MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 6

A spherical solid ball of volume V is made of a material of density

$\displaystyle { \rho }_{ 1 }$ It is falling through a liquid of density

$\displaystyle { \rho }_{ 1 }\left( { \rho }_{ 2 }<{ \rho }_{ 1 } \right)$ . Assume that the liquid applies a viscous force on the ball that is Proportional to the square of its speed v, i.e.,

$\displaystyle { F }_{ viscous }=-{ kv }^{ 2 }\left( k>0 \right)$ . The terminal speed of the ball is :

0%
$\displaystyle \sqrt { \frac { Vg\left( { \rho }_{ 1 }-{ \rho }_{ 2 } \right) }{ k } }$
0%
$\displaystyle \frac { Vg{ \rho }_{ 1 } }{ k }$
0%
$\displaystyle \sqrt { \frac { Vg{ \rho }_{ 1 } }{ k } }$
0%
$\displaystyle \frac { Vg\left( { \rho }_{ 2 }<{ \rho }_{ 1 } \right) }{ k }$

Explanation

The condition for terminal speed

$(v_t)$ is Weight = Buoyant force + Viscous force

$W = V\rho_1g$

$V\rho_1g = V\rho_2 g + k v_t^2$

$v_t =\sqrt{ \dfrac{V g(\rho_1 -\rho_2)}{k}}$

A common hydrometer has a long uniform stem. When floating in pure water, 4.5 cm of its stem lies below the surface of water. In a liquid of specific gravity 2.0, 1.5 cm of the stem of the same hydrometer is immersed. Find the specific gravity of the liquid in which the hydrometer is immersed upto 0.5 cm.

0%
2
0%
3
0%
4
0%
5

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

If the terminal speed of a sphere of gold (density =

$\displaystyle 19.5{ kg }/{ { m }^{ 3 } }$ ) is

$0.2 m/s$ in a viscous liquid(density =

$\displaystyle 1.5{ kg }/{ { m }^{ 3 } }$ ), find the terminal speed of a sphere of silver (density =

$\displaystyle 10.5{ kg }/{ { m }^{ 3 } }$ ) of the same size in the same liquid

0%
$0.4 m/s$
0%
$0.133 m/s$
0%
$0.1 m/s$
0%
$0.2m/s$

Explanation

$V_T = \dfrac{2}{9\eta}r^2(d_1 - d_2)g$

$\dfrac{V_{T_2}}{0.2} = \dfrac{10.5 - 1.5}{19.5 - 1.5}$

$V_{T_2} = 0.1 m/s$

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

$\displaystyle 1000{ kg }/{ { m }^{ 3 } }$ ), If the total pressure at the bottom ofthe tankis 3 atmosphere
( 1 atmosphere =

$\displaystyle { 10 }^{ 5 }{ N }/{ { m }^{ 2 } }$ ), then the velocity of efflux is :

0%
$\displaystyle \sqrt { 200 } { m }/{ s }$
0%
$\displaystyle \sqrt { 400 } { m }/{ s }$
0%
$\displaystyle \sqrt { 500 } { m }/{ s }$
0%
$\displaystyle \sqrt { 800 } { m }/{ s }$

Explanation

We know that velocity of efflux,

$v = \sqrt{2gh}$

At the bottom of tank pressure is 3 atmosphere. So, total pressure due to water column

$h\rho g = 2 \times 10^5$

$gh = 2 \times 10^2$

$v = \sqrt{2gh} = \sqrt{2 \times 2\times 10^2} = \sqrt{400} m/s$

Water is filled in a container upto height of

$3 m$ . A small hole of area '

$\displaystyle { A }_{ 0 }$ ' 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

$\displaystyle { { A }_{ 0 } }/{ A }=0.1$ then

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

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

0%
$\displaystyle 50{ { m }^{ 2 } }/{ { s }^{ 2 } }$
0%
$\displaystyle 50.5{ { m }^{ 2 } }/{ { s }^{ 2 } }$
0%
$\displaystyle 51{ { m }^{ 2 } }/{ { s }^{ 2 } }$
0%
$\displaystyle 52{ { m }^{ 2 } }/{ { s }^{ 2 } }$

Explanation

The square of the velocity of flux

$v^2 = \dfrac{2gh}{1 - (\dfrac{A_o}{A})^2}$

$= \dfrac{2 \times 10 \times 2.475}{1 - (0.1)^2} = 50 m^2/s^2$

The angle of contact in case of liquid depends upon which of the following?

0%
nature of liquid and solid
0%
material which exists above free surface of liquid
0%
both of them above
0%
None

$300,000 kg$ commercial airlines is flying through the air with

$150,000 N$ of thrust.
For airplanes traveling at high speeds, the drag force is related to the velocity by the following equation

${F}_{drag}=-b{v}^{2}$
where drag force is opposite to the direction of the velocity.
If the drag constant

$b$ is

$3 {kg}/{m}$ and the airplane is moving at

$200 {m}/{s}$ , how would you describe its motion?

0%
In equilibrium (cruising at constant speed)
0%
Near equilibrium, positive acceleration
0%
Near equilibrium, negative acceleration
0%
Greatly accelerating during takeoff
0%
Greatly slowing down during landing

Explanation

Drag force acting at the given moment=

$F_{drag}=-bv^2=-(3)\times (200)^2=-120000N$

Hence net force acting on the plane=

$150000N-120000N$

$=30000N$

Hence the option is accelerating at the moment.

$50kg$ skydiver falls through the air and reaches terminal velocity after some time. The drag force is a function of velocity given by

${F}_{drag}=-b{v}^{2}$
where the negative sign denotes that the drag force is opposite to the direction of the velocity.
What is the terminal velocity of the skydiver (assuming the drag constant

$b$ is

$0.2 {kg}/{m}$ )?

0%
$5\dfrac{m}{s}$
0%
$50\dfrac{m}{s}$
0%
$100\dfrac{m}{s}$
0%
$250\dfrac{m}{s}$
0%
$2500\dfrac{m}{s}$

Explanation

When the falling body through the air will reach to the terminal velocity, the drag force will be equal to weight of the body.

Thus,

$bv^2=mg$

The terminal velocity,

$v=v_t=\sqrt{\frac{mg}{b}}=\sqrt{\frac{50\times 10}{0.2}}=50 m/s$

Thus , option B will be correct.

Water is flowing in streamline motion through a horizontal tube. The pressure at a point in the tube is

$P$ where the velocity of flow is

$v$ . At another point, where the pressure is

$P/2$ , the velocity of flow is:
[Density of water

$=\rho$ ]

0%
$\sqrt { { v }^{ 2 }+\cfrac { P }{ \rho } }$
0%
$\sqrt { { v }^{ 2 }-\cfrac { P }{ \rho } }$
0%
$\sqrt { { v }^{ 2 }+\cfrac { 2P }{ \rho } }$
0%
$\sqrt { { v }^{ 2 }-\cfrac { 2P }{ \rho } }$

Explanation

As the water is flowing through the horizontal tube, so in the streamline flow of water, the sum of static pressure and dynamic pressure is constant

$P+\cfrac { 1 }{ 2 } \rho { v }^{ 2 }=\cfrac { P }{ 2 } +\cfrac { 1 }{ 2 } \rho { v }_{ 1 }^{ 2 }$

$\Rightarrow { v }_{ 1 }=\sqrt { \cfrac { P }{ \rho } +{ v }^{ 2 } }$

Two spheres of the same material, but of radii

$R$ and

$3R$ are allowed to fall vertically downwards through a liquid of density

$\rho$ . The ratio of their terminal velocities is:

0%
$1:3$
0%
$1:6$
0%
$1:9$
0%
$1:1$

Explanation

Terminal velocity of the sphere

$v = \dfrac{2 gr^2 (\sigma - \rho)}{9 \eta}$

where

$\eta$ is the viscosity of the liquid,

$\sigma$ is the density of material of sphere,

$\rho$ is the density of liquid and

$r$ is the radius of the sphere.

$\implies$

$v \propto r^2$

Given :

$r_1 = R$ and

$r_2 = 3R$

Ratio of terminal velocities

$v_1 : v_2 = r_1^2 : r_2^2$

$v_1 : v_2 = R^2 : (3R)^2$

$\implies$

$v_1:v_2 = 1:9$

Due to air a falling body faces a resistive force proportional to square of velocity

$v$ , consequently its effective downward acceleration is reduced and is given by

$a = g - kv^{2}$ where

$k = 0.002m^{-1}$ . The terminal velocity of the falling body is (in m/s)

0%
$60$
0%
$70$
0%
$80$
0%
$90$

Explanation

As velocity increases

$a$ will decrease and at some point, it will become zero and velocity become constant that is terminal velocity.

At terminal velocity-

$g-kv_{ter}^2 =0 \implies v_{ter}^2= \dfrac{g}{k}=\dfrac{9.8}{0.002}=4900 \implies v_{ter}=70\dfrac{m}{s}$

The pressure exerted by a column of liquid of height

$h$ and density

$p$ is given by the hydrostatic pressure equation equal to

0%
$pgh$
0%
$pg$
0%
$gh$
0%
Cannot be calculated

Explanation

Let area of column =

$A$

Total volume of mass =

$V\times \rho = (Ah)\rho$

Pressure exerted =

$\dfrac{F}{A} = \dfrac{Ah\rho \times g}{A} = \rho gh$

Water is flowing in streamline motion through a horizontal tube. The pressure at a point in the tube is

$p$ where the velocity of flow is

$v$ . At another point, where the pressure is

$p/2$ , the velocity of flow is [density of water =

$\rho$ ]

0%
$\sqrt{v^2 + \dfrac{p}{\rho}}$
0%
$\sqrt{v^2 - \dfrac{p}{\rho}}$
0%
$\sqrt{v^2 + \dfrac{2p}{\rho}}$
0%
$\sqrt{v^2 - \dfrac{2p}{\rho}}$

Explanation

Given :

$P_1 = p$

$v_1 = v$

$P_2 = p/2$

Using Bernoulli's equation,

$P_1 + \rho gh_1 + \dfrac{1}{2}\rho v_1^2 = P_2 +\rho g h_2 + \dfrac{1}{2}\rho v_2^2$

As water flows in a horizontal tube i.e.

$h_1 = h_2$

We get

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

$p+ \dfrac{1}{2}\rho v^2 = \dfrac{p}{2} + \dfrac{1}{2}\rho v_2^2$

$\dfrac{1}{2} \rho (v_2^2 - v^2 ) = \dfrac{p}{2}$

$\implies v_2 = \sqrt{v^2 +\dfrac{p}{\rho}}$

In the diagram the area of cross section of the pistons

$A$ and

$B$ are

$8{cm}^{2}$ and

$320{cm}^{2}$ respectively then the thrust on the piston at

$B$ is

0%
$160kgf$
0%
$320kgf$
0%
$420kgf$
0%
$80kgf$

Explanation

${ A }_{ 1 }=8{ cm }^{ 2 }\quad \quad \quad { A }_{ 2 }=320{ cm }^{ 2 }$

${ F }_{ 1 }=4kgf\quad \quad \quad { F }_{ 2 }=xN$

Since

$\dfrac { { F }_{ 1 } }{ { A }_{ 1 } } =\dfrac { { F }_{ 2 } }{ { A }_{ 2 } } \Rightarrow \dfrac { x }{ 320 } =\dfrac { 4 }{ 8 }$

$\boxed { x=160kgf }$

A cylinder is filled with non viscous liquid of density

$d$ to a height

$h_o$ and a hole is made at a height

$h_1$ from the bottom of the cylinder. The velocity of liquid issuing out of the hole is

0%
$\sqrt(2gh_o)$
0%
$\sqrt(2g(h_o-h_1))$
0%
$\sqrt(dgh_1)$
0%
$\sqrt(dgh_0)$

The velocity of the water flowing from the inlet pipe is less than the velocity of water flowing out from the spin pipe B.

0%
variation of water level in vessel will be irregular.
0%
water level will remains constant.
0%
the water level will perform periodic oscillation motions.
0%
none of the above.

In a streamline flow of a liquid

0%
Every particle has its own velocity, different from others.
0%
All particles move with a constant velocity, even if the path is curvilinear.
0%
At a point on the streamline, particle can have two velocities.
0%
At a point on the streamline, particle can have only one velocity along the tangent.

Three capillaries of lengths

$L$ ,

$\dfrac{L}{2}$ and

$\dfrac{L}{3}$ are connected in series. Their radii are

$r$ ,

$\dfrac{r}{2}$ and

$\dfrac{r}{3}$ respectively. Then if stream line flow is to be maintained and the pressure across the first capillary is

$P$ , then

0%
the pressure difference across the ends of the second capillary is $8 P$
0%
the pressure difference across the third capillary is $43 P$
0%
the pressure difference across the ends of the second capillary is $16 P$
0%
the pressure difference across the ends of the second capillary is $59 P$

A square hole of side length

$l$ is made at a depth of

$y$ and a circular hole is made at a depth of

$4y$ from the surface of water in a water tank kept on a horizontal surface. If equal amount of water comes out of the vessel through the holes per second then the radius of the circular hole is equal to

$(r, l << y)$ :

0%
$l / \sqrt2$
0%
$l / 2$
0%
$l / \sqrt {2 \pi}$
0%
$\sqrt2\pi$

Explanation

Given

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

here

${ A }_{ 1 }={ l }^{ 2 }\quad \quad \& \quad { A }_{ 2 }=\Pi { r }^{ 2 }$

${ v }_{ 1 }=\sqrt { 2gh } \quad \quad { v }_{ 2 }=\sqrt { 2g4h }$

$\sqrt { 2gh } \times { l }^{ 2 }=2\sqrt { 2gh } \times \Pi { r }^{ 2 }$

$\boxed { r=\dfrac { l }{ \sqrt { 2\Pi } } }$

The angle of contact between a glass capillary tube of length

$10 cm$ and a liquid is

$90^o$ . If the capillary tube is dipped vertically in the liquid, then the liquid

0%
Will rise in the tube
0%
Will get depressed in the tube
0%
Will rise up to $10 cm$ in the tube and will over flow
0%
Will neither rise nor fall in the tube

Explanation

The right

$\left( { 90 }^{ 0 } \right)$ angle of contact implies that cohesion force & adhesive forces are equal in magnitude. Hence we will observe no capillary rise in this case.

A water tank placed on the floor has two small holes, pinched in the vertical wall, one above the other. The holes are

$3.3 cm$ and

$4.7 cm$ above the floor. If the jets of water issuing out from the holes hit the floor at the same point on the floor, then the height of water in the tank is

0%
$3 cm$
0%
$6 cm$
0%
$8 cm$
0%
$9 cm$

Explanation

Since Range =

$2\sqrt { (H-h)h }$

Given

${ h }_{ 1 }=3.3\quad \& \quad { h }_{ 2 }=4.4$

and

${ R }_{ 1 }={ R }_{ 2 }$

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

$\left( H-{ h }_{ 1 } \right) { h }_{ 1 }=(H-{ h }_{ 2 }){ h }_{ 2 }$

$(H-3.3)\dfrac { 3.3 }{ 4.7 } =H-4.7$

0.702H - 2.317 = H - 4.7

0.297H = 2.3829

$\boxed { H=8cm }$

Water and mercury are filled in two cylindrical vessels upto same height. Both vessels have a hole in the wall near the bottom. The velocity of water and mercury coming out of the holes are

$v_1$ and

$v_2$ respectively. Thus

0%
$v_1 = v_2$
0%
$v_1 = 13.6 v_2$
0%
$v_1 = \dfrac{v_2}{13.6}$
0%
$v_1 = \sqrt {13.6} \ v_2$

Explanation

Since velocity of liquid coming out of tank is irrespective of density of material & given as velocity =

$\sqrt { 2gh }$ . Hence both will come out at same velocity.

If the surface of a liquid is plane, then the angle of contact of the liquid with the walls of container is

0%
Acute angle
0%
Obtuse angle
0%
$90^o$
0%
$0^o$

Explanation

The angle of contact with wall of container is angle b/w liquid surface & normal to the wall of container. Hence for a plane liquid angle of contact is

${ 0 }^{ 0 }$ .

A tube of length L and radius

$R$ is joined to another tube of length

$\dfrac{L}{3}$ and radius

$\dfrac{R}{2}$ . A fluid is flowing through this tube. If the pressure difference across the first tube is

$P$ , then the pressure difference across the second tube is

0%
$\dfrac{16P}{3}$
0%
$\dfrac{4P}{3}$
0%
$P$
0%
$\dfrac{3P}{16}$

Explanation

Since we know that flow rate remains constant, then

flow rate in 1 = flow rate in 2.

$\dfrac { \pi { P }_{ 1 }{ r }_{ 1 }^{ 4 } }{ 8g{ l }_{ 1 } } =\dfrac { \pi { P }_{ 1 }{ r }_{ 2 }^{ 4 } }{ 8g{ l }_{ 2 } }$

$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { { l }_{ 1 } }{ { l }_{ 2 } } \times { \left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right) }^{ 4 }\quad \quad \quad \left[ now\quad since\ { r }_{ 2 }=\dfrac { { r }_{ 1 } }{ 2 } \ \& \quad { l }_{ 2 }=\dfrac { { L }_{ 1 } }{ 3 } \right]$

So,

$\dfrac { P }{ { P }_{ 2 } } =\dfrac { 3{ L }_{ 1 } }{ { L }_{ 1 } } ={ \left( \dfrac { { r }_{ 1 } }{ { 2r }_{ 1 } } \right) }^{ 4 }=\dfrac { 3 }{ 16 }$

$\boxed { { P }_{ 2 }=\dfrac { 16 }{ 3 } P }$

For a liquid, which is rising in a capillary tube, the angle of contact is

0%
$90^o$
0%
$180^o$
0%
Acute
0%
Obtuse

In a surface tension experiment with a capillary tube water rises up to

$0.1 m$ . If the same experiment is repeated on an artificial satellite which is revolving around the earth. The rise of water in a capillary tube will be

0%
$0.1 m$
0%
$9.8 m$
0%
$0.98 m$
0%
Full length of capillary tube

$5 g$ of water rises in the bore of capillary tube when it is dipped in water. If the radius of bore capillary tube is doubled, the mass of water that rises in the capillary tube above the outside water level is

0%
$1.5 g$
0%
$10 g$
0%
$5 g$
0%
$15 g$

The height of water in a capillary tube of radius

$2 cm$ is

$4 cm$ . What should be the radius of capillary, if the water rises to

$8 cm$ in tube?

0%
$1 cm$
0%
$0.1 cm$
0%
$2 cm$
0%
$4 cm$

Explanation

It is to be remembered that

height of a capillary rise

$\propto \dfrac { 1 }{ radius\quad of\quad capillary }$

hence,

$\dfrac { { h }_{ 1 } }{ { h }_{ 2 } } =\dfrac { { r }_{ 2} }{ { r }_{ 1 } }$

$\dfrac { 4 }{ 8 } =\dfrac { { r }_{ 2 } }{ 2 } \Rightarrow \boxed { { r }_{ 2 }=1cm }$

In a capillary tube, fall of liquid is possible when angle of contact is

0%
Acute angle
0%
Right angle
0%
Obtuse angle
0%
None of these

The pressure just below the meniscus of water

0%
Is greater than just above it
0%
Is less than just above it
0%
Is same as just above it
0%
Is always equal to atmospheric pressure

The surface of water in contact with glass wall is

0%
Plane
0%
concave
0%
convex
0%
Both b and c

When a capillary tube is immersed vertically in water the capillary rise is

$3 cm$ . if the same capillary tube is inclined at angle of

$60^o$ to the vertical, the length of the water column in the capillary tube above that of the outside level is

0%
$6 cm$
0%
$1 cm$
0%
$8 cm$
0%
Zero

Explanation

Vertical Capillary rise will be constant

$lcos{ 60 }^{ 0 }={ h }_{ 1 }$

$l=2\times 3cm=6cm$

A capillary tube when immersed vertically in a liquid records a rise of

$3 cm$ . if the tube is immersed in the liquid at an angle of

$60^o$ with the vertical, then length of the liquid column along the tube will be:

0%
$2 cm$
0%
$3 cm$
0%
$6 cm$
0%
$9 cm$

Explanation

$lcos{ 60 }^{ 0 }=3cm$

$l=2\times 3cm=6cm$

1175572_602251_ans_a4dddc52e7a7465eaabb5194a52219f9.jpg

Two capillary tubes of the same material but of different radii are dipped in a liquid. The heights to which the liquid rises in the two tubes are

$2.2 cm$ and

$6.6 cm$ . The ratio of radii of the tubes will be

0%
$1:9$
0%
$1:3$
0%
$9:1$
0%
$3:1$

Explanation

Since we know that height of capillary rise is inversely proportional to radii of tube, i.e.,

height

$\propto \dfrac { 1 }{ radius }$

$\dfrac { { h }_{ 1 } }{ { h }_{ 2 } } =\dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \Rightarrow \dfrac { 2.2cm }{ 6.6cm } =\dfrac { { r }_{ 2 } }{ { r }_{ 1 } }$

So,

$\boxed { \dfrac { { r }_{ 1 } }{ { r }_{ 2 } } =3 }$

Water rises up to a height

$h_1$ in a capillary tube of radius

$r$ . The mass of the water lifted in the capillary tube is

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

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

Explanation

Since we know that mass of water rise is proportional to volume of water.

Mass

$\infty$ volume

$\dfrac { { M }_{ 1 } }{ { M }_{ 2 } } =\dfrac { { V }_{ 1 } }{ { V }_{ 2 } } =\dfrac { \pi { r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ \pi { r }_{ 2 }^{ 2 }{ h }_{ 2 } } =\dfrac { { r }_{ 1 }^{ 2 }{ h }_{ 1 } }{ { r }_{ 2 }^{ 2 }{ h }_{ 2 } } \quad \rightarrow (1)$

and for capillary tube, we know that height

$\alpha$

$\dfrac { 1 }{ radius }$

So,

$\dfrac { { h }_{ 1 } }{ { h }_{ 2 } } =\dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \quad \rightarrow (II)$

hence from (1) & (II)

$\dfrac { { M }_{ 1 } }{ { M }_{ 2 } } =\dfrac { { r }_{ 1 }^{ 2 } }{ { r }_{ 2 }^{ 2 } } \times \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } =\dfrac { { r }_{ 1 } }{ { r }_{ 2 } }$

${ M }_{ 2 }=\dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \times { M }_{ 1 }=\dfrac { 2r }{ r } \times M=2M$

$\boxed { { M }_{ 2 }=2M }$

In a capillary tube, fall of liquid is possible when angle of contact is

0%
Acute angle
0%
Right angle
0%
Obtuse angle
0%
None of these

The height of water in a capillary tube of radius

$2 cm$ is

$4 cm$ . What should be the radius of capillary, if the water rises to

$8 cm$ in tube?

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

Explanation

Since we know that height of capillary rise in inversely proportional to radius of capillary.

i.e. height

$\alpha$

$\dfrac { 1 }{ radius }$

$\dfrac { { h }_{ 1 } }{ { h }_{ 2 } } =\dfrac { { r }_{ 2 } }{ { r }_{ 1 } }$

$\dfrac { 4 }{ 8 } =\dfrac { { r }_{ 2 } }{ 2 } \Rightarrow \boxed { { r }_{ 2 }=1cm }$

For tap water and clean glass, the angle of contact is

0%
$0^o$
0%
$90^o$
0%
$140^o$
0%
$8^o$

Explanation

The angle of contact between pure water and glass is found to be

${ 0 }^{ 0 }$ as both cohesive forces and adhesive forces are equal. But due to presence of ions in tap water the angle of contact is acute from

${ 8 }^{ 0 }$ to

${ 18 }^{ 0 }$ .

For tap water and clean glass, the angle of contact is

0%
$0$
0%
$90$
0%
$140$
0%
$8$

Explanation

Angle of contact between pure water and glass is

${ 0 }^{ 0 }$ but in tap water there are some foreign particles, which reduces the cohesive force, leading to on acute angle contact between

${ 8 }^{ 0 }$ to

${ 18 }^{ 0 }$ .

If every particle of fluid has irregular flow, then flow is said to be

0%
laminar flow
0%
turbulent flow
0%
fluid flow
0%
both $A$ and $B$

Venturi relation is one of the applications of

0%
equation of continuity.
0%
Bernoulli's equation.
0%
light equation.
0%
speed equation.

A liquid is kept in a glass vessel. If the liquid solid adhesive force between the liquid and the vessel is very weak as compared to the cohesive force in the liquid, then the shape of the liquid surface near the solid should be

0%
Concave
0%
Convex
0%
Horizontal
0%
Almost vertical

Water and mercury are filled in two cylindrical vessels upto same height. Both vessels have a hole in the wall near the bottom. The velocity of water and mercury coming out of the holes are

$v_1$ and

$v_2$ respectively. Then :

0%
$v_1 = v_2$
0%
$v_1 = 13.6v_2$
0%
$v_1 = \dfrac{v_2}{13.6}$
0%
$v_1 = (13.6v_2)$

Explanation

The velocity of water coming out of the holes is independent of density or nature of substance, i.e.,

Velocity =

$\sqrt { 2gh }$

Hence

${ { v }_{ 1 }={ v }_{ 2 } }$

In a hydraulic lift, used at a service station the radius of the large and small piston are in the ratio of $20 : 1$ . What weight placed on the small piston will be sufficient to lift a car of mass $1500 kg$ ?

0%
$3.75kg$
0%
$37.5kg$
0%
$7.5kg$
0%
$75kg$

There are two identical small holes of area of cross section a on the either sides of a tank containing a liquid of density p (shown in figure). The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is :

0%
$\dfrac{2gh}{pa}$
0%
$\dfrac{pgh}{a}$
0%
$ghpa$
0%
$2pagh$

Explanation

Net force (reaction)

$F=F_B-F_A=\dfrac{dp_B}{dt}-\dfrac{dp_A}{dt}$

$=av_B p\times v_B-av_A p\times v_A$

$\therefore F = ap(v^2_B-v^2_A)$
According to Bernaulli's theorem

$P_A+\dfrac{1}{2}pv^2_A+pgh = P_B+\dfrac{1}{2}pv^2_B+0$

$\Rightarrow \dfrac{1}{2}p(v^2_B-v^2_A) = pgh$

$\Rightarrow v^2_B-v_A^2=2gh$
From Eqs (i) and (ii), we get

$F=ap(2gh)=2apgh$

A uniform capillary tube of length

$l$ and inner radius

$r$ with its upper end sealed is submerged vertically into water. The outside pressure is

${p}_{0}$ and surface tension of water is

$\gamma$ . When a length

$x$ of the capillary is submerged into water, it is found that water levels inside and outside the capillary coincide, the value of

$x$ is

0%
$\cfrac { l }{ \left( l+\cfrac { { p }_{ 0 }r }{ 4\gamma } \right) }$
0%
$l\left( l-\cfrac { { p }_{ 0 }r }{ 4\gamma } \right)$
0%
$l\left( l-\cfrac { { p }_{ 0 }r }{ 2\gamma } \right)$
0%
$\cfrac { l }{ \left( l+\cfrac { { p }_{ 0 }r }{ 2\gamma } \right) }$

Explanation

For air inside capillary,

${ p }_{ 0 }(lA)=p'(l-x)A$ where

$p'$ is pressure in capillary after being submerged

$\therefore p'=\cfrac { { p }_{ 0 }l }{ l-x }$
Now since level of water inside capillary coincides with outside

$p'-{ p }_{ 0 }=\cfrac { 2\gamma }{ r }$

$\therefore \cfrac { { p }_{ 0 }l }{ l-x } -{ p }_{ 0 }=\cfrac { 2\gamma }{ r } \Rightarrow x=\cfrac { l }{ \left( l+\cfrac { { p }_{ 0 }r }{ 2\gamma } \right) }$

Find the difference of air pressure between the inside and outside of a soap bubble

$5 mm$ in diameter, if the surface tension is

$1.6 N/m$ .

0%
$2560 N/ m^2$
0%
$3720 N /m^2$
0%
$1208 N /m^2$
0%
$950 N / m^2$

Explanation

$\rho = \dfrac{4T}{R} = \dfrac{4\times 1.6}{2.5\times 10^{-3}}$

$=2560 N/m^2$

A small spherical drop fall from rest in viscous liquid. Due to friction, heat is produced. The correct relation between the rate of production of heat and the radius of the spherical drop at terminal velocity will be

0%
$\dfrac { dH }{ dt } \propto \dfrac { 1 }{ { r }^{ 5 } }$
0%
$\dfrac { dH }{ dt } \propto { r }^{ 4 }$
0%
$\dfrac { dH }{ dt } \propto \dfrac { 1 }{ { r }^{ 4 } }$
0%
$\dfrac { dH }{ dt } \propto { r }^{ 5 }$

Explanation

Viscous force acting on spherical drop

${ F }_{ v }=6\pi \eta v$

$\therefore$ Terminal velocity,

$v=\dfrac { 2 }{ 9 } \dfrac { { r }^{ 2 }\left( \sigma -\rho \right) g }{ \eta }$
where,

$\eta =$ coefficient of viscosity of liquid

$\sigma =$ density of material of spherical drop

$\rho =$ density of liquid
Power imparted by viscous force

$=$ Rate of production of heat

$\rho =\dfrac { dH }{ dt } ={ F }_{ v }\cdot v=6\pi \eta { v }^{ 2 }$

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

$\Rightarrow \dfrac { dH }{ dt } \propto { r }^{ 5 }$

A small metal ball of mass '

$m$ ' is dropped in a liquid contained in a vessel, attains a terminal velocity '

$V$ '. If a metal ball of same material but of mass

$'8m'$ is dropped in same liquid then the terminal velocity will be.

0%
$V$
0%
$2V$
0%
$4V$
0%
$8V$

Explanation

Terminal velocity of the ball

$V = \dfrac{2 gr^2(\rho - \sigma)}{9\eta}$ ........(1)

where

$\rho$ and

$\sigma$ are the density of ball and liquid, respectively.

Mass of the metal ball

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

$\implies$

$r \propto m^{1/3}$

But from equation (1), we get

$V \propto r^2$

$\implies$

$V\propto m^{2/3}$ ......(2)

Given :

$V_1 = V$

$m_1 = m$

$m_2 =8m$

From (2) we get

$\dfrac{V_2}{V_1} = \bigg(\dfrac{m_2}{m_1}\bigg)^{2/3}$

$\dfrac{V_2}{V} = \bigg(\dfrac{8m}{m}\bigg)^{2/3} = 8^{2/3}$

$\implies$

$V_2= (2^3)^{2/3} V = 4V$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 6 - 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 6 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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