Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

Assertion is correct but Reason is incorrect

Both Assertion and Reason are incorrect

MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 10

The angle of contact between glass and mercury is :

0%
$0^{0}$
0%
$30^{0}$
0%
$90^{0}$
0%
$135^{0}$

Explanation

The angle of contact between glass and mercury is

$135^{0}$

Anomalous expansion of water is demonstrated by using

0%
Jolly's apparatus
0%
Hopes apparatus
0%
Regnaults apparatus
0%
Specific apparatus

A steel ball is dropped in a viscous liquid. The distance of the steel ball from the top of the liquid is shown below. The terminal velocity of the ball is closest to :

0%
$0.26 m/s$
0%
$0.33 m/s$
0%
$0.45 m/s$
0%
$0.21 m/s$

Explanation

For terminal velocity, slope should be constant. If we look carefully slope is constant from

$t=1.6\ sec$ to

$t=2\ sec$

$V_t=\dfrac{0.4-0.3}{2-1.6}=\dfrac{0.1}{0.4}=0.25\ m/s$

So terminal velocity is closest to

$0.26\ m/s$

The terminal velocity of solid sphere of radius

$0.1$ m moving in air in vertically downward direction, is:

$(\eta =1.8\times 10^{-5} Ns/m^2$ , density of sphere

$=1000 kg/m^3$ and

$g=10 m/s^2$ )

0%
$2\times 10^5$ m/s
0%
$1.2\times 10^2$ cm/s
0%
$4\times 10^2$ cm/s
0%
None of these

A container of height

$10\ cm$ is filled with water. There is a hole at bottom. Find the pressure difference between points

$A$ &

$B$ .

0%
$1000\ Pa$
0%
Zero
0%
$1\ Pa$
0%
$100\ Pa$

Explanation

$P_{A} - P_{B} = \rho gh$

$= 10^{3} \times 10\times 10\times 10^{-2} Pa$

$= 1000\ Pa$ .

How many cylinders of hydrogen at atmospheric pressure are require to fill a balloon whose volume is

$500\ m^3$ , if hydrogen is stored in cylinder of volume

$0.05\ m^3$ at an absolute pressure of

$15\times 10^5 \ Pa$ ?

0%
$700$
0%
$675$
0%
$605$
0%
$710$

A solid sphere falls with a terminal velocity of

$20m/s^{-1}$ in air. If it is allowed to fall in vacuum

0%
terminal velocity will be $20 m/s^{-1}$ .
0%
terminal velocity will be less than $20 m/s^{-1}$
0%
terminal velocity will be more than $20 m/s^{-1}$
0%
there will be no terminal velocity

$20^o$ C, to attain the terminal velocity how fast will an aluminium sphere of radii

$1$ mm fall through water? [Assume flow to be laminar flow and specific gravity

$(Al)=2.7$ ,

$\eta_{water}=8\times 10^{-4}$ Pa]

0%
$5$ m/s
0%
$4.7$ m/s
0%
$4$ m/s
0%
$2$ m/s

A flask containing air at

$27^oC$ at atmospheric pressure is croked up. A pressure of

$2.5\ atm$ , inside the flask would force the rock out. The temperature at which it will happen is :

0%
$67.5^oC$
0%
$577^oC$
0%
$670^oC$
0%
$750\ K$

A block of wood is floating in water in a closed vessel as shown in the figure. The vessel is connected to an air pump. When more air is pushed into the vessel, the block of wood floats with (neglect compressibility of water)

0%
larger part in the water
0%
smaller part in the water
0%
same part in the water
0%
at some instant it will sink

A hole is made at the bottom of a tank filled with water

$\left(density={10}^{3}kg/m^{3}\right)$ . If the total pressure at the bottom of the the tank is

$3atm\left(1atm=10^{5}{N/m}^{2}\right)$ , then the velocity of efflux is

0%
$\sqrt{400}m/s$
0%
$\sqrt{200}m/s$
0%
$\sqrt{600}m/s$
0%
$\sqrt{500}m/s$

Explanation

We know that velocity of the efflux is:

$v= \sqrt{\dfrac{2\Delta P}{\rho}}$

$\Delta P= pressure$

$at$

$the$ botton

$- atmospheric$

$pressure$

So we get,

$v=\sqrt{\dfrac{2\times(3-1) \times 10^{5}}{10^{3}}}$

Therefore

$v=\sqrt{400}ms^{-1}$

hence Option A is correct

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

According to Pascal's law, the pressure at bottom of each vessels containing liquid in equilibrium of rest is the same.

$P_1=P_2=P_3$

Hence

$F_1=F_2=F_3$

Thus, the reason is correct explanation for assertion.

Find the force acting at the pivoted end of the rod in terms of mass

$m$ of the rod.

0%
$\left(\sqrt{2}-1\right)mg$
0%
$\left(\sqrt{2}+1\right)mg$
0%
$\left(\sqrt{3}+1\right)mg$
0%
$\left(\sqrt{2}+3\right)mg$

Explanation

Weight of the rod

$w=LA\rho{g}$

(A=area of cross section,

$\rho$ =density of material)

Submerged length of the rod is given by

$OT=\dfrac{L}{2\sin\theta}$

Buoyant force

$B=\left(\dfrac{L}{2\sin\theta}\right)A\rho_w{g}$

from Archimedes' principle (

$\rho_w=$ density of water).

This force acts at

$P$ , where

$OP=\dfrac{OT}{2}=\dfrac{l}{4\sin\theta}$

Balancing torque about

$O$ is

$W\ OQ\ \cos\theta=B\ OP\ \cos\theta$

$\left(LA\rho{g}\right){L/2}=\left(\dfrac{L}{2\sin\theta}\right)A\rho_w{g}\left(\dfrac{L}{4\sin\theta}\right)$

$\sin^2\theta=\dfrac{\rho_w}{4\rho}$

$\Rightarrow\sin^2\theta=1/2\Rightarrow\theta=45°$

Also, balancing forces

$R+W=B$

$R=B-W$

$=\dfrac{\rho{A}\rho_w{g}}{2\sin\theta}-LA\rho{g}$

$=LA\rho{g}\left[\dfrac{1}{\sin\theta}-1\right]\left(\because\rho_w=2\rho\right)$

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

1596175_1739170_ans_4ff448b15463415091340a70bf0a4512.png

A jet of water having velocity

$=10\ m/s$ and stream cross-section

$=2\ cm^2$ hits a flat plate perpendicularly, with the water splashing out parallel to plate. The plate experiences a force of :

0%
$40\ N$
0%
$20\ N$
0%
$8\ N$
0%
$10\ N$

A tube

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

$1\ cm$ high and cross- section

$100\ cm^2$ . Water fills the system upto a height of

$100\ cm$ from the bottom of the vessel. The force exerted by the liquid at the bottom of the vessel is :

0%
$1000\ N$
0%
$990\ N$
0%
$900\ N$
0%
$100\ N$

When an iceberg floats in sea water, only one tenth of its volume is inside the sea water.

0%
True
0%
False

For a surface molecule

0%
the net force on it is zero
0%
there is a net downward force
0%
the potential energy is less than that of a molecule inside
0%
the potential energy is more than that of a molecule inside

Explanation

In Fig. (b) A molecule viewed from the top to surface of the fluid. Force exerted on a molecule by all nearby molecule are equal and opposite, so net horizontal component of force is zero.

In Fig. (a) as a molecule is viewed horizontally to the surface of fluid a vertically downward force acts on molecule due to which it does not escape from surface. Hence the right option is (b). There is a net attractive force on molecules at surface acting downwards. So the P.E of molecules at the surface is negative to lower ones but in magnitude PE (

$mgh$ ) is more than that of the lower or inside molecules. Thus this verifies option (d).

1660367_1794876_ans_69617bd04f914e9c9737d645174de659.png

A ball whose density is 0.4 × 103 kg/m falls into water from a height of 9 cm . To what depth does the ball sink

0%
9 cm
0%
6 cm
0%
4.5 cm
0%
2.25 cm

Explanation

The velocity of ball before entering the water surface

$v = \sqrt{2gh} = \sqrt{2g \times 9}$
When ball enters into water, due to upthrust of water the velocity of ball decreases (or retarded)
The retardation,

$\alpha = \frac{apparent weight}{mass of ball}$

$= \frac{V(\rho - \sigma) g}{V \rho}$

$= \left ( \frac{\rho - \sigma}{\rho} \right ) g = \left ( \frac{0.4 - 1}{0.4} \right ) \times g = - \frac{3}{2} g$
If h be the depth upto which ball sink, then,

$0 - v^2 = 2 \times \left ( - \frac{3}{2} g \right ) \times h \Rightarrow 2g \times 9 = 3gh \therefore h = 6 cm$

The atmospheric pressure at earth surface is

$P_{1}$ and inside mine is

$P_{2}.$ They are related as:

0%
$P_{1} = P_{2}.$
0%
$P_{1} > P_{2}.$
0%
$P_{1} < P_{2}.$
0%
$P_{2}$ = 0

From the adjacent figure, the correct observation is

[ KCET 2005 ]

0%
The pressure on the bottom of tank (a) is greater than at the bottom of (b).
0%
The pressure on the bottom of the tank (a) is smaller than at the bottom of (b).
0%
The pressure depend on the shape of the container
0%
The pressure on the bottom of (a) and (b) is the same

Explanation

Pressure =

$h\rho g$ i.e. pressure at the bottom is independent of the area of the bottom of the tank. It depends on the height of water upto which the tank is filled with water. As in both the tanks, the levels of water are the same, pressure at the bottom is also the same.

Radius of an air bubble at the bottom of the lake is

$r$ and it becomes

$2r$ when the air bubbles rises to the top surface of the lake. If

$P$ cm of water be the atmospheric pressure, then the depth of the lake is

0%
$2p$
0%
$8p$
0%
$4p$
0%
$7p$

Explanation

Initially the radius of the bubble is

$r$ . After reaching surface it becomes

$2r$ . The atmospheric pressure is given as,

$P_{atm}= P \space cm \space of \space water$

What we can conclude from this process is that the volume is changing in the air bubble but the temperature remains unchanged.

For isothermal process,

$P_1V_1=P_2V_2$

Let the height of water surface be

$h$ .

$(P+h)(\frac{4}{3}\pi r^3)=P(\frac{4}{3}\pi 8r^3)$

$\Rightarrow h+P=8P$

$\Rightarrow h=7P$ is our required answer.

A hydraulic lift is designed to lift heavy objects of maximum mass

$2000 kg$ . The area of cross-section of piston carrying the load is

$2.25 \times 10^{-2} m^2$ . What is the maximum pressure the smaller piston would have to bear?

0%
$0.8711 \times 106 \ N/m^2$
0%
$0.5862 \times 107 \ N/m^2$
0%
$0.4869 \times 105 \ N/m^2$
0%
$0.3271 \times 104 \ N/m^2$

Explanation

Pressure on the piston= $\dfrac{F}{A}$

Force $F=m×a$

= $2000×9.8$

Area of cross section $A=2.25×10^{-2}m^2$

Therefore the pressure $P=\dfrac{2000×9.8}{2.25×10^(-2)}$
$P=0.8711×10^6\dfrac{N}{m^2}$

A triangular lamina of area

$A$ and height

$h$ is immersed in a liquid of density

$\rho$ in a vertical plane with its base on the surface of the liquid. The thrust on the lamina is

0%
$\dfrac{1}{2}A\rho gh$
0%
$\dfrac{1}{3}A\rho gh$
0%
$\dfrac{1}{6}A \rho gh$
0%
$\dfrac{2}{3}A \rho gh$

Explanation

The Area of the lamina is A and the height is h which is immersed in a liquid of the density

$\rho$ in a vertical plane with its base on the surface of the liquid.

The thrust on the lamina = pressure at centroid

$\times$ Area

$=\dfrac{h\rho g}{3}\times A = \dfrac{1}{3} A \rho gh$

Angle of contact for a glass tube dipped in mercury is obtuse. True / False.

0%
True
0%
False

A water supply maintains a constant rate of flow for water in a hose. You want to change the opening of the nozzle so that water leaving the nozzle will reach a height that is four times the current maximum height the water reaches with the nozzle vertical. To do so, should you

0%
decrease the area of the opening by a factor of $16$
0%
decrease the area by a factor of $8$
0%
decrease the
area by a factor of $4$
0%
decrease the area by a factor of $2$
0%
give up because it cannot be done?

Figure shows aerial views from directly above two dams. Both dams are equally wide (the vertical dimension in the diagram) and equally high (into the page in the diagram). The dam on the left holds back a very large lake, and the dam on the right holds back a narrow river. Which dam has to be built more strongly?

0%
the dam on the left
0%
the dam on the right
0%
both the same
0%
cannot be predicted

A boat travelling at

$6 m/s$ in sea water has a

$250 mm$ diameter propeller that discharges the water at a velocity of

$12 m/s$ . Given that the density of seawater is

$1030 \mathrm{k}\mathrm{g}/\mathrm{m}^{3}$ . The effect of propeller hub is negligible. The magnitude of thrust produced

$F$ is (in

$N$ )

0%
$1030$
0%
$2730$
0%
$4660$
0%
$980$

Explanation

Pressure difference across the propeller hubs

$=p_{1}-p_{2}=\dfrac{1}{2}\left ( 1030 \right )\left ( 12^{2}-6^{2} \right )$

$=55620p_{a}$
So,
the thrust produce=area x pressure difference

$=\pi \times \left ( 125 \times 10^{-3} \right )^{2}\times 55620N$

$=2730.240N$

$\approx 2730N$

Ratio of area of hole to beaker is

$0.1$ Height of liquid in beaker is

$3m$ , and hole is at the height of

$52.5 cm$ from the bottom of beaker, find the square of the velocity of liquid coming out from the hole:

0%
$50 \ (m/s)^2$
0%
$50.5 \ (m/s)^2$
0%
$51 \ (m/s)^2$
0%
$42 \ (m/s)^2$

Explanation

Given : Ratio of area of hole to that of beaker

$\dfrac{A_h}{A_b} = 0.1$

From the figure,

$h = 3 - 0.525 =2.475 m$

$\therefore$ Velocity of liquid coming out of hole

$v = \sqrt{\dfrac{2gh}{1 - \frac{A_h^2}{A_b^2}}}$

$\therefore$

$v = \sqrt{\dfrac{2\times 10 \times 2.475}{1 - 0.01}} = \sqrt{50}$

$\implies$

$v^2 = 50$

$(m/s)^2$

A wide vessel with a small hole at the bottom is filled with two liquids. The density and height of one liquid are

${\rho}_{1}\ and\ {h}_{1}$ and that of the other are

${\rho}_{2}$ and

${h}_{2}$ . (Given

${\rho}_{1} > {\rho}_{2} )$ . The velocity of liquid coming out of the hole is :

0%
$v = \sqrt{2g({h}_{1} + {h}_{2})}$
0%
$v = \sqrt{2g({h}_{1}{\rho}_{1} + {h}_{2}{\rho}_{2})/ ({\rho}_{1} + {\rho}_{2})}$
0%
$v = \sqrt{2g[{h}_{1} + \dfrac{{h}_{2}{\rho}_{2}}{{\rho}_{1}}]}$
0%
$v = \sqrt{2g[\dfrac{{h}_{1}{\rho}_{1}}{{\rho}_{2}} + {h}_{2}]}$

Explanation

Applying Bernoulli's equation to points A and B

$P_0 + \rho_1gh_1 + \rho_2gh_2 +0 = P_0 + \dfrac{1}{2} \rho_1v^2$
where v is the velocity at point B

$\therefore \dfrac{1}{2} \rho_1v^2 =g( \rho_1h_1 + \rho_2h_2)$

$\therefore v_1^2 = 2g\dfrac{( \rho_1h_1 + \rho_2h_2)}{\rho_1}= 2g(h_1 + \frac{\rho_2}{\rho_1} \times h_2)$

$\therefore v_1 = \sqrt{2g(h_1 + \dfrac{\rho_2}{\rho_1} h_2)}$

Which of the following is not possible ?

Explanation

According to the Pascal's law, the pressure at any horizontal point in a continuous fluid is equal. So if 2 fluids with densities

$3 \rho\ and\ \rho$ are taken together, they cannot be at the same height. Hence, option A is wrong.
Similarly, in capillary rise problems, the fluid always forms a concave surface. Hence, option C is wrong.
Using a similar reasoning as option A, the fluid in the vessel and the hand should be on the same height. So option D is wrong as well

Figure shows a pilot-static tube to measure air speed. (attached to a manometer)
The manometer shows a difference in levels

$9 mm$ . lt contains water of density 998

$\mathrm{k}\mathrm{g}/\mathrm{m}^{3}$ . The density of air is 1.2

$\mathrm{k}\mathrm{g}/\mathrm{m}^{3}$ . The gravitational field strength is 9.81 m/s

$^2$ . The agreement between the measured and calculated values of air speed is not very good so that

$V_{calibrated}=Cv_{air}$ where C is called calibration coefficient. Then:

0%
the pressure difference is $88 Pa$ (nearly).
0%
the coefficient C is due to fact that even air is slightly viscous causing some turbulence by the hypodermic needles.
0%
in order to apply Bernoulli equation, the flow need not be streamline non-viscous and incompressible.
0%
if calibrated flow meter records air speed of $8.8 m/s$ , the factor C is $0.73.$

Explanation

For(A):Pressure difference

$=\rho _{water}g.(\Delta h)$

$=(998)(9.81)(9\times 10^{-3})$

$=88.11N/m^{2}=88.11Pa$

$\simeq 88Pa$
For(B): Coefficient C is also called Mach factor and also takes account of compressibility of air.
For(C): Bernoulli equation applies strictly to streamlined, non-viscous, incompressible flow.
For(D):

$p_{s}=$ static pressure above the left hand manometer arm

$p_{T}$ Total pressure at the head of the right hand manometer arm

$p_{T}$

$-$

$p_{s}+\dfrac{1}{2}\rho _{air}v^{2}$

$p_{T}-p_{s}=\rho _{water}.g.\Delta h=\dfrac{1}{2}\rho _{air}v^{2}$

$\Rightarrow v=\sqrt{\dfrac{2\rho _{water}.g.(\Delta h)}{\rho _{air}}}$

$=\sqrt{\dfrac{2\times 88Pa}{1.2kg/m^{3}}}$

$=12.11m/s$

Coefficient,

$C=\dfrac{v_{calibrated}}{v_{air}}=\dfrac{8.8m/s}{12.11m/s}$

$\simeq 0.73$ (to two significant figure)

The difference in the pressure between the air just over the wing

$\mathrm{P}_{1}$ and that under the wing

$\mathrm{P}_{2}$ is:

0%
$2.8\times 10^{3}$ Pa
0%
$0.723\times 10^{2}$ Pa
0%
$1.46\times 10^{4}$ Pa
0%
Zero

Explanation

Compression of the streamlines means that the stream tube above the wing
has a smaller cross-sectional area than that in front of the plane and from the continuity equation , the velocity v' of the air must therefore be greater above the wing.
Area ${A}'= \dfrac{8}{10}A$ $\Rightarrow \dfrac{A}{{A}'}=\dfrac{10}{8}$
From continuity equation,
$Av={A}'{v}'$
$\therefore {v}' =\dfrac{A}{{A}'}v= \dfrac{10}{8} \times 200= 250 \: ms^{-1}$
The greater velocity (v',as obtained above) implies lower pressure than the normal pressure of the air in front of the plane. Given that the flow-lines under the wing are not compressed at all. The pressure under the wing is just the normal pressure of the air in front of the wing.

From Bernoulli's equation , with both points 1 (for p1) and 2 (for p2) at effectively same elevation.
$P_a + \dfrac{1}{2}\rho_av_1^2 = P_b + \dfrac{1}{2}\rho_bv_2^2$
Given:
$v_2=200\: ms^{-1}$
$\therefore P_2-P_1 = \dfrac{1}{2}\rho_a( v_1^2-v_2^2)= \dfrac{1}{2}\times 1.3( 250^2-200^2)= 1.46 \times 10^4\: Pa$

The hydraulic press shown in the figure is used to raise the mass

$m$ through a height of

$0.05 cm$ by performing

$500 J$ of work at the small piston. The diameter of the large piston is 10 cm while that of the smaller one is

$2 cm.$ The mass

$M$ is

0%
$10^{4} kg$
0%
$10^{3} kg$
0%
$100 kg$
0%
$10^{5}kg$

Bantu slips into a large lake and he doesn't know swimming. But he is a great fan of Superman. He shouted for help remembering Superman. As usual, Superman arrives on the top of a cliff and, due to some reason, Superman lost his flying power immediately after arrival on cliff. Due to shortage of time, somehow, Superman manages a strong and long straw and decided to drink whole water of lake to save Bantu.(Data:Atmospheric pressure

$=1.2 \times 10^{5} Pa, g=10m/s^{2}$ , density of water

$= 1000kg/m^{3})$ Assume Superman has infinite power and ability to drink whole water. Which of the following statements is/are true?
(1) Superman cannot save Bantu by this way.
(2) Superman can drink some water but not whole water.
(3) Superman will save Bantu by drinking whole water.

0%
Only (1) & (2)
0%
Only (3)
0%
Only (1)
0%
All (1), (2) & (3) are wrong

Explanation

The water can rise in the straw only upto a level which would make the pressure at the bottom of the straw equal to the atmospheric pressure. Thus the maximum height of the liquid in the straw is given as

$\rho g h=1000\times 10\times h=P_{atm}=1.2\times 10^5$
or
h=12 m.
As the straw is more than 12.5 m, Superman cannot save Bantu.

Equal volumes of two immiscible liquids of densities

$\delta$ and 3

$\delta$ are filled in a vessel. Two small holes are punched at depth

$h/3$ and

$4h/3$ from upper surface of lighter liquid. If

$V_{1}$ and

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

$V_{1}/V_{2}$ is :

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

Explanation

Applying Bernoulli's theorem to pints A( just at the surface of the liquid) and B( just outside the first hole in the vessel) taking the position of the hole as reference

$P_0 + \rho_1g\dfrac{h}{3} + \dfrac{1}{2}\rho_1 v^2 = P_0 + 0 + \dfrac{1}{2}\rho_1v_1^2$ .....(1)

$P_0$ - atmospheric pressure

$v$ - velocity of the liquid on the surface

$v_1$ - velocity of the liquid at the first hole

Applying Bernoulli's theorem to pints A( just at the surface of the liquid) and C( just outside the second hole in the vessel) taking the position of the hole as reference

$P_0 +\rho_1 gh + \rho_2g\dfrac{h}{3} + \dfrac{1}{2}\rho_1 v^2 = P_0 + 0 + \dfrac{1}{2}\rho_2v_2^2$

Putting

$\rho_2= 3 \rho_1$ ,

the above eqn becomes

$P_0 +\rho_1 gh + 3 \rho_1g\dfrac{h}{3} + \dfrac{1}{2}\rho_1 v^2 = P_0 + \dfrac{1}{2} 3\rho_1v_2^2$ ......(2)

$v_2$ - velocity of the liquid at the second hole

$v=0$ since the liquid is at rest on on the surface

Putting

$v=0$ in eqns (1) we get

$v_1= \sqrt{\dfrac{2gh}{3}}$

Putting

$v=0$ in eqns (2) we get

$v_2= \sqrt{\dfrac{4gh}{3}}$

$\therefore \dfrac{v_1}{v_2} = \sqrt{\dfrac{2}{4}}= \sqrt{\dfrac{1}{2}}$

Water is filled upto height H units in a tank placed on the ground and whose side walls are vertical. A hole is made in one of the vertical wall such that the emerging stream of water strikes the ground at the maximum range. If the level in the tank is changing at the rate of R units per second, at that instant the rate at which range will be changing will be:

0%
R/2 units per second
0%
R units per second
0%
2R units per second
0%
Zero

Explanation

According to to Torricelli's theorem, velocity of efflux at point A ( shown in the figure ) is
$v_A=\sqrt{(2gh)}$
After emerging from the orifice the water adopts parabolic path. if it takes time t secs in falling a vertical height (H=h) and covers a horizontal range $R_1$ , then
$H-h = \dfrac{1}{2}gt^2$
$\Rightarrow t= \sqrt{\dfrac{2(H-h)}{g}}$
$R_1= v_At= \sqrt{(2gh)} * \sqrt{\dfrac{2(H-h)}{g}}= 2\sqrt{h(H-h)}$ ....(1) where $R_1$ is the range.
The above results show that Range is same whether the hole is made at distance h from the top or at (H-h) from the top.
$\therefore h= H-h$
$\Rightarrow h=\dfrac{1}{2}H$ ....(2)
If h is changing at the rate $\dfrac{dh}{dt}=R$
differentiating eqn (1) wrt t, we get
$\dfrac{dR_1}{dt}= (h(H-h))^{-\dfrac{1}{2}}(\dfrac{dh}{dt}(H-h) + h( -\dfrac{dh}{dt}))=(h(H-h))^{-\dfrac{1}{2}} (R(H-h) -Rh)$ ....(3)
Substituting $h=\dfrac{H}{2}$ from eqn(2) in eqn(3) we get
$\dfrac {dR_1}{dt} = (\dfrac{H}{2})^{-1}R(H-2h)= (\dfrac{H}{2})^{-1}R(H-H)=0$

A tank is filled up to a height 2H with a liquid and is placed on a platform of height H from the ground. The distance x from the ground where a small hole is punched to get the maximum range R is

0%
H
0%
1.25 H
0%
1.5 H
0%
2 H

Explanation

We get the distance travelled in the y direction as

$x=\dfrac {1}{2}gt^2\Rightarrow t=\sqrt {\dfrac {2x}{g}}$
Velocity of efflux

$v=\sqrt{2gh}=\sqrt {2g(3H-x)}$
Hence we get the range as

$R=vt=2\sqrt {x(3H-x)}$
For range to be maximum

$\dfrac {dR}{dx}=0$
Thus we get

$\dfrac{dR}{dx}=2(x(3H-x))^{-1/2}(3H-2x)$
Which gives

$x=\dfrac {3}{2}H$

In a container water is filled upto certain height with a hole at its bottom. A bird is coming towards free surface of the liquid with velocity

$\overrightarrow V_b=(12\hat i-10\hat j)m/s$ and fish is rising upwards with velocity

$\overrightarrow V_{fish}(3\hat j)m/s$ .

The speed of bird as seen by fish under normal incidence condition at the moment when water level in container is

$20 m$ from its bottom is: (given hole area

$=1 cm^2$ ; surface area of liquid

$=20 cm^2)$

0%
$20 m/s$
0%
$40 m/s$
0%
$10 m/s$
0%
$12 m/s$

Explanation

$\upsilon = \sqrt {2gh}=\sqrt {2\times 10\times 20}=20 m/s$

$A\times u=a\upsilon$

$u=\dfrac {1}{20}\times 20=1m/sec$

$V_{APP}=9\times \dfrac {4}{3}=12m/sec$
So Velocity of bird w.r. t. fish

$=16\hat j+12\hat i = 20m/sec$

A long capillary glass tube of uniform diameter of

$1mm$ is filled completely with water and then held vertically in air. It is now opened at both ends. Find the length of the water column remaining in the glass tube. Surface tension of water is

$0.075 N/m.$

$(g=10m/s^2$ and density of water is

$10^3 kg/m^3)$ :

0%
Zero
0%
1.5 cm
0%
3 cm
0%
6 cm

Explanation

Refer to the figure.

Pressure at a

$P_a= P_0-\dfrac{2T}{r}$ .....(1)

Pressure at b

$P_b=P_0 + \dfrac{2T}{r}$ .....(2)

Also,

$P_b -P_a = \rho gh$

Substituting

$P_a$ and

$P_b$ in

$\therefore 2 \times \dfrac{2T}{r} = \rho gh$

$\Rightarrow h= \dfrac{4T}{r\rho g}= \dfrac{4 \times 0.075}{0.5 \times 10^{-3} \times 10^3 \times 10}=0.6 \times 10^{-1} \:m=6 \:cm$

In the figure, the cross-sectional area of the smaller tube is

$a$ and that of the larger tube is

$2a$ . A block of mass

$m$ is kept in the smaller tube having the same base area

$a$ , as that of the tube. The difference between water levels of the two tubes is

0%
$\dfrac {P_0}{\rho g}+\dfrac {m}{a\rho}$
0%
$\dfrac {P_0}{\rho g}+\dfrac {m}{2a\rho}$
0%
$\dfrac {m}{a\rho}$
0%
$\dfrac {m}{2\rho a}$

Explanation

By Pascal's Law, pressure is same at the same horizontal level in a liquid.
Let the heights of fluid above this level be

$h$ and

$h_1$ as shown in the figure.
Pressure from part 1 is

$P_0+h_1\rho g+\dfrac{mg}{a}$
Pressure from part 3 is

$P_0+h\rho g$
Equating these two we have

$P_0+h_1\rho g+\dfrac{mg}{a}= P_0+h\rho g$

$\therefore h-h_1=\dfrac{m}{a\rho}$
Therefore difference in height of liquid in the two arms is

$\dfrac{m}{a\rho}$

A capillary tube with inner cross-section in the form of a square of side a is dipped vertically in a liquid of density

$\rho$ and surface tension

$\sigma$ which wet the surface of capillary tube with angle of contact

$\theta$ . The approximate height to which liquid will be raised in the tube is : (Neglect the effect of surface tension at the corners of capillary tube)

0%
$\dfrac{2\sigma cos\theta }{a\rho g}$
0%
$\dfrac{4\sigma cos\theta }{a\rho g}$
0%
$\dfrac{8\sigma cos\theta }{a\rho g}$
0%
None of these

Explanation

Upward force by capillary tube on top surface of liquid is equal to the product of the perimeter and surface tension.

$F_{up}= 4a\sigma \cos\theta$ ... ..(1)

In equilibrium, surface tension force is equal to the weight of the liquid column of height h, h being the height to to which the liquid rises

$W= a^2\rho gh$ .....(2)

Equating (1) and (2)

$4a\sigma \cos\theta = a^2\rho gh$

$\therefore h= \dfrac{4\sigma \cos\theta}{a\rho g}$

The figure shows a crude type of atomizer. When bulb A is compressed, air flows swiftly through tube BC causing reduced pressure in the particles of the vertical tube. The liquid rises in the tube, enters

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

$P_a+P$ , where

$P$ is the gauge pressure and

$P_a$ is the atmospheric pressure,

$v$ is the speed of air in BC, find how large would

$v$ need to be to cause the liquid to rise to BC.

(Density of air

$=1.3 kg/m^3$ .)

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

Explanation

Absolute pressure in bulb

$=$ Gauge pressure + Atmospheric pressure

$=P+P_a$
Using Bernoulli's equation,

$P_a+P=P_{BC}+\displaystyle\frac {\rho_a v^2}{2}$
where

$\rho_a$ is the density of air.

$\therefore P_{BC}=P_a+P=\left (\displaystyle\frac {1.3}{2}\right )v^2$

$P_{BC}=P_a-\rho gh$

So equating these two values for

$P_{BC}$ , we get:

$v=\sqrt{(\displaystyle\frac {P+\rho gh}{{0.65}})}$

In a cylindrical water tank here are two small holes Q and P on the wall at a depth of

$h_1$ from the upper level of water and at a height of

$h_2$ from the lower end of the tank, respectively, as shown in the figure. Water coming out from both the holes strike the ground at the same point. The ratio of

$h_1$ and

$h_2$ is

0%
$1$
0%
$2$
0%
$> 1$
0%
$< 1$

Explanation

The two streams strike at the same point on the ground.

$R_1=R_2=R$

$u_1t_1=u_2t_2$
where

$u_1$ velocity of efflux at

$Q=\sqrt {(2gh_1)}$ and

$u_2=$ velocity of efflux at

$P=\sqrt {[2g(H-h_2)]}$

$t_1=$ time of fall of water stream through Q is

$\sqrt {\displaystyle\frac {2(H-h_1)}{g}}$

$t_2=$ time of fall of the water stream through

$P=\sqrt {\displaystyle\frac {2h_2}{g}}$
Putting theses values is Eq. (i), we get

$\sqrt{2gh_1}(\sqrt{\displaystyle\frac{2(H-h_1)}{g}})=\sqrt{2g(H-h_2)}\sqrt{\displaystyle\frac{2h_2}{g}}$

$(H-h_1)h_1=(H-h_2)h_2$
or

$[H-(h_1+h_2)][h_1-h_2]=0$

$H=h_1+h_2$ is irrelevant because the holes are at two different heights. Therefore,

$h_1=h_2$ or

$h_1/h_2=1$

A hole is made at the bottom of a tank filled with water (density

$=10^3 kg/m^3)$ . If the total pressure at the bottom of the tank is

$3 atm$ (

$1 atm=10^5 N/m^2$ ), then the velocity of efflux is

0%
$\sqrt {400}m/s$
0%
$\sqrt {200}m/s$
0%
$\sqrt {600}m/s$
0%
$\sqrt {500}m/s$

Explanation

Applying Bernoulli's equation to two points one on the surface and the other just outside the hole at the bottom,

$P_{atm} + \rho gh + \dfrac{1}{2}\rho v^2 = P_{atm} + 0 + \dfrac{1}{2}\rho v'^2$

Given,

$v=0$

Pressure difference between top and bottom is

$2 atm$ i.e equivalent of

$20m$ of water column

$\therefore h= 20$

$\dfrac{1}{2}\rho v'^2 = \rho gh$

$\therefore \dfrac{1}{2} \times 1 \times v'^2 = 10 \times 20$

$\Rightarrow v' = \sqrt{400}$

A cylindrical vessel of

$90\ cm$ height is kept filled up to the brim. It has four holes

$1, 2, 3$ and

$4$ which are, respectively, at heights of

$20\ cm,\ 30\ cm,\ 40\ cm$ and

$50\ cm$ from the horizontal floor

$PQ$ . The water falling at the maximum horizontal distance from the vessel comes from:

0%
hole number $4$
0%
hole number $3$
0%
hole number $2$
0%
hole number $1$

Explanation

The maximum horizontal distance from the vessel comes from hole numbers 3 and 4. Now,

$v=\sqrt {2gh}$ , where

$h$ is the height of the hole from the top.

Let say height of vessel is '

$H$ '.
The vertical distance travelled by the water is given as

$H-h=\dfrac{1}{2}gt^2$
We get the time of descent as

$t=\sqrt{\dfrac{2(H-h)}{g}}$
Horizontal distance,

$x=vt=\sqrt {2gh}\sqrt {\dfrac {2(H-h)}{g}}=2\sqrt {h(H-h)}$
Substituting the values we get the maximum distance for hole number 3 and 4.

Equal volume of two immiscible liquids of densities

$\rho$ and

$2\rho$ are filled in a vessel as shown in the 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%
$\displaystyle\frac {1}{2}$
0%
$\displaystyle\frac {1}{4}$
0%
$\displaystyle\frac {1}{\sqrt 2}$

Explanation

Using the velocity of efflux for the upper hole we get

$v_1=\sqrt {2g\left (\displaystyle\frac {h}{2}\right )}=\sqrt {gh}$
For the second hole we apply the Bernoulli's equation, thus we get

$\rho gh+2\rho g\left (\displaystyle\frac {h}{2}\right )=\displaystyle\frac {1}{2}(2\rho )v_2^2$

$\Rightarrow v_2=\sqrt {2gh}$

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

Figure shows a cylindrical vessel of height

$90 cm$ filled up to the brim. There are four holes in the vessel as shown. The liquid falling at maximum horizontal distance from the vessel would come from

0%
hole 1
0%
hole 2
0%
hole 3
0%
hole 4

Explanation

R would be maximum if

$hh'$ is maximum.

$1.\ hh'=70\times 20=1400$

$2.\ hh'=60\times 30=1800$

$3.\ hh'=50\times 40=2000$

$4.\ hh'=40\times 50=2000$
Clearly, the liquid coming out from holes 3 and 4 would fall at maximum horizontal distance.

A cylindrical vessel of a very large cross-sectional area is containing two immiscible liquids of density

$\rho_1=600 kg/m^3$ and

$\rho_2=1200 kg/m^3$ as shown in the figure.
A small hole having cross-sectional area

$5 cm^2$ is made in right side vertical wall as shown. Take atmospheric pressure as

$p_0=10^5 N/m^2, g=9.8m/s^2$ . For this situation, mark out the correct statements(s). (Take cross-sectional area of the cylindrical vessel as

$1000 cm^2$ . Neglect the mass of the vessel)

0%
If the surface on which the vessel is placed is smooth, then a rightward force of magnitude 2N is to be applied on the vessel to maintain its static equilibrium.
0%
If the surface on which the vessel is placed is smooth, then no force is needed to maintain its static equilibrium.
0%
If the surface on which the vessel is placed is rough $(\mu=0.04)$ , then the minimum force (horizontal) needed to be applied on the vessel to maintains its static equilibrium is zero.
0%
If the surface on which the vessel is placed is rough ( $\mu=0.04$ ), then the maximum force (horizontal) needed to be applied on the vessel to maintain its static equilibrium is 19.8 N.

Explanation

The velocity with which the liquid comes out is given by the expression using Bernoulli's theorem:

$P_0 + \rho_1g \times 30 \times 10^{-2} +\rho_2g \times 10 \times 10^{-2} = P_0 + \dfrac{1}{2}\rho_2v^2$ ....(1)
Given

$\rho_1 = 600\: kg /m^3$
Given

$\rho_2 = 1200\: kg/m^3$
Substituting in eqn(1) we get,

$600 \times g\times 30 \times 10^{-2} +1200 \times g \times 10 \times 10^{-2} = \dfrac{1}{2} \times 1200 \times v^2$

$\therefore v^2 = 4.9$

$\therefore v_2 = 2.21 \: m/s$

Due to transfer of liquid, change in momentum takes place and hence a force has been exerted on the liquid in backward direction, this force is given by :

$F=\rho_2av^2= 1200 \times 5 \times 10^{-4} \times 4.9\approx 3\:N$ , where

$a$ is the cross-sectional area of hole.

Normal contact force between the vessel and the surface is

$N= mg$

$N = \rho_1 h_1Ag + \rho_2 h_2Ag$
Substituting the values we get

$N=420\:kg\:m/s^2$
Limiting frictional force:

$f_L= \mu \times N= 0.04 \times 420=16.8\: N$
As

$f_L$ > F, the minimum force necessary to maintain the static equilibrium is 0. For the maximum value of the applied force,

$F_{max} =f_L + F=16.8 +3 = 19.8\:N$

A tank in filled with water of density

$10^3 kg/m^3$ and oil of density

$0.9\times 10^3 kg/m^3$ . The height of water layer is

$1 m$ and that of the oil layer is

$4 m$ . The velocity of efflux from an opening in the bottom of the tank is

0%
$\sqrt {85}m/s$
0%
$\sqrt {88}m/s$
0%
$\sqrt {92}m/s$
0%
$\sqrt {98}m/s$

Explanation

Let

$d_w$ and

$d_0$ be the densities of water and oil, respectively. Then the pressure at the bottom of the tank is

$h_wd_wg+h_0d_0g$ .

Let this pressure be equivalent to pressure due to water of height h. Then,

$hd_wg=h_wd_2g+h_0d_0g$

$\therefore h=h_w+\displaystyle\dfrac {h_0d_0}{d_w}=1+\dfrac {4\times 0.9}{1}$

$=1+3.6=4.6 m$

According to Toricelli's theorem,

$v=\sqrt {2gh}=\sqrt {2\times 10\times 4.6 m/s}=\sqrt {92}m/s$

A cylindrical vessel contains a liquid of density

$\rho$ up to height h. The liquid is closed by a piston of mass

$m$ and area of cross section

$A$ . There is a small hole at the bottom of the vessel. The speed

$v$ with which the liquid comes out of the hole is

0%
$\sqrt {2gh}$
0%
$\sqrt {2\left (gh+\frac {mg}{\rho A}\right )}$
0%
$\sqrt {2\left (gh+\frac {mg}{A}\right )}$
0%
$\sqrt {2gh+\frac {mg}{A}}$

Explanation

Applying Bernoulli's theorem at points 1 and 2, difference in pressure energy between 1 and 2

$=$ difference in kinetic energy between 1 and 2.
Hence,

$h\rho g+\displaystyle\frac {mg}{A}=\frac {1}{2}\rho v^2$
or

$v=\sqrt {2gh+\displaystyle\frac {2mg}{\rho A}}=\sqrt {2\left (gh+\displaystyle\frac {mg}{\rho A}\right )}$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 10 - 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 10 - MCQExams.com

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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