The residual pressure of a vessel at $${27}^{o}C$$ is $$1\times{10}^{-11}N/{m}^{2}$$. The number of molecules per cc in this vessel is nearly

$${10}^{-11}\times 6\times{10}^{23}$$

$$2.68\times{10}^{-19}\times {10}^{-11}$$

MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 14

The residual pressure of a vessel at

27^{o} C

${27}^{o}C$ is

1 \times 10^{- 11} N / m^{2}

$1\times{10}^{-11}N/{m}^{2}$ . The number of molecules per cc in this vessel is nearly

0%
$2400$
0%
$2.4\times{!0}^{8}$
0%
${10}^{-11}\times 6\times{10}^{23}$
0%
$2.68\times{10}^{-19}\times {10}^{-11}$

Water is filled in a container up to a height of

3 m

$3m$ . A small hole of area

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

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

$A$ . If

$\dfrac{a}{A}=0.1$ then

$v^2$ (in

$m^2s^{-2}$ ) is (where v is the velocity of water coming out of the hole)

0%
$50$
0%
$51$
0%
$48$
0%
$52$

A drop of liquid of density

$\rho$ is floating half-immersed in a liquid of density

$d$ . If

$\rho$ is the surface tension the diameter of the drop of the liquid is

0%
$\sqrt {\dfrac {\sigma}{g(2\rho - d)}}$
0%
$\sqrt {\dfrac {3\sigma}{g(2\rho - d)}}$
0%
$\sqrt {\dfrac {6\sigma}{g(2\rho - d)}}$
0%
$\sqrt {\dfrac {12\sigma}{g(2\rho - d)}}$

Explanation

The equation for the surface tension of the drop is:

$2\pi r\sigma + \dfrac {1}{2} \times \dfrac {4}{3} \pi r^{3} dt = \dfrac {4}{3} \pi r^{3}\rho g$

$2\pi r\sigma = \dfrac {\pi r^{3} g}{3} [4\rho - 2d]$

$r^{2} = \dfrac {3\times 2\pi \sigma}{\pi g(2\rho - 2d)}$

$r^{2} = \dfrac {3\sigma}{g(2\rho - d)}$

$r = \sqrt {\dfrac {3\sigma}{g(2\rho - d)}}$

$Diameter = 2r = \sqrt {\dfrac {12\sigma}{g(2\rho - d)}}$ .

Two identical holes of each of cross sectional area

$A$ are opened on the opposite sides of a wide vertical vessel filled with water.The difference in vertical height of the two holes is

$h$ . The resultant force of reaction of water flowing out of vessel is-

0%
$h\rho gA$
0%
$\cfrac{1}{2}h\rho g A$
0%
$2h\rho gA$
0%
$\cfrac{1}{4}h\rho g A$

The piston shown in the figure is moved downwards such that the vapour volume is decreased by 246.3cc. How many grams of

${ H }_{ 2 }O$ is condensed? Assume aqueous tension at

$\quad { 27 }^{ 0 }C$ is 360 mm Hg)

0%
1.8 gms
0%
0.09 gms
0%
0.9 gms
0%
0.18 gms

Figure shows a large open tank. Water emerging out from holes

$O_1$ and

$O_3$ strikes the ground at same point but from

$O_2$ has maximum range. The height of

$O_2$ from ground

0%
$\frac{{{h_1} - {h_2}}}{2}$
0%
$\frac{{{h_2} + {h_1}}}{2}$
0%
$\frac{{\sqrt {{h_1}{h_2}} }}{2}$
0%
$2 {{\sqrt {{h_1}{h_2}} }}$

A thin square plate of side

$5\ cm$ is suspended vertically from a balance so that lower side just dips into water with side to surface. When the plate is clean

$(\theta = 0^{\circ})$ , it appears to weigh

$0.044\ N$ . But when the plate is greasy

$(\theta = 180^{\circ})$ , it appears to weigh

$0.03\ N$ . The surface tension of water

0%
$3.5 \times 10^{- 2}N/m$
0%
$7.0 \times 10^{- 2}N/m$
0%
$14.0 \times 10^{- 2}N/m$
0%
$1.08 N/m$

Explanation

Difference in apparent weights is due to differences in forces of surface tension. Due to

$180^{\circ}$ , the force surface tension in one case is opposite to the force of surface tension in the other case.

$\therefore 2\times \sigma_{w}\times \dfrac {10}{100} = 0.004 - 0.03$
or

$\sigma_{w}\times \dfrac {0.014}{1}\times 5 N/m = 0.07\ N/m$ .

Water flows through a pipe. The area of cross - section at one place

$A_1$ = 10

$cm^{2}$ , velocity of flow 1 m/s and pressure 2000 pa. At another place area

$A_2$ = 5

$cm^{2}$ . What is the pressure at area

$A_2$ = 5

$cm^2$

0%
500 Pa
0%
100 Pa
0%
400 Pa
0%
200 Pa

A constant of

$20 cm$ maintained in the contains water of

$1 kg$ as shown in the figure. A small orifice area

${ 10 }^{ -2 }{ m }^{ 2 }$ is made at the bottom of the vertical wall of the container the ejected water is directed as shown in the figure. Assuming the mass of container is negligible, the net force on the container is

0%
$Zero$
0%
$2\,N$
0%
$11\,N$
0%
$18\,N$

A cylindrical vessel of height

$500$ mm has an orifice (small hole) at its bottom. The orifice is initially closed and water is filled in it up to height

$H$ . Now the top is completely sealed with a cap and the orifice at the bottom is opened. Some water comes from the orifice and the water level in the vessel becomes steady with height of water column being

$200$ mm. Find the fall in height (in mm) of water level due to opening of the orifice.

0%
6mm
0%
12mm
0%
16mm
0%
18mm

An ideal liquid of density

$\rho$ is filled in a horizontally fixed syringe fitted with piston. There is no friction between the piston and the inner surface of the syringe. Cross-section area of the syringe is A An orifice is made at one end of the syringe. When the piston is pushed into the syringe, the liquid comes out of the orifice and then following a parabolic path falls on the ground.
With what velocity does the liquid come out of the orifice?

0%
$\sqrt { \dfrac { F }{ \rho A } }$
0%
$\sqrt { \dfrac { 2F }{ \rho A } }$
0%
$\sqrt { \dfrac { F+2\rho ghA }{ \rho A } }$
0%
$\sqrt { \dfrac { F+\rho ghA }{ \rho A } }$

An ideal liquid of density

0%
$\sqrt { \dfrac { F+\rho ghA }{ \rho A } }$
0%
$\sqrt { \dfrac { 2F}{ \rho A } }$
0%
$\sqrt { \dfrac { 2F+\rho ghA }{ \rho A } }$
0%
$\sqrt { \dfrac { 2(F+\rho ghA) }{ \rho A } }$

A test tube filled with water is being spun around in an ultracentrifuge with angular velocity

$\omega$ . The test tube is lying along a radius and the free surface of the water is at radius

$r_0$ (Figure). The length of the test tube is

$l$ density of the water. Ignore gravity and ignore atmospheric pressure. Choose the correct statement.

0%
The pressure at the mid point of test tube is given by $\dfrac {1}{8}\rho l\ \omega^2(l+4r_0)$
0%
The pressure at the mid point of test tube is given by $\dfrac {1}{2}\rho l\ \omega^2 \left (\dfrac {l}{2}+r_0 \right)$
0%
The pressure at the mid point of test tube is given by $\dfrac {1}{4}\rho l\ \omega^2 \left (\dfrac {l}{2}+r_0 \right)$
0%
The pressure remains constant through out the tube

An object of volume

$0.5 m^3$ and density

$500 kg/m^3$ is glued to the flat bottom of a vessel containing water. The flat base of the object glued to the bottom has an area

$100 cm^2$ and density of water is

$1.0 g/cm^3$ . If the maximum force that the glue can withstand is

$2000 N$ , to what maximum height above the base can water be filled in the vessel so that the object is not torn apart from the base?

0%
$5 m$
0%
$10 m$
0%
Up to any height
0%
Insufficient information

A solid sphere moves at a terminal velocity of

$20ms^{-1}$ in air at a place where

$g = 9.8ms^{-2}$ . The sphere is taken in agravity=free hall having air at the same pressure and pushed down at a speed of

$20ms^{-1}$ .

0%
Its initial acceleration will be $9.8ms^{-1}$ downward
0%
Its initial acceleration will be $9.8ms^{-1}$ upward
0%
The magnitude of acceleration will decrease as the time passes.
0%
It will eventually stop

A cylindrical vessel is filled with water to a height

$H$ . A vessel has two small holes in the side, from which water is rushing out horizontally and the two streams strike the ground at the same point. If the lower hole

$Q$ is

$h$ height above the ground, then the height of hole

$P$ above the ground will be

0%
$2h$
0%
$H/h$
0%
$H - h$
0%
$H/2$

Water is flowing continuously from a tap of area

$10^{-4} m^2$ . The water velocity as it leaves the top is

$1 m/s$ .
Find out area of the water stream at a distance

$0.15 m$ below the top.

0%
$0.5 \times 10^{-4} m^2$
0%
$1 \times 10^{-4} m^2$
0%
$2 \times 10^{-4} m^2$
0%
$0.25 \times 10^{-4} m^2$

Explanation

$A_1 V_1 = A_2 V_2$

$V_2 = \sqrt{V_1^2 + 2 gh}$

$V_2 = \sqrt{1^2 + 2 \times 10 \times 0.15}$

$V_2 = 2 \, m/s$

$10^{-4} \times 1 = A_2 \times 2$

$A_2 = 0.5 \times 10^{-4} m^2$ .

A cylinder of mass m and density

$\rho$ hanging from a string is lowered into a vessel of cross-sectional area s containing a liquid of density

$\sigma (< \rho)$ until it is fully immersed. The increase in pressure at the bottom of the vessel is?

0%
$\dfrac{m\rho g}{\sigma s}$
0%
$\dfrac{mg}{s}$
0%
$\dfrac{m\sigma g}{\rho s}$
0%
Zero

A solid ball of density

$\rho_1$ and radius

$r$ falls vertically through a liquid of density

$rho_2$ . Assume that the viscous force acting on the ball is

$F=krv$ , where

$k$ is a constant and

$v$ its velocity. What is the terminal velocity of the ball ?

0%
$\dfrac{4\pi gr^2( \rho_1-\rho_2)}{3k}$
0%
$\dfrac{2\pi r( \rho_1-\rho_2)}{3gk}$
0%
$\dfrac{2\pi g( \rho_1+\rho_2)}{3gr^2k}$
0%
$None\ of\ these$

The area of cross section of the two vertical arms of a hydraulic press are

$1\ cm^2$ and

$10\ cm^2$ respectively. A force of

$10\ N$ applied, as shown in the figure, to a tight fitting light piston in the thinner arms balances a force

$F$ applied to the corresponding piston in the thicker arm. Assuming, that the levels of water in both the arms are the same, we can conclude :

0%
$F=100\ N$
0%
$F=25\ N$
0%
$F=50\ N$
0%
$F$ , as applied, cannot keep piston in equilibrium

Determine the velocity

$v_{B}$ of the water issuing out at B.

0%
$v_B=\sqrt{2gh_{2}}$
0%
$v_B=\sqrt{2g\left(h_{1}+h_{2}\right)}$
0%
$v_B=\sqrt{2g\left(h_{2}-h_{1}\right)}$
0%
$v_B=\sqrt{gh_{2}}$

Explanation

As long as water fills the tube (as shown in the figure) and points

$A$ and

$B$ are open to the atmosphere, the velocity at

$B$ will be given by Torricelli's theorem.

Hence,

$v_B=\sqrt{2gh_2}$ where

$h_2$ is the difference in the levels

$A$ and

$B$ .

1596383_1739547_ans_113c3c61e91d4ecba9bdcee8c561d609.png

The velocity of efflux is

0%
$10ms^-1$
0%
$20ms^-1$
0%
$4ms^-1$
0%
$35ms^-1$

Explanation

Since area of a hole is very small in comparison to base area.

$A$ of the cylinder, velocity of liquid inside the cylinder is negligible. Let velocity of efflux be

$v$ and atmospheric pressure be

$P_0$ .

Consider two points A (inside the cylinder) and B (just outside the hole) in the same horizontal line as shown in the figure in question.

Pressure at

$A, P_A=P_0+h\rho_2g+\left(h-y\right)\rho_1{g}$

Pressure at

$B$ ,

$P_B=P_0$

According to Bernoulli's theorem,

Pressure energy at

$A=Pressure\ energy\ at\ B+ Kinetic\ energy\ at\ B$

$\therefore P_A=P_B+\dfrac{1}{2}\rho_1v^2$

$\therefore v=4ms^-1$

A bent tube is lowered into water stream as shown in the figure. The velocity of the stream relative to the tube is equal to

$v=2\ m/s$ . The closed upper end of the tube locates at height

$h_0=10\ cm$ above free surface of water has a small orifice. To what height

$h$ will the water get spurt ?

0%
$5\ cm$
0%
$10\ cm$
0%
$20\ cm$
0%
$40\ cm$

A cylindrical vessel filled with water upto a height

$h$ weight

$w$ and is resting on a horizontal plane. The side wall of the vessel has a plugged circular hole touching the bottom. The coefficient of friction between the bottom of the vessel and the plane is

$\mu$ . The minimum diameter of the hole so that the vessel just begins to move on the floor after the removal of the plug is :

0%
$\sqrt{\dfrac{\pi \mu w}{\rho h g}}$
0%
$\sqrt{\dfrac{ \mu w}{2\pi \rho h g}}$
0%
$\sqrt{\dfrac{ \mu w}{\pi \rho h g}}$
0%
$\sqrt{\dfrac{ 2\mu w}{\pi \rho h g}}$

Water is filled in a cylindrical to a height of 3 m. The ratio of the cross-sectional area of the orifice and the beaker is 0.1 he square of the speed of the liquid coming out from the orifice is (g = 10 m/s) [IIT-JEE, 2004]

0%
50 m/s
0%
50.5 m/s
0%
51 m/s
0%
52 m/s

Explanation

.elt A = cross-section of tank
a = cross-section hole
V = velocity with which level decreases

$\upsilon$ = velocity of efflux
From equation of continuity

$a \upsilon = AV \Rightarrow = \frac{a \upsilon} {A}$
By using Bernoulli's theorem for energy per unit volume
Energy per unit volume at a point A = Energy per unit volume at point B

$P + \rho gh + \frac{1} {2} \rho V^{2} = P + 0 + \frac{1} {2} \rho V^{2}$

$\Rightarrow \upsilon^{2} = \frac{2gh} {1 - \left ( \frac{a} {A} \right )^{2}} = \frac{2 \times 10 \times \left (3 - 0.525 \right )} {1 - \left ( 0.1 \right )^{2}} = 50 \left ( m/sec \right )^{2}$
1719349_1812490_ans_2295a3cec2e04880bf3c3b7a60381395.png

1719349_1812490_ans_2295a3cec2e04880bf3c3b7a60381395.png

A liquid is kept in a cylindrical vessel which is being rotated about a vertical axis through the centre of the circular base. If the radius of the vessel is

$r$ and angular velocity of rotation is

$\omega$ , then the difference in the heights of the liquid at the centre of the vessel and the edge is

0%
$\frac{r \omega} {2g}$
0%
$\frac{r^{2} \omega^{2}} {2g}$
0%
$\sqrt{2gr\omega}$
0%
$\frac{omega^{2}} {2gr^{2}}$

Explanation

Using Bernoulli's theorem,

$P_A + \frac{1} {2} dv^{2}_A + dgh_A = P_B + \frac{1} {2} dv^{2}_B + dgh_B$
Here,

$h_A = h_B$

$\therefore P_A + \frac{1} {2} dv^{2}_A = P_B + \frac{1} {2} dv^{2}_B$
Now,

$v_A = 0, v_B = r \omega$ and

$P_A - P_B = hdg$

$\therefore hdg = \frac{1} {2}dr^{2} \omega^{2}$ or

$h = \frac{r^{2} \omega^{2}} {2g}$
1719340_1812481_ans_c6d9b695d6f74e22afa96a78d661e425.png

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 14 - 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 14 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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