Waterfalls from a tap with $${A_o} = 4{{\text{m}}^{\text{2}}},A = 1{{\text{m}}^{\text{2}}}\;{\text{and}}\;h = 2{\text{m}}$$, then velocity v is

$$6.5\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$

$$2.5\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$

$$4.5\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$

$$1.5\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$$

MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 9

The apparent depth of water in a cylindrical water tank of diameter

$2R$ cm is reducing at the rate of

$x$ cm/min when water is being drained out at a constant rate. The amount of water drained in

$cc/minute$ is: (

${n_1}$ = refractive index of air,

${n_2}$ = refractive index of water)

0%
$\dfrac{{x\pi {R^2}{n_1}}}{{{n_2}}}$
0%
$\dfrac{{x\pi {R^2}{n_2}}}{{{n_1}}}$
0%
$\dfrac{{2\pi {R^{}}{n_1}}}{{{n_2}}}$
0%
$\pi {R^2}x$

Explanation

We know that

$\dfrac{Real Depth}{Apparent\ depth}=\dfrac{{n}_{2}}{{n}_{1}}$

$\Rightarrow \$ Real Depth

$=\left(\dfrac{{n}_{2}}{{n}_{1}}\right)$ Apparent Depth

Rate of change in real depth

$=\left(\dfrac{{n}_{2}}{{n}_{1}}\right)$ rate of change in apparent depth

$=\left(\dfrac{{n}_{2}}{{n}_{1}}\right)x$

$\dfrac{dv}{dt}=\dfrac{d\left(A\right)h}{dt}=\dfrac{Adh}{dt}=\left[n{\left(R\right)}^{2}\right]\left[\dfrac{{n}_{2}}{{n}_{1}}x\right]$

Option

$B$ is correct

A cylindrical tank

$I m$ in radius rests on a platform

$S m$ high water,initially the tank is filled with water upto a height of

$S m$ . A v.mg whose area is

$10 ^ { 4 } m ^ { 2 }$ is removed from an office on the side of the tank at the bottom .Calculate

0%
initial speed with which the water flows from the orifice.
0%
initial speed with which the water strikes the ground and
0%
time taken to empty the tank to half its original value.
0%
Does the time to empty the tank upon the height of stand?

Water is streaming downward from a faucet opening with an area of

$3.0\times 10^{ -5 }{ m }^{ 2 }$ . It leaves the faucet with a speed of 5.0 m/s. The cross sectional area of the stream 0.50 m below the faucet is :

0%
$1.5\times 10^{ -5 }{ m }^{ 2 }$
0%
$2.0\times 10^{ -5 }{ m }^{ 2 }$
0%
$2.5\times 10^{ -5 }{ m }^{ 2 }$
0%
$3.0\times 10^{ -5 }{ m }^{ 2 }$

Explanation

Given,

$A_1=3\times 10^{-5}m^2$

$v_1=5m/s$

$P_1=P_2=1atm$

$h_1-h_2=0.5m$

$A_2=?$

From the equation of continuity,

$A_1v_1=A_2v_2$

$A_2=\dfrac{v_1}{v_2}A_1$

$A_2v_2=5\times 3\times 10^{-5}$

$A_2v_2=1.5\times 10^{-4}$ . . . . . . . . .(1)

From the Bernaullies Equation,

$P_1+\rho gh_1+\dfrac{1}{2}\rho v_1^2=P_2+\rho gh_2+\dfrac{1}{2}\rho v_2^2$

$\rho gh_1+\dfrac{1}{2}\rho v_1^2=\rho gh_2+\dfrac{1}{2}\rho v_2^2$

$v_2^2=2g(h_1-h_2)+v_1^2$

$v_2^2=2\times 10\times 0.5+5^2$

$v_2^2=10+25=35$

$v_2=5.9m/s$

From equation (1),

$A_2v_2=1.5\times 10^{-4}$

$A_2=\dfrac{1.5\times 10^{-4}}{v_2}=\dfrac{1.5\times 10^{-4}}{5.9}$

$A_2=2.5\times 10^{-5}m^2$

The correct option is C.

Which of the following devices directly measures the rate of flow of liquid?

0%
Venturimeter
0%
Simple barometer
0%
Hydraulic lift
0%
Both $(1)$ & $(2)$

A block of wood floats in water with

$\dfrac{4}{5}$ th of its volume submerged, but it just floats in another liquid. The density of liquid is? (in kg

$/m^3$ )

0%
$750$
0%
$800$
0%
$1000$
0%
$1250$

Explanation

${ ρ }_{ 1 }=ρ×\frac { v4 }{ 5v } =54e$

${ ρ }_{ 2 }=ρ×v=ρ$

${ ρ }_{ 1 }=54{ ρ }_{ 2 }$

${ ρ }_{ 2 }=45\times 1000=800kg{ m }^{ -3 }$

Consider two shoulder bags, one with a thin strap, and the other with a wide strap. If both the bags are equally heavy, the bag with the wide strap will be more comfortable to carry because

0%
the thin strap has greater inertia per unit length
0%
the thin strap causes greater momentum to be exerted
0%
the wide strap allows friction to be spread over a bigger area, thus reducing force between shoulder and the strap
0%
the wide strap enables the weight of the bag to be spread over a bigger area, thus reducing the pressure on the shoulder.

Explanation

Pressure is defined as

$P =\dfrac{\text { Force }}{\text { Area }}$

Hence, pressure is inversely proportional to area.

Since, wider strap has more area, than the thinner strap, so less pressure will be applied on shoulder.

Hence, the bag with wider strap is more comfortable.

The correct option is D.

A small spherical ball of steel through a viscous medium with terminal velocity v. If a ball of twice the radius of the first one but of the same mass is dropped through the same method, it will fall with a terminals velocity (neglect buoyancy)

0%
$\dfrac { v }{ 2 }$
0%
$\dfrac { v }{ \sqrt { 2 } }$
0%
$v$
0%
$2v$

Water enters a horizontal pipe of non uniform cross section with a velocity of

$0.4$

$m s ^ { - 1 }$ and leaves the other end with a velocity of

$0.6$

$m s ^ { - 1 }$ .pressure of water at the first end is

$1500$

$N m ^ { - 2 }$ ,then pressure at the other end is

0%
$1000$ $N m ^ { - 2 }$
0%
$1200$ $N m ^ { - 2 }$
0%
$1400$ $N m ^ { - 2 }$
0%
$1600$ $N m ^ { - 2 }$

Explanation

Bernoulli's equation for a horizontal pipe can be used to find the pressure at the second point of the pipe, which is

$\dfrac 12 \rho V_1^2+P_1=\dfrac 12\rho V_2^2+P_2$

where

$\rho$ is the density of the water

$=1000kg/m^3$

$V_1$ is the velocity at first point

$=0.6\,m/s$

$P_1$ is the pressure at first point

$=1500N/m^2$

$V_2$ is the velocity of second point

$=0.4m/s$

$P_2$ is the pressure at second point

$=?$

Putting these values in the Bernoulli's equation,

$\dfrac 12(1000)(0.6)^2+1500=\dfrac 12(1000)(0.4)^2+P_2$

$P_2=500[(0.6)^2-(0.4)^2]+1500=1600\,Nm^{-2}$

A sphere of mass M and radius R is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to

0%
$R^{2}$
0%
1/R
0%
R
0%
$1/R^{2}$

Explanation

$\textbf{Step 1: Draw FBD}$

$\textbf{Step 2: Apply stoke's law}$

${\textbf{Correct option: A}}$

Water enters a pipe of diameter

$3.0\ cm$ with a velocity of

$3.0$ to m/s. The water encounters a constriction where its velocity is

$12\ m/ s$ . What is the diameter of the constricted portion of the pipe?

0%
$0.33\ cm$
0%
$0.75\ cm$
0%
$1.0\ cm$
0%
$1.5\ cm$

Water is flowing through a tube whose one end is closed with a valve. Gauge pressure inside the tube is

$5 \times 10 ^ { 3 } \mathrm { N } / \mathrm { m } ^ { 2 }.$ When the valve is opened, this gauge pressure falls to

$0.5 \times 10 ^ { 5 } \mathrm { N } / \mathrm { m } ^ { 2 } .$ If density of water is

$10 ^ { 3 } \mathrm { kg } / \mathrm { m } ^ { 3 }$ then the then the initial speed of water flowing in the tube is

0%
$20 \mathrm { m } / \mathrm { s }$
0%
$15 \mathrm { m } / \mathrm { s }$
0%
$25 \mathrm { m } / \mathrm { s }$
0%
$30 \mathrm { m } / \mathrm { s }$

Velocity of efflux

0%
$1\ ms^{-1}$
0%
$2\ ms^{-1}$
0%
$3\ ms^{-1}$
0%
$4\ ms^{-1}$

Explanation

In the given figure,

$\begin{array}{l} 600\, ks/{ m^{ 3 } }\, \, \, -\, \, 60cm \\ 900\, ks/{ m^{ 3 } }\, \, \, -\, \frac { { 60\times 2 } }{ 3 } =40\, cm \\ v=\sqrt { 2gh } \\ =\sqrt { 2\times 10\times \frac { { 80 } }{ { 100 } } } \\ =4\, m{ s^{ -1 } } \end{array}$

So,

Option

$D$ is correct answer.

1215062_1259679_ans_517b2db7bf764314bd14d44046127b33.png

A drop of liquid having radius 2 cm has a terminal velocity 20 cm/sec, the terminal velocity of a drop of 1 mm radius will be :

0%
40 cm/s
0%
20 cm/s
0%
10 cm/s
0%
5 cm/s

A cylinder with a movable piston contains air under a pressure

${ p }_{ 1 }$ and a soap bubble of radius 'r' The which the air should be compressed by slowly pushing the pistc size by half will be: ( The surface tension is S, and temperature T is malntalned constant)

0%
$\left[ { 8p }_{ 1 }+\dfrac { 24S }{ r } \right]$
0%
$\left[ { 4p }_{ 1 }+\dfrac { 24S }{ r } \right]$
0%
$\left[ { 2p }_{ 1 }+\dfrac { 24S }{ r } \right]$
0%
$\left[ { 2p }_{ 1 }+\dfrac { 12S }{ r } \right]$

During the melting of a slab of ice at

$273\ K$ at atmospheric pressure:

0%
Positive work is done by the ice-water on the atmosphere
0%
Positive work is done on the ice-water by the atmosphere
0%
The internal energy of the ice-water increases
0%
The internal energy of the ice-water system decreases

Explanation

There is a decrease in volume during melting of an ice slab at

$273\ K$ . Therefore, negative work is done by ice-water system on the atmosphere or positive work is done on the ice-water system by the atmosphere. Hence option (b) is correct. Second, heat is absorbed during melting (i.e.,

$dQ$ is positive) and as we have seen, work done by ice-water system is negative (

$dW$ is negative.) Therefore, from the first law of thermodynamics,

$dU = dQ - dW$ , with change in internal energy of ice-water system,

$dU$ will be positive or internal energy will increase.

Which of the following devices directly measures the rate of flow of liquid?

0%
$Venturimeter$
0%
$Simple \ barometer$
0%
$Hydraulic \ lift$
0%
Both (A) & (B)

A cylinderical vessel is filled with equal amount of weight of mercury and water.The overall height of the two layer is 29.2 cm,specific gravity of mercury is 13.6.Then the pressure of the liquid at the bottom of the vessel is:

0%
$29.2 cm$ of water
0%
$\dfrac{29.2}{13.6} cm$ of mercury
0%
$4 cm$ of mercury
0%
$15.6 cm$ of mercury

Explanation

${ M }_{ Hg }={ M }_{ w }$

${ P }_{ Hg }\left( A\times { L }_{ Hg } \right) ={ P }_{ w }\left( A\times { L }_{ w } \right)$

$\therefore$

${ L }_{ w }=13.6{ L }_{ Hg }$

${ L }_{ w }+{ L }_{ Hg }=29.2$

$\Rightarrow { L }_{ Hg }=2cm\quad \quad { L }_{ w }=27.2cm=2cm$ of

$Hg$

Total pressure

$=4cm$ of

$Hg$ .

Figures below show water flowing through a horizontal pipe from left to right. Note that the pipe in the middle is narrower. Choose the most appropriate depiction of water levels in the vertical pipes.

A steel ball of mass 'm' falls in a viscous liquid with a terminal velocity

$4 ms^{-1}$ . If another steel ball of mass 8m falls through the same liquid then its terminal velocity is

0%
$4 ms^{-1}$
0%
$8 ms^{-1}$
0%
$16 ms^{-1}$
0%
$2 ms^{-1}$

Explanation

$V = \dfrac{2}{q}\dfrac{{{r^2}g\left( {s - \sigma } \right)}}{\eta }$

$\because$ it is independent of mass

$.$

$\therefore$ velocity remains unchanged

Hence,

option

$(A)$ is correct answer.

The radius of one arm of a hydraulic lift is three times the radius of the other arm. What force should be applied on the narrow arm so as to lift

$50kg$ at the wider arm?

0%
60 N
0%
55.6 N
0%
26.7 N
0%
30 N

A beaker containing water is placed on the platform of a spring balance. The balance reads

$1.5kg$ . A stone of mass

$0.5kg$ and density

$500kg/{m}^{3}$ is completely immersed in water without touching the walls of the beaker. What will be the balance reading now?

0%
$2kg$
0%
$2.5kg$
0%
$1kg$
0%
$3kg$

${A_o} = 4{{\text{m}}^{\text{2}}},A = 1{{\text{m}}^{\text{2}}}\;{\text{and}}\;h = 2{\text{m}}$ , then velocity v is

0%
$2.5\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$
0%
$6.5\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$
0%
$4.5\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$
0%
$1.5\;{\text{m}}{{\text{s}}^{{\text{ - 1}}}}$

The volume of an air bubble increases by

$x$ % as it raises from the bottom of a water lake to its surface. If the water barometer reads

$H$ . The depth of the lake is :

0%
$\dfrac {H}{100x}$
0%
$\dfrac {Hx}{100}$
0%
$Hx \times 100$
0%
$\dfrac {H}{x}100$

Water flows along a horizontal pipe of non-uniform cross-section. 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%
$0.62$ m of Hg
0%
$1.49$ m of Hg
0%
$0.5$ m of Hg
0%
$1$ m of Hg

A solid cone of height

$H$ and base radius

$H/2$ floats in a liquid of density

$\rho$ . It is hanging from the ceiling with the help of a string. The force by the fluid on the curved surface of the cone is (

${P}_{0}=$ atmospheric pressure)

0%
$\pi{H}^{2}\left( \cfrac { { P }_{ 0 } }{ 4 } +\cfrac { \rho gH }{ 3 } \right)$
0%
$\pi{H}^{2}\left( \cfrac { { P }_{ 0 } }{ 4 } +\cfrac { \rho gH }{ 6 } \right)$
0%
$\cfrac{\pi{H}^{2}}{4}\left( \cfrac { { P }_{ 0 } }{ 4 } + { \rho gH }{ } \right)$
0%
$\cfrac{\pi{H}^{2}}{4}\left( { { P }_{ 0 } }{ } + { \rho gH }{ } \right)$

A container is filled with liquid upto height H as shown in figure. There are 2 holes in the container such that water coming out from them have same range. If the range of water coming out of theholes is half of maximum range, then the distance

$x$ between these holes is

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

The surface tension and bulk modulus of elasticity of water are S and B respectively. Then the ratio

$\frac{B}{S}$ is dimensionally equivalent to the dimension of -

0%
Length
0%
Wave number
0%
$(area)^{-1}$
0%
Force

Water is flowing continuously from a tap haveing an internal diameter $8\times{ 10 }^{ -3}m$ . The water velocity as it leaves the tap is $0.4{ ms }^{-1 }$ . The diameter of the water stream at a distance $2\times{ 10 }^{ -1}m$ below the tap is close to:

0%
$9.6\times{ 10 }^{ -3}m$
0%
$3.6\times{ 10 }^{ -3}m$
0%
$5.0\times{ 10 }^{ -3}m$
0%
$7.5\times{ 10 }^{ -3}m$

From a tap of cross-sectional area

${ 10 }^{ -4 }{ m }^{ 2 }$ a stream of water runs down If the flow is streamlined and the pressure is assumed to be constant throughout, and the cross- section area of the stream at a point 0.15cm below the tap is

${ 5\times 10 }^{ -5 }{ m }^{ 2 }$ , then the velocity of stream near the tap is

0%
$2{ ms }^{ -1 }$
0%
$1{ ms }^{ -1 }$
0%
$3{ ms }^{ -1 }$
0%
$0.5{ ms }^{ -1 }$

If the velocity head of a stream of water is equal to

$40\ cm$ then its speed of flow is approximately

0%
$1.0\ m/s$
0%
$2.8\ m/s$
0%
$140\ m/s$
0%
$10.3\ m/s$

Explanation

Velocity head is defined as

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

so the velocity

$v=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.4}=2.8m/s$

A cylindrical tank of height

$H$ is completely filled with water. On its vertical side, there are two tiny holes. one above the middle at a height

$h_1$ and the other below the middle at depth

$h_2$ . If the jets of water from the holes meet at the same point at the horizontal plane through the bottom of the tank then the ratio

$\frac {h_1}{h_2}$ is:

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

An empty plastic box of mass m is found to accelerate up at the rate of g/6 when placed deep inside water. how much sand should be placed inside the box so that it may accelerate down at the rate of g/6

0%
2m/5
0%
m/5
0%
4m/5
0%
2m/3

An open vessel containing the liquid upto a height of 15m. A small hole is made at height of 10 m from the base of the vessel , then the initial velocity of efflux is

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

0%
1 m/s
0%
$10\sqrt {2 } m/s$
0%
5 m/s
0%
10 m/s

Explanation

Velocity of Efflux

$= \sqrt{2gH}$

where,

$H =$ Height of open surface =

$15-10= 5\;m$

So, Velocity

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

Therefore, D is correct option.

The discharge velocity at the pipe exit as shown in the figure is

0%
$\sqrt{2gh}$
0%
$\sqrt{2gH}$
0%
$\sqrt{2g(H-h)}$
0%
$\sqrt{2g(h+H)}$

The lower end of a glass capillary tube is dipped in water. Water rises to a height of 8 cm. The angle of contact is

${ 0 }^{ \circ }$ . the tube is then broken at a height of 6 cm. The height of water column and angle of contact will be.

0%
$6m,{ sin }^{ -1 }\left( \dfrac { 3 }{ 4 } \right)$
0%
$6m,{ sin }^{ -1 }\left( \dfrac { \sqrt { 7 } }{ 4 } \right)$
0%
$4m,{ sin }^{ -1 }\left( \dfrac { \sqrt { 7 } }{ 8 } \right)$
0%
$6m,{ 0 }^{ \circ }$

Explanation

we know that

$h = {(2T cos θ) / (ρrg)}$

As radius and content of tube is same.

$T, r, ρ, g$ = constant

$∴ {h / (cosθ)}$ = constant

$∴ (h1 / h2) = {(cosθ_1) / (cosθ_2)}$

Initial height =

$h1 = 8 cm$

As capillary is broken at height 6 cm, meaning that water will rise to height 6 cm hence

$h2 = 6 cm$

$(h1 / h2) = {(cos 0) / (cos θ_2)}$ --------- initial angle = 0°

$∴ (8/6) = {1 / (cos θ_2)}$

$cos θ_2 = 3/4$

$θ_2 = cos^{–1} (3/4)$

∴ height =

$6 \,cm$ & angle =

$cos^{–1} (3/4)$

$sin^{–1} (\sqrt7/4)$

A steel ball of density 7800 kg/

${ m }^{ 3 }$ falls through water of density 1000 kg/

${ m }^{ 3 }$ and coefficient of viscosity 0.01 poise. Find its terminal velocity (approx) if the radius is 1 cm.

0%
1500 m/s
0%
500 m/s
0%
1000 m/s
0%
None of these

Two tubes

$A$ and

$b$ of length

$100 \mathrm { cm }$ and

$50 \mathrm { cm }$ and radii

$0.2 \mathrm { mm }$ and

$0.1 \mathrm { mm}$ respectively joined in series. If a liquid is passing through the two tubes entering A at a pressure

$Hg$ and leaving

$B$ at a
pressure of

$71 cm Hg,$ the pressure at the junction of

$A$ and

$B$

0%
$70 cm Hg$
0%
$80 cm Hg$
0%
$81 cm Hg$
0%
zero

If pressure at half the depth of a lake is equal to $\cfrac {2}{3}$ pressure at the bottom of the lake then what is the depth of the lake $[RPET=2000]$

0%
$10$ m
0%
$20$ m
0%
$60$ m
0%
$30$ m

Explanation

The Pressure at bottom of the lake =

$P_0+h\rho g$
The Pressure at half the depth of a lake =

$P_0+\dfrac{h}{2}\rho g$
According to given condition

$P_0+\dfrac{1}{2}h\rho g=\dfrac{2}{3}(P_0+h\rho g)\Rightarrow \dfrac{1}{3}P_0 = \dfrac{1}{6}h\rho g$

$h=\dfrac{2P_0}{\rho g}=\dfrac{2\times10^5}{10^3\times10}=20m$

Radius of a soap bubble is $r$ , surface tension of soap solution is $T$ . Then without increasing the temperature, how much energy will be needed to double its radius.

0%
$4\pi {r^2}T$
0%
$2\pi {r^2}T$
0%
$12\pi {r^2}T$
0%
$24\pi {r^2}T$

Explanation

As we know ,

without increasing temperature increasing radius will be done by performing some work on the bubble,

$W=8\pi T(R_{2}^{2}-R_{1}^{2})$

$=8\pi T[(2r)^{2}-(r)^{2}]=24\pi r^{2}T$

By sucking through a straw, a student can reduce the pressure in his lungs to 750 mm of Hg (density

$13.6 gcm^{-3}$ ) using the straw, he can drink water from a glass up to a maximum depth of

0%
$10.2 cm$
0%
$75.3 cm$
0%
$13.6 cm$
0%
$1.96 cm$

Explanation

Normal pressure is

$760 mm\,Hg$ so the boy can create a depression of

$10 mm\,Hg$ in his lungs.

According to the law of flotation

Weight of floating body

$=$ Weight of displaced liquid

Wt. Of water displaced from glass

$=$ Wt of air of lung reduced (interm of mmHg)

Wt of water displaced from glass = density of water

$(d_1) \times$ volume of water displaced

$(v_1)$

Wt of air reduce from lung (interm of mmHg)

$=$ density of Hg

$(d_2)\times$ volume of air displaced

$(v_2)$

Now

$d_1 \times v_1 = d_2 \times v_2$

And

$v_1 = d_2 \times \dfrac{v_2}{d_1} = 13.6 g/cm –3 \times \dfrac{10mm}{ 1g/cm} –3 = 136mm = 13.6cm$

A cube of edge length 5 cm is place inside a liquid. The pressure at the center of a face is 12 pa. Find the force exerted by the liquid on this face.

0%
0.03 N
0%
0.08 n
0%
0.06 n
0%
0.3 N

When water is flowing through a pipe with a speed

$v$ , then its power is proportional to

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

Explanation

Using Bernoulli's equation.

Energy per unit volume before

$=$ Energy per unit volume after

${ p }_{ 1 }+\dfrac { 1 }{ 2 } { pv }_{ 1 }^{ 2 }+{ pgh }_{ 1 }={ p }_{ 2 }+\dfrac { 1 }{ 2 } { pv }_{ 2 }^{ 2 }+{ pgh }_{ 2 }$

${ A }_{ 2 }<{ A }_{ 1 }$

${ V }_{ 2 }>{ V }_{ 1 }$

${ p }_{ 2 }<{ p }_{ 1 }$

Hence power

$\alpha { v }^{ 2 }$ .

1224139_1451476_ans_40d4ed981a63440e87ce1ce82f7a2df4.png

Vessel shown in the figure has two sections of areas of cross section

$A{}_{1}$ and

${A}_{2}$ . A liquid of density

$\rho$ fills both sections up to a height

$h$ in each. Neglect atmospheirc pressure

0%
The pressure at the base of the vessel is $2h\rho g$
0%
The force exerted by the liquid on the base is $2h\rho {A}_{2}g$
0%
The weight of the liquid is less thatn $2hg\rho {A}_{2}$
0%
Walls of the vessel at the level $x$ exert a downward force $hg\rho({A}_{2}-{A}_{1})$ on the liquid.

Explanation

$\begin{array}{l} pressure\, at\, the\, bottom\, =\rho g\Delta H=2\rho gh \\ Force\, at\, bottom=2\rho gh\times { A_{ 2 } } \\ weight\, of\, liquid\, is\, less\, than\, 2\rho gh\times { A_{ 2 } } \\ \Delta F=\rho gh{ A_{ 2 } }-\rho gh{ A_{ 1 } }=\rho gh\left( { { A_{ 0 } }-{ A_{ 1 } } } \right) . \end{array}$

Hence, the option

$D$ is the correct answer.

A block of sided

$0.5 m$ is

$30\%$ submerged in a liquid of density

$1 \, gm/cc$ . Then find mass of an object placed on block for complete submergence.

0%
$87.3 kg$
0%
$85.3 kg$
0%
$82.3 kg$
0%
$80.3 kg$

Explanation

Initial condition

$B = mg$

$\rho\dfrac{3}{10}Vg = mg$ . ...(1)

$\rightarrow (1000 \,kg/m^3) \dfrac{3}{10} (0.5m)^3 = m$

$m = 37.5 \,kg$
finally

$\rho vg = (m + M)g$ ...(2)
From equation (1) & (2)

$\dfrac{(m + M)g}{mg} = \dfrac{\rho Vg}{\rho\dfrac{3}{10}Vg}$

$1 + \dfrac{M}{m} = \dfrac{10}{3}$

$\dfrac{M}{m} = \dfrac{7}{3}$

$M = \dfrac{7}{3} (37.5)kg = 87.3 \,kg$

The density of Saturn is

$687 kg/m^3$ . It is kept in a lake filled with water. Then:

0%
It will float in water
0%
It will get submerged in water completely and sink to the bottom of the lake
0%
It will get submerged in water completely but will not sink to the bottom
0%
None of these

Venturimetre is used to measure

0%
Flow speed of a fluid
0%
Atmospheric pressure
0%
Specific gravity
0%
Pressure in a closed vessel

The working of venturimeter is based on

0%
Torricelli's law
0%
pascal's law
0%
Bernoulli's theorem
0%
Archimede's principle

The angle of contact for the pair of pure water with clean glass is

0%
acute
0%
obtuse
0%
$90^0$
0%
$0^0$

A film of water is formed between two straight parallel wires each

$10\mathrm { cm }$ long and at separation

$0.5\mathrm { cm } .$ Calculate the work required to increase

$1\mathrm { mm }$ distance between the wires. Surface tension of water

$= 72 \times 10 ^ { - 3 } \mathrm { N } / \mathrm { m } .$

0%
$1.44 \times 10 ^ { - 3 }$
0%
$1.44 \times 10 ^ { - 7 }$
0%
$1.44 \times 10 ^ { - 5 }$
0%
$1.44 \times 10 ^ { - 4 }$

Explanation

Given:

Length

$L=10$ cm

Width

$W=0.5$ cm

Surface tension

$S=72 \times { 10^{ -3 }}$

Increase distance by

$\Delta l=1$ mm

change in area

$\Delta A=(L\times W)=10(.1)\times 10^{-4}$

$\begin{array}{l} E=S\Delta A \\ =72\times { 10^{ -3 } }\times 2\times 1\times { 10^{ -4 } } \\ =1.44\times { 10^{ -5 } }J \\ Ans.\, \, (C) \end{array}$

Calculate the energy spent in spraying a drop of mercury of radius

$R$ into

$n$ droplets

all of same size. Given the surface tension of mercury is

$T .$

0%
$4 \pi R ^ { 2 } T \left( n ^ { 1 / 3 } - 1 \right)$
0%
$4 \pi r ^ { 2 } T \left( n ^ { 1 / 3 } - 1 \right)$
0%
$4 \pi r ^ { 2 } \left( n ^ { 1 / 3 } - 1 \right)$
0%
None of these

Explanation

$\begin{array}{l} n\cdot \frac { 4 }{ 3 } \pi { r^{ 3 } }=\frac { 4 }{ 3 } \pi { R^{ 3 } } & \\ \Rightarrow n{ r^{ 3 } }={ R^{ 3 } } & ...(i) \\ E=\left( { n\cdot 4\pi { r^{ 2 } }-4\pi { R^{ 2 } } } \right) T & \\ =4\pi T\left( { n\cdot { { \left( { \frac { { { R^{ 3 } } } }{ n } } \right) }^{ \frac { 2 }{ 3 } } }-{ R^{ 2 } } } \right) & \\ =4\pi T{ R^{ 2 } }\left( { { n^{ 1/3 } }-1 } \right) & \\ Ans.\, (A) & \end{array}$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 9 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.