MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 8

Figure 4.191 shown water filled in a symmetrical container. Four pistons of equal area

$A$ are used at the four openings to keep the water in equilibrium. Now an additional force

$F$ is applied at each piston. The increase in the pressure at the center of the container due to this addition is

0%
$\dfrac { F }{ A }$
0%
$\dfrac { 2F }{ A }$
0%
$\dfrac { 4F }{ A }$
0%
$0$

Explanation

As at each piston additional force

$F$ is introduced so change in pressure will be the net additional force upon total area upon which it acts:

$\Delta P= \dfrac{\Delta F}{Total\, area }$

$=\dfrac{4F}{4A}=\dfrac{F}{A}$

so, net increase in pressure at cenftre

$= \dfrac{F}{A}$

The height at which liquid will stand in the open end of the pipe is

0%
$4\ m$
0%
$3.5\ m$
0%
$4.5\ m$
0%
$5.6\ m$

Explanation

Let atmospheric pressure be

$P_0$

Then absolute pressure of air

$P + P_0 = 4 atm$

Pressure at A,

$P_A = P + P_0 + \rho gh$ from equation of continuity and Bernoulli

$P_B = P_C$

Apply Bernoullis at A & E

$P + P_0 + \rho gh = P_C + \dfrac{PV_C^2}{2} = P_0 + \dfrac{PV_E^2}{2}$

From hydrostatics

$P_C = P_0 + \rho gh'$

From continuity

$A_C V_C = A_{\varepsilon} V_{\varepsilon}$

So we get

$V_{\varepsilon} = 13.4 m/s$

$V_C = 6.7 m/s , h' = 4.5 m$

Discharge rate

$= A_{\varepsilon} V_{\varepsilon} = 3 \times 10^{-4} \times 13.4 m^3 /s$

$= 4.02 \times 10^{-3} m^3/s$

1345075_986406_ans_8ca7408353714215a1423b397cf3a5fd.png

A body

$B$ is capable of remaining stationary inside a liquid at the position shown in Fig.4.205 (a). If the whole system in gently placed on smooth inclined plane (Fig.4205 (b)) and is allowed to side down, then

$(O\ <\ \theta \ <\ 90^{o})$

0%
The body will move up (relative to liquid)
0%
The body will move down (relative to liquid)
0%
The body will remain stationary (relative to liquid)
0%
The body will move up for some inclination $\theta$ and will move down for another inclination $\theta$

Air is streaming past a horizontal airplane wing such that its speed is

$90\ ms^{-1}$ at the lower surface and

$120\ ms^{-1}$ over the upper surface. If the wing is

$10\ m$ long and has a average width of

$2\ m$ , the difference of pressure on the two sides and the gross lift on the wing respectively, are (density of

$air=\ 1.3\ kg\ { m }^{ -3 }$ )

0%
$5\ Pa$
0%
$95\ Pa$
0%
$4098\ Pa$
0%
$4095\ Pa$

The horizontal distance

$x$ travelled by the liquid initially is

0%
$\sqrt { \left( H-3h \right) h }$
0%
$\sqrt { 3Hg }$
0%
$\sqrt { \left( 3H-4h \right) h }$
0%
$2h$

Explanation

Applying Bernoulli's equation at

$A$ &

$B$ -

${ P }_{ 0 }+dg\dfrac { H }{ 2 } +2d.g.\left( \dfrac { H }{ 2 } -h \right) ={ P }_{ 0 }+\dfrac { 1 }{ 2 } \left( 2d \right) { V }^{ 2 }$

$\dfrac { dgH }{ 2 } +dgH-2dgh=d{ V }^{ 2 }$

$\Rightarrow \dfrac { 3dHg }{ 2 } -2dgh={ V }^{ 2 }d$

$\Rightarrow \dfrac { 3Hg }{ 2 } -2gh={ V }^{ 2 }$

$\Rightarrow V=\sqrt { \left( 3H-4gh \right) \dfrac { g }{ 2 } }$

$x=V(t)$

$\Rightarrow x=V\sqrt { \dfrac { 2h }{ g } }$

$\Rightarrow x=\sqrt { \left( 3H-4h \right) \dfrac { g }{ 2 } } \sqrt { \dfrac { 2h }{ g } }$

$\Rightarrow x=\sqrt { \left( 3H-4h \right) .h }$

1168133_988349_ans_a35debcadcb2409aba3e94dc869993d1.png

A tank with a small hole at the bottom has been filled with water and kerosene (specific gravity

$0.8$ ). The height of water is

$3\ m$ and that of kerosene

$2m$ . When the hole opened the velocity of fluid coming out from it is nearly: (take

$g=10\ {ms}^{-2}$ and density of water

${10}^{3}\ kg\ {m}^{-3}$ )

0%
$9.6\ {ms}^{-1}$
0%
$8.5\ {ms}^{-1}$
0%
$7.6\ {ms}^{-1}$
0%
$10.7\ {ms}^{-1}$

Explanation

Applying Bernoulli's law between the kerosene - water interface and the hole :

$P_{0}+(\rho _{w}gh_{1}+\rho _{k}gh_{2})+\dfrac{1}{2}\rho _{w}v_{1}^{2}=P_{0}+0+\dfrac{1}{2}\rho _{w}v_{2}^{2}$ (hole is at base so

$h=0$ )

$\Rightarrow 3\times 10^{3}g+2\times 0.8\times 10^{3}g=\dfrac{1}{2}\times 10^{3}\times v_{2}^{2}$ (

$v_{1}\approx 0$ as flow is very slow at interface)

$\Rightarrow v_{2}^{2}=9.2g$

$\Rightarrow v_{2}=\sqrt{92}$ (given :

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

$\Rightarrow v_{2}=9.6 m/s$

The height

${h}_{m}$ at which the hole should be punched so that the liquid travels the maximum distance is

0%
$\dfrac { 2H }{ 3 }$
0%
$\dfrac { 3H }{8 }$
0%
$\dfrac { 4H }{ 3 }$
0%
$\dfrac { 5H }{ 3 }$

Explanation

$x = v\sqrt{\dfrac{2h}{g}}$

$= \sqrt{\dfrac{(3H-4h)g}{2} \times \dfrac{2h}{g}}$

$= \sqrt{(3H-4h)h}$

$x^2 = 3Hh - 4h^2$

$\dfrac{dx^2}{dh} = 3H-8h= 0$

$\Rightarrow h = \dfrac{3H}{8}$

Calculate the pressure inside a small air bubble of radius

$r$ situated at a depth

$h$ below the free surface of liquids of densities

${\rho}_{1}$ and

${\rho}_{2}$ and surface tension

${T}_{1}$ and

${T}_{2}$ . The thickness of the first and second liquids are

${h}_{1}$ and

${h}_{2}$ , respectively. Take atmosphere pressure

$={P}_{0}$ .

0%
${ P }_{ 0 }+{ \rho }_{ 1 }{ gh }_{ 1 }+{ \rho }_{ 2 }g\left( h-{ h }_{ 1 } \right) -\dfrac { 2{ T }_{ 2 } }{ r }$
0%
${ P }_{ 0 }+{ \rho }_{ 1 }{ gh }_{ 1 }+{ \rho }_{ 2 }g\left( h-{ h }_{ 1 } \right) +\dfrac { 2{ T }_{ 2 } }{ r }$
0%
${ P }_{ 0 }-{ \rho }_{ 1 }{ gh }_{ 1 }+{ \rho }_{ 2 }g\left( h-{ h }_{ 1 } \right) +\dfrac { 2{ T }_{ 2 } }{ r }$
0%
None of these

Explanation

$P_c=P_B+\dfrac{2T}{r}$

$P_B=P_A+\rho_1gh_1+\rho_2g(h-h_1)$

where

$P_A=P_0$

$\boxed{P_c=P_0+\rho_1gh_1+\rho_2g(h-h_1)+\dfrac{2T_2}{r}}$

1345091_987385_ans_85efd41d075d4c2dabe2d1b2994e005b.png

Water is filled in a container up to height

$3\ m$ . A small hole of area

$a$ is punched in the wall of the container at a height

$52.5\ cm$ from the bottom. The cross sectional area of the container is

$A$ . If

$a/A =0$ ; then

${v}^{2}$ is (where

$v$ is the velocity of water coming out of the hole)

0%
$50$
0%
$51$
0%
$48$
0%
$51.5$

Explanation

Given

$\cfrac{a}{A}=0\\ \Rightarrow V_x=\sqrt{2gH}\\ V^2=2gH=2\times10\times2.5=50$
1157572_986681_ans_04a54cb202254f6097ad30fe0266965f.png

In the diagram shown, the difference In the two tubes of the manometer is

$5\ cm$ , the cross section of the tube at

$A$ and

$B$ is

$6\ {mm}^{2}$ and

$10\ {mm}^{2}$ respectively. The rate at which liquid flows though the tube is

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

0%
$7.5\ cc/s$
0%
$8.0\ cc/s$
0%
$10.0\ cc/s$
0%
$12.5\ cc/s$

A cylindrical vessel of cross-section

$A$ contains water to a height

$h$ . There is a hole in the bottom of radius

$a$ . The time in which it will be emptied is

0%
$\dfrac { 2A }{ { \pi a }^{ 2 } } \sqrt { \dfrac { h }{ g } }$
0%
$\dfrac { \sqrt { 2 } A }{ { \pi a }^{ 2 } } \sqrt { \dfrac { h }{ g } }$
0%
$\dfrac { 2\sqrt { 2 } A }{ { \pi a }^{ 2 } } \sqrt { \dfrac { h }{ g } }$
0%
$\dfrac { A }{ { \sqrt { 2 } \pi a }^{ 2 } } \sqrt { \dfrac { h }{ g } }$

A jar contains a gas and a few drops of water at TK. The pressure in the jar is

$830$ mm of Hg. The temperature of the jar is reduced by

$1\%$ . The saturated vapour pressure of water at the two temperatures are

$30$ and

$25$ mm of Hg. Calculate the new pressure in the jar.

0%
$917$ mm of Hg.
0%
$817$ mm of Hg.
0%
$1017$ mm of Hg.
0%
$777$ mm of Hg.

Mark out the correct statement(s).

0%
Net force acting on the base of the vessel $>$ weight of the liquid inside the vessel.
0%
Net force acting on the base of the vessel $=$ weight of the liquid inside the vessel.
0%
Net pressure force acting on the liquid $=$ weight of the vessel.
0%
Both (a) and (c) are correct.

A section of pipe is shown in the figure. Area of cross and height of A and B are

$0.1{m}^{2}, 1 m$ and

$0.06 {m}^{2}$ ,

$2m$ respectively. find The pressure difference between A and B if velocity at B is

$5 m/s$ and liquid flowing is water.

0%
$20 kpa$
0%
$18 kpa$
0%
$16 kpa$
0%
$22 kpa$

Explanation

From Bernoulli's equation

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

$Av=$ constant by equation of continuity

${ p }_{ A }-{ p }_{ B }=\rho g({ h }_{ B }-{ h }_{ A })+\dfrac { 1 }{ 2 } \rho({ v }_{ B }^{ 2 }-{ v }_{ A }^{ 2 })$

$=\rho g[2-1]+\dfrac { 1 }{ 2 } \rho [25-{ \left( 5\times \dfrac { 0.06 }{ 0.1 } \right) }^{ 2 }]$

$=\rho g+\dfrac { \rho }{ 2 } [25-9]=\rho [10+8]$

$\Delta p=18\times { 10 }^{ 3 }pa$

$\Delta p=18\quad kpa$

The U tube having identical limbs contains mercury (density

${ \rho }_{ m }$ ) to a level as shown in the figure. If the left limbs is filled to the top with water (

${ \rho }_{ w }$ ), then the rise of mercury level in the right limb will be

0%
12ρm
0%
$\dfrac { l{ \rho }_{ w } }{ 2{ \rho }_{ m }+{ \rho }_{ w } }$
0%
$\dfrac { { \rho }_{ m }l }{ { \rho }_{ w } }$
0%
$\dfrac { l{ \rho }_{ w } }{ 2{ \rho }_{ m }-{ \rho }_{ w } }$

Explanation

The correct option is D.

Given,

U tube contains mercury density

$\rho_m$

So,

The pressure at the bottom should be the same due to both the limbs.

$\rho_A=\rho_wg(1+x)$ And

$\rho_B=\rho_mg(2x)$

Since

$\rho_A=\rho_B$

Thus,

$x=\dfrac{1\rho_w}{2\rho_m-\rho_w}$

969455_1017410_ans_aac7c14c4c3a4b45b1e28d8c82ba3bde.png

A small sphere of radius r falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity is proportional to:

0%
${r^5}$
0%
${r^2}$
0%
${r^3}$
0%
${r^4}$

Explanation

Let r is the radius of sphere and ${{v}_{t}}$ is its terminal speed. Then the weight of sphere is balanced by the buoyant force and viscous force such that:

Weight,

$w=mg$

$\because \rho =\dfrac{m}{V}$

$m=\dfrac{4}{3}\pi {{r}^{3}}\rho g........(1)$

$so,$

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

Buoyant force,

${{F}_{B}}=\dfrac{4}{3}\pi {{r}^{3}}\sigma g..........(2)$

Where, $\sigma$ is density of water.

Viscous force, $F=6\pi \eta rvt..............(3)$

Where, $\eta$ is viscosity.

From equation (1) (2) and (3)

$w={{F}_{B}}+{{F}_{v}}$

$\dfrac{4}{3}\pi {{r}^{3}}\rho g=\dfrac{4}{3}\pi {{r}^{3}}\sigma g+6\pi \eta rvt$

${{v}_{t}}=\dfrac{2}{9}\dfrac{{{r}^{2}}\left( \rho -\sigma \right)g}{\eta }.......(4)$

The rate of production of heat when the sphere attains its terminal velocity is

equal to work done by the viscous forces.

$W=\dfrac{dQ}{dt}={{F}_{v}}\times {{v}_{t}}$

$W=6\pi \eta rv_{t}^{2}$

$W=6\pi \eta r{{\left( \dfrac{2}{9}\dfrac{{{r}^{2}}\left( \rho -\sigma \right)g}{\eta } \right)}^{2}}$

$\dfrac{dQ}{dt}\propto {{r}^{5}}$

Water flow through a vertical tube of variable cross-section. The area of cross-section at

$A$ and

$B$ are

$6$ and

$3\ mm^{2}$ respectively. If

$12\ cc$ of water enters per second through

$A$ , find the pressure difference

$\left|P_{A}-P_{B} \right|$

$(g=10\ m/s^{2})$ . The separation between the cross-sections at

$A$ and

$B$ is

$100\ cm$ .

0%
$0.4\ \times 10^{5}\ dyne/cm^{2}$
0%
$2.29\ \times 10^{5}\ dyne/cm^{2}$
0%
$5.9\ \times 10^{5}\ dyne/cm^{2}$
0%
$3.9\ \times 10^{5}\ dyne/cm^{2}$

Water if flowing steadily through a horizontal pipe of non-uniform cross-section. If the velocity of water at a point where cross-section is

$0.02 m^3$ is

$2m/s$ , what is the velocity of water at another point where the cross-section is

$0.01 m^3$ ?

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

Explanation

According to torricellis theorem.

${ A }_{ 1 }{ V }_{ 1 }={ A }_{ 2 }{ V }_{ 2 }$

where

$A-$ Area of cross section

$V-$ Velocity of liquid

So,

$0.02\left( 2 \right) ={ V }_{ 2 }\left( 0.01 \right)$

$\Rightarrow \quad { V }_{ 2 }=4m/s$

A container is filled with water, accelerating with acceleration

$10 m/s^2$ , long

$+ve$ X-axis on a smooth horizontal surface. The velocity of efflux of water at a point

$P$ at the bottom of the tank and near its left most corner is

0%
$4.43 m/s$
0%
$5.48 m/s$
0%
$4 m/s$
0%
$3 m/s$

Explanation

Due to acceleration liquid profile gets tilted as shown

$tan\theta =\dfrac { a }{ g } =1$

$\theta ={ 45 }^{ 0 }$

So,

$x=\dfrac { 1 }{ 2 } m$

Now, height of the liquid in left most section

$=\left( 1+\dfrac { 1 }{ 2 } \right) m=\dfrac { 3 }{ 2 } m$

So,

$V=\sqrt { 2gh }$

$V=\sqrt { 2\times 10\times \dfrac { 3 }{ 2 } }$

$V=\sqrt { 30 }$ m/s

$V=5.48$ m/s

1106853_1020773_ans_6a470a99959147c88f042cc7575919b1.png

During the rocket propulsion, gases ejecting with velocity

$1\ km/s$ relative to rocket. The rate of fuel consumption is

$\dfrac {m}{10}\ kg/s$ , where

$m$ is the instantaneous mass of the rocket. If air resistance varies according to equation

$f=0.15\ mv$ , then terminal velocity of the rocket is:-

0%
$300\ m/s$
0%
$600\ m/s$
0%
$7.92\ m/s$
0%
$11.2\ m/s$

Explanation

Thrust force must balance air drag to attain terminal velocity.

$\cfrac { m }{ 10 } \times 1000=0.15mv\\ v=\cfrac { 100 }{ 15 } \times 100\\ v=600m/s$

990841_1021477_ans_18ade4b2b03a4564aafaf3f2038be60a.jpg

If ball of steel (density

$\rho=7.8\ {cm}^{-3}$ ) attains a terminal velocity of

$10\ cm {s}^{-1}$ when falling in a tank of water (coefficient of viscosity

${ \eta }_{ water }=8.5\times { 10 }^{ -4 }\ Pa.s$ ) then its terminal velocity in glycerin (

$\rho =1.2\ g{ cm }^{ -3 }\ \eta =13.2\ Pa.s$ ) would be nearly:

0%
$1.5\times { 10 }^{ -5 }\ cm{ s }^{ -1 }$
0%
$1.6\times { 10 }^{ -5 }\ cm{ s }^{ -1 }$
0%
$6.25\times { 10 }^{ -4 }\ cm{ s }^{ -1 }$
0%
$6.45\times { 10 }^{ -4 }\ cm{ s }^{ -1 }$

Explanation

Terminal velocity of ball when falling in liquid is given by

$(\rho_{b}-\rho_{l})Vg=6\pi\eta rv$ ...eq(1)

1. case

when ball falling in tank of water of density

$\rho_{water}=1gcm^{-3}$ ,

$\rho_{ball}=7.8gcm^{-3}$

$\eta_{water}=8.5\times10^{-4}Pa.s$ and terminal velocity

$v=10cms^{-1}$

Put all the above value in eq(1)

$\Rightarrow (7.8-1)Vg=6\pi\times8.5\times10^{-4}r\times10$ ...eq(2)

2. case

when ball falling in tank of glycerin

$\rho_{ball}=7.8gcm^{-3}$

$\eta_{glycerin}=13.2Pa.s$ and terminal velocity

$v_{1}$

put all the value in eq(1)

$\Rightarrow (7.8-1.2)Vg=6\pi\times13.2r\times v_{1}$ ...eq(3)

divide eq(3) b eq(1)

$\dfrac{7.8-1.2}{7.8-1}=\dfrac{13.2\times v_{1}}{8.5\times10^{-4}\times10}$

$\Rightarrow v_{1}=\dfrac{6.6\times8.510^{-4}\times10}{6.8\times13.2}=6.25\times10^{-4}cms^{-1}$

terminal velocity of ball in glycerin is

$6.25\times10^{-4}cms^{-1}$

Hence C option is correct.

When an air bubble of radius

$r$ rises from the bottom to the surface of a lake, the radius becomes

$\dfrac {5r}{4}$ . Taking the atmospheric pressure to be equal to

$10\ m$ height of water column, the depth of the lake would approximately be (ignore the surface tension and the effect of temperature):

0%
$11.2\ m$
0%
$8.7\ m$
0%
$9.5\ m$
0%
$10.5\ m$

Explanation

$\quad Given\quad that\quad temperature\quad and\quad surface\quad tension\quad \\ \quad \quad does\quad not\quad play\quad any\quad role\quad in\quad the\quad process\\ \quad \quad Let\quad the\quad temperature\quad remain\quad constant\quad throughout\quad the\quad process\\ \quad \quad so\quad { p }_{ 1 }{ v }_{ 1 }={ p }_{ 2 }{ v }_{ 2 }\quad since\quad T=c\\ \quad \quad { p }_{ 1 }=\quad { p }_{ atm }+\rho gh\quad \quad \quad \quad ;\quad h=\quad depth\quad of\quad the\quad lake\\ \quad \quad { p }_{ 2 }=\quad { p }_{ atm }=10\rho g\\ { \quad v }_{ 1 }=\frac { 4 }{ 3 } \pi { r }^{ 3 }\quad ;\quad { v }_{ 2 }=\frac { 4 }{ 3 } \pi { \frac { 5r }{ 4 } }^{ 3 }\quad putting\quad the\quad value\quad in\quad { p }_{ 1 }{ v }_{ 1 }={ p }_{ 2 }{ v }_{ 2 }\quad we\quad get\\ \quad h=\quad 9.53m\quad$

An open U-tube contains mercury. When

$13.6 cm$ of water is pourd into one of the arms of the tube then the mercury rise in the other arm from its initial level is:

0%
$1 cm$
0%
$0.5 cm$
0%
$10 cm$
0%
$5 cm$

Explanation

$(2x+h)P_m g=p_m h\ g+P_w\ g\ 13.6$

$2P_m\ g\ x=P/w\ g\ 13.6 \left(\dfrac {P_m}{P_w}=13.6\right)$

$\boxed {x=1/2}cm$

1432518_1018784_ans_233148b75fb6425bace2a60618811587.png

An open U-tube contain mercury when

$13.6cm$ of water is poured into one of the arms of the mercury rise in the other its initial level is

0%
$1cm$
0%
$0.5cm$
0%
$10cm$
0%
$5cm$

The lower end of glass capillary tube is dipped in water.Water rises to a height of

$8cm$ .The Tube is then broken at a height of

$6cm$ . The height of water column and angle of contact will be

0%
$6m\sin - 1\dfrac{3}{4}$
0%
$6m\cos - 1\dfrac{3}{4}$
0%
$4m\sin - 1\dfrac{1}{2}$
0%
$6m\tan - 1\dfrac{3}{4}$

A water drop of radius

$1.5\ mm$ is falling from height

$1\ km$ having drag constant

$0.5$ density of water drop is

$1000\ kg/{m}^{3}$ and density of air is

$1.29\ kg/{m}^{3}$ . Find the terminal velocity.

0%
$8.7\ m/s$
0%
$7.8\ m/s$
0%
$5.6\ m/s$
0%
$4.3\ m/s$

A wooden piece

$5\, N$ in weight and

$5cm \times 3cm \times 2cm$ in size lies on

$5cm \times 2cm$ face. The pressure exerted by it in

$N\, per\, cm^2$ is:

0%
$150$
0%
$50$
0%
$0.5$
0%
$15$

Explanation

force on wooden piece

$=N=mg$

$N=5N$

$P= \dfrac {N}{A} =\dfrac {5N}{10cm^2} =0.5N/cm^2$

A hydraulic lift is used to lift a car of mass

$2000\ kg$ . The cross-sectional area of the larger piston on which the car is supported is

$10^{-1} m^{2}$ . The smaller pistion is

$5m$ below the bigger piston. The cross-sectional area of the smaller piston is

$10^{-2} m^{2}$ . What weight placed on smaller piston will be sufficient to keep the car in equilibrium?

0%
$250\ kg$
0%
$170\ kg$
0%
$50\ kg$
0%
$350\ kg$

In a car lift, compressed air exerts a force

$F$ on a small piston having a radius of

$5\ cm$ . This pressure is transmitted to a second piston of radius

$15\ cm$ . If the mass of the car to be lifted is

$1350\ kg$ , what is

$F$ ?

0%
$1.5\ \times 10^{3}N$
0%
$2.5\ \times 10^{3}N$
0%
$3.5\ \times 10^{3}N$
0%
$4.5\ \times 10^{3}N$

Explanation

pascal's law states that

${ F }_{ 1 }{ d }_{ 1 }={ F }_{ 2 }{ d }_{ 2 }$

${ P }_{ 1 }=\cfrac { { F }_{ 1 } }{ { A }_{ 1 } } ,{ P }_{ 2 }=\cfrac { { F }_{ 2 } }{ { A }_{ 2 } }$

Let

${ F }_{ 1 }=F$

${ F }_{ 2 }=1350\times 10N$

${ d }_{ 1 }=10cm$

${ d }_{ 1 }=30cm$

$F\times 10=13500\times 30$

$F=\cfrac { 405000 }{ 10 }$

$F=4.5\times { 10 }^{ 3 }N$

The motion of a body is given by the equation

$\dfrac{dv}{dt} = 6-3v:$ where

$v$ is in

$m/s$ . If the body was at rest at

$t=0$
(i) the terminal speed is

$2\ m/s$
(ii) the magnitude of the initial acceleration is

$6\ m/s$
(iii) the speed varies with time as

$v=2(1-e^{-3t})m/s$
(iv) The speed is

$1\ m/s$ , when the acceleration is half the initial

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

Explanation

$\dfrac{dv}{dt}=6-3v$

$\int{\dfrac{dv}{6-3v}}=\int{dt}$

$-\dfrac{1}{3}\log \left| 2-v \right|=dt$

$-\log \left| 2-v \right|=3t$

$\left| 2-v \right|={{e}^{-3t}}$

$2-{{e}^{-3t}}=v$

Initial acceleration is $0$ .

At $t=0$ , terminal velocity is $v=2m/s$

Hence, condition 1 and 3 are correct.

The velocity of air over the upper surface of the wing of an aerplane is

$40m/s$ and that on the lower surface is

$30m/s$ . If the area of the wing is

$5m^{2}$ and the mass of the wing is

$300\ kg$ , the net force acting on the wing is (Density of air

$=1.3kg/m^{3}$ and

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

0%
$725N$ upward
0%
$725N$ downward
0%
$2275N$ upward
0%
$2275\ N$ downward

Explanation

consider the wing be of negligible height

from Bernoulli's equation

Pressure developed across top and bottom of th ewing is

$P=\cfrac{1}{2}P({V}_{1}^{2}-{V}_{2}^{2})$

$=\cfrac{1}{2}\times 1.3\times ({40}^{2}-{30}^{2})=455Pa$

This pressure provides an upthrust

$F$ of:

$F=P\times A=455\times 5=2275N$ (A: Area of wing)

Thus net force on the wing

$N=mg-F$

$=300\times 10-2275$

$N=725N$ (downward)

1407284_1091980_ans_bee0093750494d158927522bbf72e079.PNG

An orifice of area

$10sq.mm$ is made at the bottom of a vessel containing water to a height of

$10m$ above the orifice. What is the force on the vessal

0%
$9.98N$
0%
$9.8N$
0%
$1.96N$
0%
$19.6N$

The volume of an air bubble is doubled as it rises from the bottom of lake to its surface. The atmospheric pressure is

$75\ cm$ of mercury. The ratio of density of mercury to that of lake water is

$\dfrac{40}{3}$ . The depth of the lake in metre is :

0%
$10$
0%
$15$
0%
$20$
0%
$25$

Explanation

Let

$v'$ be the volume of bubble at the surface and

$v$ at depth

$h$

Ratio of volume of air bubble at top to bottom

$v'/v=2 \Rightarrow \ v'=2v$

From idle gass equation

$p=p'\left(\dfrac{v'}{v} \right)=2p'$

$\therefore p'=P_o$ so

$p=2P_o$

Ratio of density of mercury to that of lake water,

$\rho_{hg} / \rho_w=40 /3$

Pressure at

$h$ depth

$=P_o+ \rho_w g h$

$\Rightarrow 2P_o=P_o+ \rho_w g h$

$\Rightarrow P_o= \rho_w g h$

$\Rightarrow \rho_{hg} g h_{hg}= \rho_{w} g h_{w}$

$h_w=\dfrac{\rho_{hg} h_{hg}}{\rho{w}}$

$h_w=\dfrac{40\times .75}{3}$

$h=10\ m$

two equal drops of water each of radius

$r$ are falling through air with a steady velocity

$8cm/s$ . the two drops combine to form a big drop. The terminal velocity of big drop will be:

0%
$8(2)^{\dfrac{2}{3}}cm/s$
0%
$16(2)^{\dfrac{2}{3}}cm$
0%
$4(2)^{\dfrac{2}{3}}cm$
0%
$32cm/s$

Explanation

$V=\dfrac{mg}{6\pi hr \theta}$

$S=\dfrac{mg}{6\pi hr v}$

$2\left[\dfrac{4}{3}\pi r^3\right]g=\dfrac{4}{3}\pi R^3g$

$R=2^{\dfrac{9}{3}}r$

$V_1=\dfrac{2mg}{6\pi hR}$

$V_1=\dfrac{2mg}{6\pi hr}$

$=2\times 8$

$V_1=8\times (2)^{2/3}cm/sec$

A man of mass 100 kg stands on a wood plank of area

$4\, m^2$ . What is the pressure exerted on the floor? Assume the area of a human foot to be

$200\, cm^2$ .

0%
500 N
0%
25 N
0%
50000 N
0%
250 N

The base area of a boat is

$2m^{2}$ . A man weighing

$76kg$ weight steps in to the boat. Calculate the depth in to which the boat sinks further in to water.

0%
$1.2\ cm$
0%
$2.5\ cm$
0%
$3.8\ cm$
0%
$4.2\ cm$

A liquid is coming out from the orifice of tank and falls up to a maximum horizon distance of

$6m$ . The height

$h$ is equal to

0%
$1.5m$
0%
$3.0m$
0%
$4.5m$
0%
$6.0m$

A tank full of water has a small hole at its bottom. Let

${t_1}$ be the time taken to empty first one thbird of the tank and

${t_2}$ be the time taken to empty second one third of the tank and

${t_3}$ be the time taken to empty rest of the tank then

0%
${t_1} = {t_2} = {t_3}$
0%
${t_1} > {t_2} > {t_3}$
0%
${t_1} < {t_2} < {t_3}$
0%
${t_1} > {t_2} < {t_3}$

Explanation

As the water level is decreasing the pressure at the bottom due to water (pgh) is decreasing. Because of this, inner pressure and o'cater pressure at the small hole are having decreasing difference. Due to this lowering pressure difference, the velocity of efflux will decrease. Thus it will take more time to sap out same amount of water at lower water level than that at higher.

So, $t_{3}> t_{2}> t_{1}$ Option - C is correct.

Consider a soap film on a rectangular frame of wire of area

$4 \times 4 cm^2$ . If the area of the soap film is increased to

$4 \times 5cm^2$ , the work done in the process will be (The surface tension of the soap film is

$3 \times 10^{-2} N/m$ )

0%
$12 \times 10^{-6} J$
0%
$2.4 \times 10^{-5} J$
0%
$60 \times 10^{-6} J$
0%
$96 \times 10^{-6} J$

Explanation

Consider the problem

Given,

initial area

$\begin{array}{l} { A_{ 1 } }=4cm\times 4cm=16c{ m^{ 2 } } \\ =16\times { 10^{ -4 } }{ m^{ 2 } } \end{array}$

Final area

$\begin{array}{l} { A_{ 2 } }=4cm\times 5cm=20c{ m^{ 2 } } \\ =20\times { 10^{ -4 } }{ m^{ 2 } } \end{array}$

Surface tension

$T = 3 \times {10^{ - 2}}N/m$

Work done

$=?$

In case of rectangular frame the water wets it from two sides,

Hence, the change in area

$=dA=2$

$\begin{array}{l} \left( { { A_{ 2 } }-{ A_{ 1 } } } \right) =2\times \left( { 20\times { { 10 }^{ -4 } }-16\times { { 10 }^{ -4 } } } \right) \\ \therefore dA=2\left( { { A_{ 2 } }-{ A_{ 1 } } } \right) =2\times 4\times { 10^{ -4 } } \\ =8\times { 10^{ -4 } }{ m^{ 2 } } \end{array}$

Work done

$=$ change in surface energy

$\begin{array}{l} =TdA=3\times { 10^{ -2 } }\times 8\times { 10^{ -4 } } \\ =2.4\times { 10^{ -5 } }J \end{array}$

Hence, Option

$B$ is the correct answer.

Some amount of water is poured on top face of a horizontal glass place. Excess amount of water spills over the edge a little amount of water there in a thin uniform layer a portion of which is shown in the figure. Surface tension of water is

$\sigma$ the vertical thickness of the water layer far away the edge is t, the point in the plane on the diagram on the liquid where the liquid protrudes the most is D. The vertical depth of this point from the horizontal free surface of the water layer is d, p is density of water and g is acceleration due to gravity. Mark the correct relation(s)

0%
$t = \sqrt { \dfrac { 2 \sigma ( 1 + \sin \theta ) } { \rho g } }$
0%
$t = \sqrt { \dfrac { 2 \sigma ( 1 - \cos \theta ) } { \rho g } }$
0%
$d = \sqrt { \dfrac { 2 \sigma } { \rho g } }$
0%
$d = \sqrt { \dfrac { 2 \sigma ( 1 - \cos \theta ) } { p g } }$

Air is streaming past a horizontal air plane wing such that its speed in 120 m/s over the upper surface and 90 m/s at the lower surface. If the density of air is 1.3 kg per

$metre^3$ and the wing is 1o m long and has an average width of 2m , then the difference of the pressure on the two sides of the wing of

0%
4095.0 Pascal
0%
409.50 Pascal
0%
40.950 Pascal
0%
4.0950 Pascal

Explanation

Using the Bernoulli's theorem in the upper and lower surface of the wings in the airplane

$P_{1}-P_{2}=\dfrac{1}{2} \rho\left(v_{2}^{2}-v_{1}^{2}\right)$

$=\dfrac{1}{2} \times 1.3 \times\left[(120)^{2}-(90)^{2}\right]$

$=4095 \mathrm{N} / \mathrm{m}^{2}$ or Pascal

What principle law explains the working of the hydraulic brakes in automobiles?

0%
Bernoulli's Law
0%
Posieulles principle
0%
Pascal's law
0%
Archimedes principle

If the surface tension of a liquid increases proportional to the n th power of its density then what is the value of n -

0%
$1$
0%
$2$
0%
$3$
0%
$9$

Find the depression of the meniscus in the capillary tube of diameter

$0.4\ mm$ , dipped in a beaker containing mercury. (Density of mercury

$=13.6\times 10^{3}\ kg/m^{3}$ ) ; surface tension of the mercury is

$0.49\ N/m$ ; angle of contact is

$135^{o}$ ).

0%
$0.025\ cm$
0%
$0.021\ cm$
0%
$0.020\ cm$
0%
$0.027\ cm$

Explanation

$\text { We know that } \\$

$h =\left|\frac{2 T \cos \theta}{\rho g r}\right| \\$

$H =\frac{2 \times 0.49 \times 1 / \sqrt{2}}{13.6 \times 10^{3} \times 10 \times 0.2 \times 10^{-3}}=0.025 \mathrm{~cm} \\$

$\text { Answerl:- (A) } 0.025 \mathrm{~cm}$

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

0%
density of liquid
0%
height of the liquid column above orifice
0%
acceleration due to gravity
0%
None of these

A large tank filled with water has two holes in the bottom, one with twice the radius of the other. In steady flow the speed of water leaving the larger hole is______ the speed of the water leaving the smaller.

0%
twice
0%
four times
0%
half
0%
the same as

Explanation

Velocity of water surface, ${{v}_{1}}=0$

From Bernoulli equation

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

${{P}_{atm}}+\dfrac{\rho v_{1}^{2}}{2}+\rho g{{h}_{1}}={{P}_{atm}}+\dfrac{\rho v_{2}^{2}}{2}+\rho g{{h}_{2}}$

$\dfrac{\rho v_{2}^{2}}{2}=\rho g({{h}_{1}}-{{h}_{2}})$

${{v}_{2}}=\sqrt{2g({{h}_{1}}-{{h}_{2}})}$

Velocity of water is $\sqrt{2g({{h}_{1}}-{{h}_{2}})}$

Speed of water from hole is independent on area.

Hence, In steady flow the speed of water leaving the larger hole is the same as the speed of the water leaving the smaller.

1060024_1170448_ans_99ba55b5d62e4eb781dd71ca95e3e10b.png

A liquid enters a tube of variable cross section with a velocity

$3\operatorname { ms } ^ { - 1 }$ through the wider end and leaves from the narrow and whose radius is half of the wider one. The velocity with which it leaves the tube is:

0%
$1.2 \operatorname { ms } ^ { - 1 }$
0%
$12 \operatorname { ms } ^ { - 1 }$
0%
$10 \operatorname { ms } ^ { - 1 }$
0%
$20 \operatorname { ms } ^ { - 1 }$

Explanation

The equation of continuity,

$A_1v_1=A_2v_2$

$\pi(2r)^2 v_1=\pi r^2 v_2$

$4\pi r^2v_1=\pi r^2 v_2$

$v_2=4\times 3 m/s$

$v_2=12m/s$

The correct option is B.

1443224_1165507_ans_3c3d831c13d8444ea45ba77f9b963d65.png

Water flows through a horizontal tube as shown in figure. If the difference of heights of water column in the vertical tubes in

$2\ cm,$ and the areas of cross-section at

$A$ and

$B$ are

$4 \mathrm\ { cm } ^ { 2 }$ and

$2 \mathrm\ { cm } ^ { 2 }$ respectively, find the rate of flow of water across any action

0%
$126 \mathrm\ { cm } / \mathrm { s }$
0%
$136 \mathrm\ { cm } / \mathrm { s }$
0%
$146 \mathrm\ { cm } / \mathrm { s }$
0%
$156 \mathrm\ { cm } / \mathrm { s }$

Explanation

$\text { given, } h =2 \mathrm{~cm} \\$

$A_{1} =4 \mathrm{~cm}^{2} \\$

$A_{2} =2 \mathrm{~cm}^{2} \\$

$Acc. \text { to } \text { Continuity theorem } \\$

$A_{1} v_{1}=A_{2} v_{2} \\$

$4 \cdot v_{1} =2 \cdot v_{2} \\$

$v_{2} =2 v_{1}$

Acc.to Bernoullis theorem

$P_{0}+\rho g h+\frac{1}{2} \rho v_{1}^{2}=P_{0}+0+\frac{1}{2} \rho v_{2}^{2}$

$g \times 2+\frac{1}{2} v_{1}^{2}=\frac{4 v_{1}^{2}}{2}$

$4000=3 v_{1}^{2}$

$v_{1}=36.5 \mathrm{~cm} / \mathrm{s}$

rate of flow

$=36.5 \times 4$

$=146 \mathrm{~cm}^{3} / \mathrm{s}$

Answer:-

$(C)$

The velocity of air over the upper surface of the wing of an airplane is

$40m/s$ and that on the lower surface is

$30m/s$ . If the area of the wing is

$5m^{2}$ and the mass of the wing is

$300\ kg$ , the net force acting on the wing is (Density of air

$=1.3kg/m^{3}$ and

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

0%
$725N$ downward
0%
$725N$ upward
0%
$2275N$ upward
0%
$2275\ N$ downward

Explanation

REF.Image

Consider the wing be of negligible height.

From Bernoulli's equation;

pressure develop across top & bottom of the wing is:

$P=\frac{1}{2}P(V_{1}^{2}-V_{2}^{2})$

$=\frac{1}{2}\times 1.3\times (40^{2}-30^{2})= 455$ Pa

This pressure provides an upthrust F of:

$F= P\times A= 455\times 5= 2275N$ [A: Area of wing]

Thus Net force on the wing.

$N= Mg-F$

$= 300\times 10-2275$

$\Rightarrow N= 725N$ (Ans)

[Downward]

1106787_1179502_ans_ed21ef64d0774605866e39d01d76e129.JPG

A large vessel open at the top contains a liquid. If a hole is made at depth of

$5\ m$ from the free surface, the velocity of efflux is

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

0%
$5\ ms^{-1}$
0%
$10\ ms^{-1}$
0%
$15\ ms^{-1}$
0%
$20\ ms^{-1}$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 8 - 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 8 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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