed of the ball is

0%
$\sqrt {\dfrac {Vg(\rho_{1} - \rho_{2})}{k}}$
0%
$\dfrac {Vg\rho_{1}}{k}$
0%
$\sqrt {\dfrac {Vg\rho_{1}}{k}}$
0%
$\dfrac {Vg(\rho_{1} - \rho_{2})}{k}$

Explanation

$Fv=k{ v }^{ 2 }$

For terminal velocity, force on the ball must be zero

$k{ v }^{ 2 }+v\rho _{ 2 }g=v\rho _{ 1 }g\\ v=\sqrt { \cfrac { vg\left( { \rho }_{ 1 }-{ \rho }_{ 2 } \right) }{ K } }$

Depth of sea is maximum at Mariana Trench in West Pacific Ocean. Trench has maximum depth of about

$11km$ . At bottom of trench water column above it exerts

$1000$ atm pressure. Percentage change in density of sea water at such depth will be around
(Given,

$B=2\times {10}^{9}N{m}^{-2}$ and

${p}_{atm}=1\times {10}^{5}N{m}^{-2}$ )

0%
about $5$ %
0%
about $10$ %
0%
about $3$ %
0%
about $7$ %

A vertical tank, open at the top, is filled with a liquid and rests on a smooth horizontal surface. A small hole is opened at the center of one side of the tank. The area of a cross-section of the tank is

$N$ times the area of the hole, where

$N$ is a large number. Neglect mass of the tank itself. The initial acceleration of the tank is:

0%
$\dfrac {g}{2N}$
0%
$\dfrac {g}{\sqrt {2}N}$
0%
$\dfrac {g}{N}$
0%
$\dfrac {g}{2\sqrt {N}}$

Explanation

Let the height of the tank is L, density \rho and area is A

velocity of efflux=v

$\Rightarrow v=\sqrt{2g\dfrac{L}{2}}=\sqrt{gL}$ ...eq(1)

when the liquid flows out of the container horizontally, a force is exerted on the container.

force

$F=\rho\times$ (area of the hole)

$\times v^2$

$m_{0}a=\rho \times \dfrac{A}{N}\times v^2$ \therefore a is acceleration

$\rho = \dfrac{m_{0}}{AL}$ ...eq(2)

from eq(1) and eq(2)

$\Rightarrow m_{0}a=\dfrac{m_{0}}{AL}\times\dfrac{A}{N}\times gL$

$\Rightarrow a=\dfrac{g}{N}$

Hence C option is correct .

A sealed tank contains water to a height of 11 m and air at 3 atm. Water flower out from the bottom of a tank through a small hole. The velocity of efflux is (g=

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

0%
18.1 ${ ms }^{ -1 }$
0%
24.2 ${ ms }^{ -1 }$
0%
20.4 ${ ms }^{ -1 }$
0%
28.6 ${ ms }^{ -1 }$

Explanation

According to Bernaulis theoram

$0.01\times { 10 }^{ 5 }\times 3+\rho gh=\cfrac { 1 }{ 2 } \rho { v }^{ 2 }\\ 3.03\times { 10 }^{ 5 }+1000\times 10\times 11=\cfrac { 1 }{ 2 } \times 1000{ v }^{ 2 }\\ v=\sqrt { \cfrac { 4.13\times { 10 }^{ 5 }\times 2 }{ 1000 } } \\ v=28.6m/s$

Pressure applied to an enclosed fluid is transmitted undiminished to every point of the fluid and the walls of containing vessel. This law was first formulated by

0%
Reynolds
0%
Bernoulli
0%
Pascal
0%
Torricelli

Two syringes of different cross section (without needle) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1 cm and 3 cm respectively. If a force of 10 N is applied to the smaller piston then the force exerted on the larger piston is

0%
30 N
0%
60 N
0%
90 N
0%
100 N

Explanation

Since pressure is transmitted undiminished throughout the water

$\therefore \displaystyle \frac{F_1}{A_1} =\frac{F_2}{A_2}$
where

$F_1$ and

$F_2$ are the forces on the smaller and on the larger pistons respectively and

$A_1$ and

$A_2$ are the respective areas.

$\therefore \displaystyle F_2=\frac{A_2}{A_1}F_1 = \frac{\pi (D_2/2)^2}{\pi (D_1/2)^2} F_1 = \left ( \frac{D_2}{D_1} \right )^2 F_1$

$\displaystyle = \frac {(3 \times 10^{-2}\,m)^2}{(1 \times 10^{-2}\,m)^2} \times 10\,N = 90\,N$

When the flow parameters of any given instant remain same at every point, then flow is said to be

0%
laminar
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steady state
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turbulent
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quasistatic

The pressure of water in a water pipe when tap is opened and closed in respectively

$3\times 10^{5}N/m^{2}$ and

$3.5\times 10^{5} N/m^{2}$ . Determine the velocity of flow when tap is open?

0%
$10\ m/s$
0%
$5\ m/s$
0%
$20\ m/s$
0%
$15\ m/s$

Explanation

$\Delta P=\cfrac { 1 }{ 2 } \rho { v }^{ 2 }\\ \left( 0.5\times { 10 }^{ 5 } \right) =\cfrac { 1 }{ 2 } \times 1000\times { v }^{ 2 }\\ v=10m/s$

In a wind tunnel experiment, the pressures on the upper and lower surfaces of the wings are 0.90 x

$10^5$ Pa and 0.91 x 10

$^5$ Pa respectively. If the area of the wing is 40 m

$^2$ the net lifting force on the wing is

0%
2 x 10 $^4$ N
0%
4 x 10 $^4$ N
0%
6 x 10 $^4$ N
0%
8 x 10 $^4$ N

Explanation

Given data,

Pressure of the upper surface

$P_{2}=0.91\times10^{5}Pa$

Pressure of the lower surface

$P_{1}=0.90\times10^{5}Pa$

Area

$=40m^{2}$

Force

$=?$

Pressure difference which provides the lift

= pressure difference xx Area of the wing

$F=(P_2−P_1)\times A$

$F=(0.91\times 10^{5}−0.90\times 10^{5})Pa\times ms^{−2}$

$F=0.01\times10^{5}Pa\times 40m^{2}$

$F=4\times 10^{4}N.$

Hydraulic brakes are based on

0%
Pascal's law
0%
Torricelli's law
0%
Newton's law
0%
Boyle's law

In the figure shown an ideal liquid is flowing through the tube which is of uniform area of cross-section. The liquid has velocities

$v_A$ and

$v_B$ , and pressures

$P_A$ and

$P_B$ at points A and B respectively. Then it going to be

0%
$v_B>v_A$
0%
$v_B=v_A$
0%
$P_B<P_A$
0%
$P_B=P_A$

Explanation

For a streamline flow of an ideal liquid,

$v_A = v_B$ . The pressure at B = pressure at A + pressure due to column of liquid of height

$AB$ .

A rain drop of radius 0.3 mm falls through air with a terminal velocity of 1 m

$s^{-1}$ . The viscosity of air is

$18 \times 10^{-5}$ poise. The viscous force on the rain drop is then

0%
$1.018 \times 10^{-2}$ dyne
0%
$2.018 \times 10^{-2}$ dyne
0%
$3.018 \times 10^{-2}$ dyne
0%
$4.018 \times 10^{-2}$ dyne.

Explanation

Here,

$r = 0.3 mm = 0.03 cm,$

$v = 1 ms^{-1} = 100 cm s^{-1},$

$\eta = 18 \times 10^{-5}$ poise.
According to stokes law, force of viscosity on rain drop is

$F = 6 \pi \eta r v$

$= 6 \times 3.142 \times 18 \times 10^{-5} \times 0.03 \times 100$ dyne

$= 1.018 \times 10^{-3}$ dyne.

A steam of water flowing horizontally with a speed of

$25 ms^{-1}$ gushes out of a tube of cross-sectional area

$10^{-3}m^2$ , and hits at a vertical wall nearby. What is the force exerted on the wall by the impact of water?

0%
$125\ N$
0%
$625\ N$
0%
$-650\ N$
0%
$-1125\ N$

Explanation

Here,

$u=25 ms^{-1}, v= 0, t= 1s, A=10^{-3}m^2$
Density of water,

$rho=1000 kg m^{-3}$

Mass of water gushed out per second, m =

$\dfrac{volume\times density}{time}=\dfrac{area\times distance\times density}{time}$

$Area \times velocity \times density$

$=A v\rho = 10^{-3}\times 25\times1000=25 kg$
Force exerted on the wall by the impact of water,

$F=ma= m(\dfrac{v-u}{t})= 25\times (\dfrac{-2}{1})= 625 N$

F'=-F= 625 N

Spherical balls of radius R are falling in a viscous fluid of velocity v. The retarding viscous force acting on the spherical ball is

0%
directly proportional to R but inversely proportional to v.
0%
directly proportional to both radius R and velocity v.
0%
inversely proportional to both radius R and velocity v
0%
inversely proportional to R but directly proportional to velocity v

Explanation

Retarding force acting on a ball falling into a viscous fluid

$F=6\pi\eta Rv$

Where

$R=$ Radius of ball

$v=$ velocity of ball

$\eta=$ coefficient of viscosity

$\therefore F\propto R$ and

$F\propto v$ or in words,

Retarding force is directly proportional to both

$R$ and

$v$

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

0%
zero
0%
equal to g
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less than g
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more than g

Applications of Bernoulli's theorem can be seen in the

0%
dynamic lift of Aeroplane
0%
hydraulic press
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helicopter
0%
none of these

Torricelli's barometer used mercury but Pascal duplicated it using French wine of density 989 kgm

$^{-3}$ . in that case, the height of the wine column for normal atmospheric pressure is (Take the density of mercury =13.6 x 10

$^3$ kg m

$^{-3}$ )

0%
5.5 m
0%
10.5 m
0%
9.8 m
0%
15 m

Explanation

$P= h\rho g$
Height of mercury column in Torricelli's barometer

$=0.76 m \ of \ Hg,$

$\rho _{mercury} =13.6 \times 10^3 kg m^{-3}$

$\rho _{french\, wine}= 984 kg m^{-3}$

$0.76 \times (13.6 \times10m^3) \times 9.8 = h \times 984 \times 9.8$

$h = \displaystyle \frac{ 0.76 \times 13.6 \times 10m^3 \times 9.8}{984 \times 9.8} = 10.5 m$

The velocity of water in river is

$18 km h^{-1}$ near the surface. If the river is 5 m deep, then the Shearing stress between the surface layer and the bottom layer is then(coefficient of viscosity of water,

$\eta = 10^{-3} Pa \,\, s$ )

0%
$10^{-2} N m^{-2}$
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$10^{-3} N m^{-2}$
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$10^{-4} N m^{-2}$
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$10^{-5} N m^{-2}$

Explanation

As the velocity of water at the bottom of the river is zero,

$\displaystyle dv=18 km h^{-1} =18 \times \frac{5}{18} = 5 m s^{-1}$
Also,

$dx = 5 m$ ,

$\eta = 10^{-3} Pa \,s$
Force of viscosity,

$\displaystyle F =\eta A \frac{dv}{dx}$

$\therefore$ Shearing stress, =

$\displaystyle \frac{F}{A} = \eta \frac{dv}{dx}$

$\displaystyle = \frac{10^{-3} \times 5}{5} = 10^{-3} N m^{-2}$

A solid sphere falls with a terminal velocity v in air. If it is allowed to fall in vacuum,

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terminal velocity of sphere = v
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terminal velocity of sphere < v
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terminal velocity of sphere > v
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sphere never attains terminal velocity

A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is

$2.0 \times 10^{-5} kg m^{-1} s^{-1}$ , the terminal velocity of the drop will be:
(The density of water =

$10^3 kg m^{-3}$ and

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

0%
$1.0 \times 10^{-4} m s^{-1}$
0%
$2.0 \times 10^{-4} m s^{-1}$
0%
$2.5 \times 10^{-4} m s^{-1}$
0%
$5.0 \times 10^{-4} m s^{-1}$

Explanation

Here,

$r = 0.0015 mm = 0.0015 \times 10^{-3} m$

$\eta =2.0 \times 10^{-5} kg m^{-1} s^{-1}$

$\rho = 1.0 \times 10^3 kg m^{-3}$

$g= 10ms^{-2}$
Neglecting the density of air, the terminal velocity of the water drop is

$\displaystyle v_T = \dfrac{2}{9} \dfrac{r^2 \rho g}{\eta} = \dfrac{2 \times (0.0015 \times 10^{-3})^2 \times 1.0 \times 10^3 \times 10}{9 \times 2.0 \times 10^{-5}}$

$v_T=2.5 \times 10^{-4}m s^{-1}$

At what velocity does water emerge from an orifice in a tank in which gauge pressure is

$3\times 10^5N m^{-2}$ before the flow starts? (Take the density of water= 1000 kg m

$^{-3}$ .)

0%
$24.5 m s^{-1}$
0%
$14.5 m s^{-1}$
0%
$34.5 m s^{-1}$
0%
$44.5 m s^{-1}$

Explanation

Here,

$P = 3 \times 10^5 Nm^{-2},$

$\rho = 1000 kgm^{-3},$

$g = 9.8 ms^{-2}$
As

$P = h\rho g$

$\therefore \displaystyle h = \frac{P}{\rho g}= \frac {3 \times 10^5}{1000 \times 9.8} m$

Velocity of efflux,

$\displaystyle v = \sqrt {2gh} = \sqrt \frac {2 \times 9.8 \times 3 \times 10^5}{1000 \times 9.8}$

$= \sqrt {600} = 24.495 \,ms^{-1} = 24.5 \,ms^{-1}$

Which of the following device is used to measure the rate of flow of liquid through a pipe?

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Thermometer
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Barometer
0%
Manometer
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Venturimeter

Two capillaries of same length and radii in the ratio 1: 2 are.connected in series. A liquid flows through them in streamlined condition. If the pressure across the two extreme ends of the combination is 1 m of water, the pressure difference across first capillary is

0%
9.4 m
0%
4.9 m
0%
0.49 m
0%
0.94 m

Explanation

Here,

$l_1 = l_2 = 1m$ and

$\displaystyle \frac{r_1}{r_2} = \frac{1}{2}$

$V = \displaystyle \frac{ \pi P_1 r_1^4 }{8 \eta l} = \frac{ \pi P_2 r_2^4 }{8 \eta l}$ or

$\displaystyle \frac{ P_1 }{P_2 } = \left( \frac{ r_2 }{ r_1} \right)^4 = 16$

$\therefore P_1 = 16 P_2$

Since, both tubes are connected in series, hence pressure difference across the combination is

$P = P_1 + P_2$

$\Rightarrow$

$\displaystyle 1 = P_1 + \frac{P_1}{16}$

$\displaystyle P_1 = \frac{16}{17} = 0.94 m$

Which of the following diagrams does not represent a streamline flow?

A large cylindrical tank has a hole of area

$A$ at its bottom and water is poured into the tank through a tube of cross-section area

$A$ ejecting water at the speed

$v$ . Which of the following is true?

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Water level in tank keeps on rising
0%
No water can be stored in the tank
0%
Water level will rise to height $v^{2}/2g$ and then stop
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The water level will be oscillating

Explanation

In the beginning ,the water escaping the vessel at the rate

$\sqrt{2gh}$ and filling it at the rate of

$vA$

So the rate of change in the water level of the vessel at any instant is

(

$v-\sqrt{2gh})A$

where 'h' is the instantaneous height

Since h is quite small in the beginning, the water is escaping the vessel at a slow rate but filling at a much larger rate as such (

$v-\sqrt{2gh})A$ ,is positive (increase in water level)

soon after the filling of water occurs, the height increases and the rate of escape become larger

But when the rate =rate of filling ,then an equilibrium state is reached and there is no net change in the height of the water

the equilibrium is reached when (

$v-\sqrt{2gh})A$ =0

$\rightarrow$

$h=\dfrac{v^2}{2g}$

The velocity of the liquid coming out of a small hole of a vessel containing two different liquid of densities

$2\rho$ and

$\rho$ shown in the figure is

0%
$\sqrt { 6gh }$
0%
$2\sqrt { gh }$
0%
$2\sqrt { 2gh }$
0%
$\sqrt { gh }$

Explanation

Pressure at A

$={ \rho }_{ atm }+\rho g\left( 2h \right)$

Applying Bernoulli's theorem at A and B

${ \rho }_{ atm }+2\rho gh+\rho g\left( 2h \right) +\cfrac { 1 }{ 2 } \rho { v }_{ 2 }^{ 2 }={ \rho }_{ atm }+2\rho g\left( 0 \right) +\cfrac { 1 }{ 2 } 2\rho { v }^{ 2 }$

$4\rho gh={ v }^{ 2 }\rho \quad \left( \because { v }_{ 2 }=0 \right)$

$v=2\sqrt { gh }$

1145258_985138_ans_c0d8366477e14642be29db395c315748.png

A tube

$1\ { cm }^{ 2 }$ in cross section is attached to the top of a vessel

$1\ cm$ high and of cross section

$100\ { cm }^{ 2 }$ . Water is poured into the system filling it to depth of

$100\ cm$ above the bottom of the vessel as shown in the figure. Take

$g=10\ { m\ s }^{ -2 }$ . Find the correct statement.

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The force exerted by water against the bottom of the vessel is $100\ N$ .
0%
The weight of water in the system is $1.99\ N$
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Both $(a)$ and $(b)$ are correct.
0%
Neither $(a)$ nor $(b)$ is correct.

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 7 - 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 7 - MCQExams.com

0:0:2

Practice Class 11 Engineering Physics Quiz Questions and Answers

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