Match column I with column II : List I List II Magnus energy Pascal's Law Loss of Energy Archimedes' principle Pressure is same at same level in a liquid Viscous force Hydraulic Machines Lifting of asbestos roofs

A$$\rightarrow$$4, B$$\rightarrow$$3, C$$\rightarrow$$1, D$$\rightarrow$$1

B$$\rightarrow$$4, C$$\rightarrow$$2, D$$\rightarrow$$1, A$$\rightarrow$$1

C$$\rightarrow$$4, A$$\rightarrow$$2, B$$\rightarrow$$1, D$$\rightarrow$$1

D$$\rightarrow$$4, A$$\rightarrow$$2, B$$\rightarrow$$1, C$$\rightarrow$$1

MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 3

The pressure that will be built up by a compressor in a paint-gun when a stream of liquid paint flows out with a velocity of

$25 ms^{-1}$ ( density of paint is

$0.8 gm -cm^{-3}$ ) is : ( in

$Nm^{-2}$ )

0%
$2.5 \times 10^{2}$
0%
$2.5 \times 10^{3}$
0%
$2.5 \times 10^{5}$
0%
$5 \times 10^{5}$

Explanation

Applying Bernoulli's Theorem just inside and just outside the hole from which the point is flowing,

$p_{1}+0=\dfrac{1}{2}\rho v^{2}$

$\Rightarrow p_{1}=\dfrac{1}{2}\times 800\times \left ( 25 \right )^{2}$

$p_{1}=2.5\times 10^{5}N/m^{2}$

A plane is in a level flight at a constant speed and each of its two wings has an area of

$25 \ m^{2}$ . If the speed of air is

$180 \ km h^{-1}$ over the lower wing and

$234 \ km h^{-1}$ over the upper wing surface, the plane's mass is : (Take density of air

$=1 kg m^{-3}$ .)

0%
$4.400 gm$
0%
$4400 gm$
0%
$44.0 kg$
0%
$4400 kg$

Explanation

Applying bernouillis theorem above and below the wing,

$p_{1}+\dfrac{1}{2}f v_{1}^{2}=p_{2}+\dfrac{1}{2} f v_{2}^{2}$

$\Rightarrow p_{1}-p_{2}=\dfrac{1}{2}\left ( 65^{2} -50^{2}\right )$

$\Rightarrow p_{1}-p_{2}=862.5pa$

So, upthrust on wings=weight of plane

$\Rightarrow \dfrac{862.5\times 2\times 25}{10}=$ mass of plane

$\Rightarrow Mass=4312kg\approx 4400kg$

A tank containing water has an orifice on one vertical wall. If the centre of the orifice is 4.9 m below the surface of water in the tank, the velocity of discharge is :

0%
0.98 m/s
0%
9.8 m/s
0%
98 m/s
0%
9.8 cm/s

Explanation

Velocity of discharge at the orifice =

$\sqrt{2gh}$ [Toricelli's Theorem]

$=\sqrt{2 \times 9.8 \times 4.9}$

$=9.8 m/s$

A cylindrical vessel contains a liquid of density

$\rho$ upto a 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 with which the liquid comes out of the vessel is:

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

Explanation

Applying Bernoulli theorem across 1 and 2,

$\dfrac{2mg}{A\rho }+2\rho gh=\dfrac{1}{2}\rho v^{2}$

$v_{2}=\dfrac{2mg}{A\rho }+2gh$

$v^{2}=2g\left ( h+\dfrac{m}{\rho A} \right )$

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

Two hail stones with radii in the ratio of

$1:2$ fall from a great height through the atmosphere. Then their terminal velocities are in the ratio of:

0%
$1:2$
0%
$2:1$
0%
$1:4$
0%
$4:1$

Explanation

$\textbf{Hint}$ : Use the formula of terminal velocity

$\textbf{Step 1}$ :

We know

Terminal velocity

$V=\dfrac{2r^2(\rho-\rho_0)g}{9 \eta}$

Here we can see that

$V \propto r^2$

$\textbf{Step 2}$

We can write

$\dfrac{V_1}{V_2}=\dfrac{r_1^2}{r_2^2}$

So, ratio of their terminal velocities of the given two hail stones is

$=\dfrac{r^2}{(2r)^2}$

$=\dfrac{r^2}{4r^2}=\dfrac{1}{4}$

So, the ratio is

$1:4$

Thus option C is correct.

There is a hole at the bottom of a large open vessel.If water is filled upto a height

$h$ , it flows out in time

$t$ .If water is filled to a height

$4h$ , it will flow out in time is

0%
$4t$
0%
$t/4$
0%
$t/2$
0%
$2t$

Explanation

$v-u = -gt$

$\Rightarrow 0 - \sqrt{2gh} = -gt$

$t = \sqrt{\dfrac{2h}{g}}$

For height

$h^{1} = 4h t^{1} = \sqrt{\dfrac{2 (4h)}{g}} = 2t$

A tank with vertical walls is mounted so that its base is at a height of

$1.2 m$ above the horizontal ground. The tank is filled with water to a depth

$2.8m.$ A hole is punched in the side wall of thetank at a depth

$x \ m$ below the surface of water to have maximum range of the emerging stream.Then the value of

$x$ in meter is

0%
$4.0$
0%
$1.6$
0%
$2.0$
0%
$2.3$

Explanation

Velocity of ejection at a depth x is

$\sqrt{2gx}$
time of flight

$= \sqrt{\dfrac{2H}{9}} = \sqrt{\dfrac{2(1.2+2.8-x)}{9}}$

$= \sqrt{\dfrac{2(4-x)}{9}}$

$\therefore$ Horizontal Range

$R = Vt = \sqrt{4 x (4-x)}$

$= 2\sqrt{x(4-x)}$

$\dfrac{dR}{dx} = 0 \Rightarrow x= \dfrac{4}{2} = 2m$

A wind - powered generator converts wind energy into electrical energy. Assume that the generator converts a fixed fraction of the wind energy intercepted by its blades into electrical energy. For wind speed

$V$ , the electrical power output will be proportional to

0%
$V$
0%
$V^{2}$
0%
$V^{3}$
0%
$V^{4}$

Explanation

Applying bernoullis eqn at point 1 and 2,

$p_{0}+\dfrac{1}{2}f v^{2}=p_{r},\left [ p_{0}=0\right ]$

So,

force on the wing

$=p_{1}\times A=\dfrac{1}{2}f v^{2}$

So,

work done on wing

$=\dfrac{1}{2}f v^{2}.\dfrac{v}{k}=Kv^{3}$

So,

power output in wing

$\propto v^{3}$

Water stands at a height of

$100 cm$ in a vessel whose side walls are vertical. A , B and C are holes at height

$80cm, 50 cm,$ and

$20 cm$ respectively from the bottom of the vessel. The correct system of water flowing out is :

Explanation

At any height h, velocity of water ejection

$= \sqrt{2gh}$
time of flight

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

$\therefore$ range

$R = vT = \sqrt{2gh} \times \sqrt{\dfrac{2(H-h)}{g}} = 2\sqrt{h(H-h)}$

$\therefore$ for

$h = 20cm , 80cm$

$R$ is same.

Therefore

$R$ is maximum for

$h = \dfrac{H}{2}$

There are two holes

$O_{1}$ and

$O_{2}$ in a tank of height H. The water emerging from

$O_{1}$ and

$O_{2}$ strikes the ground at the same points,as shown in fig. Then:

0%
$H=h_{1}+h_{2}$
0%
$H=h_{2}-h_{1}$
0%
$H=h_{1}h_{2}$
0%
$H=h_{2}/h_{1}$

Explanation

At any height h below the surface , velocity of ejection

$V = \sqrt{2gh}$

time of flight

$t = \sqrt{2g (H-h)}$

$\therefore$ Range

$R = vt = 2\sqrt{h (H-h)}$

$R_{1} = R_{2} \Rightarrow h_{1}(H-h_{1}) = h_{2} (H-h_{2})$

$\Rightarrow h_{1} = H-h_{2}$

$h_{1}+h_{2} =H$

A horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity 1ms

$^{-1}$ when pressure is 50kPa at a point. If the velocity of flow has to be 2 ms

$^{-1}$ at some other piont,the pressure at that point should be:

0%
50 kPa
0%
100kPa
0%
48.5 kPa
0%
24.25 kPa

Explanation

Applying Bernoulli's equation at point 1 and point 2,

$\Rightarrow P_{1}+\dfrac{1}{2}\rho V_{1}^{2}=p_{2}+\dfrac{1}{2}\rho V_{2}^{2}$

$\Rightarrow 50\times 10^{3}+\dfrac{1}{2}\times 10^{3}\left ( 1 \right )^{2}=p_{2}+\dfrac{1}{2}\times 10^{3}\times 4$

$\Rightarrow 5\times 10^{4}+500-2000=p_{2}$

$\Rightarrow p_{2}=48.5kp{a}$

Air of density

$1.3 kg m^{-3}$ blows horizontally with a speed of

$108 km h^{-1}$ . A house has a plane roof of area

$40 m^{2}$ . The magnitude of aerodynamic lift on the roof is.

0%
$2.34 \times 10^{4}$ Dynes
0%
$0.234 \times 10^{4} N$
0%
$2.34 \times 10^{4} N$
0%
$23.4 \times 10^{4} N$

Explanation

Applying bernoullis theorem above and below the roof,

$p_{1}+\dfrac{1}{2}+v_{1}^{2}=p_{2}$

$\Rightarrow p_{2}-p_{1}=\dfrac{1}{2}+v_{1}^{2}=\dfrac{1}{2}\times 1.3\times \left ( 30 \right )^{2}$

$\Rightarrow$ difference in pressure

$=585 N/m^{2}...........2$
So,
Magnitude of aerodynamic life

$=1\times Area$

$=585\dfrac N{m^{2}}\times 40m^{2}$

$2.34\times 10^{4}N$

In a plant, sucrose solution of coefficient of viscosity

$0.0015 N.m^{-2}$ is driven at a velocity of

$10^{-3} m s^{-1}$ through xylem vessels of radius

$2\mu m$ and length

$5 \mu m$ . The hydrostatic pressure difference across the length of xylem vessels in

$Nm^{-2}$ is :

0%
$5$
0%
$8$
0%
$10$
0%
$15$

Explanation

Fow rate

$q=v\times \pi r^2$

also

$q=\dfrac{\pi Pr^4}{8\mu l}=v\pi r^2$

now

$v=10^{-3}m/s\ ,\mu=0.0015N/m^2\ ,r=2\times 10^{-6}m$

on solving above equation and putting value in SI unit we get

$P=15 N/m^2$

Two identical tall jars are filled with water to the brim. The first jar has a small hole on the side wall at a depth

$h/3$ and the second jar has a small hole on the side wall at a depth of

$2h/3$ , where h is the height of the jar. The water issuing out from the first jar falls at a distance

$R_{1}$ from the base and the water issuing out from the second jar falls at a distance

$R_{2}$ from the base. The correct relation between

$R_{1} \ and \ R_{2}$ is

0%
$R_{1} > R_{2}$
0%
$R_{1} < R_{2}$
0%
$R_{2} = 2 \times R_{1}$
0%
$R_{1} = R_{2}$

Explanation

At any height h from surface, velocity

$V = \sqrt{2gh}$
Time of flight

$t = \sqrt{\dfrac{2(H-h)}{g}}$
where

$H$ is the total height of jar.

$\therefore$ Range

$R = V\times t$

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

Given,

$h_{1} = \dfrac{h}{3} \ and \ H=h$

$\Rightarrow$

$R_1 = 2\sqrt{\dfrac{h}{3}(h-\dfrac{h}{3})}$

$= \dfrac{2\sqrt{2}h}{{3}}$

For

$\ h_{2} = \dfrac{2h}{3}\ and \ H=h$

$\Rightarrow$

$R_2 = 2\sqrt{\dfrac{2h}{3}(h-\dfrac{2h}{3})}$

$= \dfrac{2\sqrt{2}h}{{3}}$

$\therefore R_{1} = R_{2}$

A tank with vertical walls is mounted so that its base is at a height

$H$ above the horizontal ground. The tank is filled with water to a depth '

$h$ ' . A hole is punched in the side wall of the tank at a depth '

$x$ ' below the water surface. To have maximum range of the emerging stream,the value of

$x$ is

0%
$\dfrac{H+h}{4}$
0%
$\dfrac{H+h}{2}$
0%
$\dfrac{H+h}{3}$
0%
$\dfrac{3(H+h)}{4}$

Explanation

Horizontal velocity of ejection, when a vessel is filled with

water to height h is

$\sqrt{2gh}$

$\therefore$ velocity of ejection

$= \sqrt{2g(h-x)}$

Time taken will be

$\sqrt{\dfrac{2h}{g}} = \sqrt{\dfrac{2(H+h-x)}{g}}$

$\therefore range = 2\sqrt{(h-x)(H+h-x)} [\because R = ut]$

range is max , when

$\dfrac{d R}{d x} = 0$

$\Rightarrow x = \dfrac{H+h}{2}$

A water tank standing on the floor has two small holes vertically one above the other punched on one side. The holes are

$h_{1}$ cm and

$h_{2}$ cm above the floor. How high does water stand in the tank, when the jets from the holes hit the floor at the same point?

0%
$(h_{2}-h_{1})$
0%
$(h_{2}+h_{1})$
0%
$(h^{2}_{2}-h^{2}_{1})$
0%
$\dfrac{h_{2}}{h_{2}-h_{1}}$

Explanation

Time taken by water at (1) to reach the ground =

$\sqrt{\dfrac{2h_{2}}{g}}$

velocity of water at (1) =

$\sqrt{2gH}$

So,

$d= \dfrac{\sqrt{2gH \times 2 P_{n}}}{g}$ -------------(1)

Similarly for point ,

Time taken =

$\sqrt{\dfrac{2h_{1}}{g}}$

velocity =

$\sqrt{2g (H +h_{2}-h_{1})}$

So,

$d= \sqrt{2g(H+h_{2}-h_{1}) \times \dfrac{2h_{1}}{g}}$ ----------(2)

Equation (1) and (2),

$\Rightarrow (H+h_{2}-h_{1})h_{1}=Hh_{2}$

$\Rightarrow Hh_{1}+h_{2}h_{1}-h_{1}^{2}=Hh_{2}$

$\Rightarrow H(h_{1}-h_{2})=h_{1}(h_{1}-h_{2})$

So,

$\Rightarrow H=h_{1}$

Height of water =

$h_{1}+h_{2}$

In the experimental arrangement shown in figure, the areas of cross-section of the wide and narrow portions of the tube are 5 cm

$^{2}$ and 2 cm

$^{2}$ respectively. The rate of flow of water through the tube is 500 cm

$^{3}$ s

$^{-1}$ . The difference of mercury levels in the U-tube is:

0%
$0.97 \ cm$
0%
$1.97 \ cm$
0%
$0.67 \ cm$
0%
$4.67 \ cm$

Explanation

$Q = 500 cm^{3}/s$
Using continuity equation we have $Q=A_1v_1=A_2v_2$
or
$500=5(v_1)=2(v_2)$
Thus we get
$v_1=100 cm/s=1 m/s$ and $v_2=250 cm/s=2.5 m/s$

Applying Bernoullis therom, In point (1) and (2),
$\Rightarrow P_{1}-P_{2}=\dfrac{1}{2}\rho(v_2^2-v_1^2)=\dfrac{1}{2}1000(2.5^2-1^2)=2625 N/m^{2}$
$\Rightarrow P_{1}-P_{2}=2625= \rho gh=13.6\times 1000\times 9.8\times h$
$\Rightarrow h = 1.97 \ cm$

Water from a tap emerges vertically down with an initial speed of

$1.0\ ms^{-1}$ . The cross sectional area of tap is

$10\ cm^{2}$ . Assume that the pressure is constant through out the stream of water,and that the flow is steady, the cross sectional area of the steam

$0.15\ m$ below the tap is

0%
$5\times 10^{-4}\ m^{2}$
0%
$5 \times 10^{-5}\ m^{2}$
0%
$5.0 \times 10^{-5}\ m^{2}$
0%
$50\times 10^{-4}\ m^{2}$

Explanation

Given initial velocity

$u = 1 m s^{-1}$
Area

$A = 10 c m^{2} = 10 \times 10^{-4} m^{2}$
Velocity at height 0.15 m below the tap is:

$v^{2} = u^{2} + 2 gh$

$= (1)^{2} + 2\times 10\times 0.1 s$

$= 4$

$\Rightarrow v = 2 m s^{-1}$

We know that by continuity Theoram

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

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

$5\times 10^{-4} = A^{2}$

A square hole of side a is made at a depth

$h$ below water surface and to the side of a water container another circular hole of radius

$r$ is made to the same container at a depth of

$4h$ . It is found that volume flow rate of water through both the holes is found to be same then:

0%
$r=\dfrac{a}{2\sqrt{\pi }}$
0%
$a=\dfrac{r}{2\sqrt{\pi }}$
0%
$a=\dfrac{r}{\sqrt{2\pi }}$
0%
$r=\dfrac{a}{\sqrt{2\pi }}$

Explanation

$\rho ^{1}A_{1}v_{1} = \rho ^{2}A_{2}v_{2}$ as volume rate is constant
velocity by toricellis therom is

$\sqrt{2gh}$

$\therefore a^{2}\left ( \sqrt{2gh} \right ) = \pi r^{2}\left ( \sqrt{2g\left ( 4h \right )} \right )$

$\Rightarrow 2\pi r^{2} = a^{2}$

$r = \dfrac{a}{\sqrt{2\pi }}$

A cylinder of height

$20m$ is completely filled with water. The velocity of efflux of water (in

$ms^{-1}$ ) through a small hole on the side wall of the cylinder near its bottom is:

0%
$10 ms^{-1}$
0%
$20 ms^{-1}$
0%
$25.5 ms^{-1}$
0%
$5 ms^{-1}$

Explanation

Given that

Pressure at the top surface

$=P_1 =p_{atm}$

Pressure at the hole,

$P_2= p_{atm}$

Here height of the cylinder ,

$h=20\ m$

From Bernoulli's Theorem we have:

Sum of all mechanical energies remains constant at a point over all points.

$\implies P +\rho g h+ \dfrac12 \rho v^2=constant$

where

$\rho$ is the density of the fluid and

$v$ is the velocity at that point .

So, applying Bernoulli Theorem at top and bottom of the cylinder we get :

$P_{atm}+ \rho gh = P_{atm} +\dfrac12 \rho v^2$

$v=\sqrt{2gh}=\sqrt{2\times 20\times 10 }= 20 ms^{-1}$

Two drops of small radius are falling in air with constant velocity

$5 cms^{-1}$ . If they coalesce, then the terminal velocity will be

0%
$10 cms^{-1}$
0%
$2.5 cms^{-1}$
0%
$5 \times \sqrt[3]{4}cms^{-1}$
0%
$5.\sqrt{2}cms^{-1}$

Explanation

Let the radius of smaller drops =r

$r^{3}=V$
radius of larger coslesce drop=

$2^{1/3}r$

$V$ of smaller drops =

$=\dfrac{2g(T-s)r^{2}}{9 \eta}=5$ ----------(1)

$V$ of larger drops=

$=\dfrac{2g(T-s)r^{2}}{9 \eta} 2^{2/3}$ [from(1)]

$=5 \times 2 ^{2/3} cm/s$

$= 5 \times \sqrt[3]{4} cm/s$

There are two holes, each of cross-sectional area

$a$ , on the opposite sides of a wide rectangular tank containing a liquid of density

$\rho$ . When the liquid flows out of the holes, the net force on the tank is :

(

$h$ is the vertical distance between the two holes )

0%
$2a \rho gh$
0%
$4a\rho gh$
0%
$0.5a\rho gh$
0%
$\rho gh$

Explanation

For 1st hole,
Velocity of water from P =

$\sqrt{2gH}=V_{1}$
Q = AV
Force exerted by water on the tank =

$\rho aV_{1}^{2}$

$=2\rho agH$ -----------------(1)
Similarly for 2nd hole,
Velocity=

$\sqrt{2g(H+h)}=V_{2}$
Q =

$aV_{2}^2$
From exerted =

$2\rho ag(H+h)$ ---------------(2)
Net force on the tank

$=2\rho ag(H+h)-2\rho ag(H)$

$=2a\rho gh$

The velocity head of a stream of water is 40 cm. Then, the velocity of flow of water is :

0%
$2.8$ $ms^{-1}$
0%
$5.6$ $ms^{-1}$
0%
$7.2$ $ms^{-1}$
0%
$9.8$ $ms^{-1}$

Explanation

Velocity head of water

$= 40\ cm$

$=\dfrac{V^{2}}{2g}$

$\Rightarrow V^{2} = \dfrac{40}{100} \times 2 \times 10$

$\Rightarrow V^{2}= 8\\ \Rightarrow V= \sqrt{8}\ m/s$

$\Rightarrow V= 2.8\ m/s$

Two equal drops of water are falling through air with a steady velocity of

$10\ cm/s$ . If the drops recombine to form a single drop then the terminal velocity is:

0%
$2^{\frac{2}{3}}$ x $5\ cm/s$
0%
$2^{\frac{2}{3}}$ x $10\ cm/s$
0%
$2^{\frac{2}{3}}$ x $15\ cm/s$
0%
$2^{\frac{2}{3}}$ x $4\ cm/s$

Explanation

From the initial Terminal velocity given,
Let radius of drop =

$r$ .
So,

$V= 10 cm/s= \dfrac{2 g (T-\sigma )r^{2}}{9 \eta}$
So,If two drops combine
As,

$\dfrac{4}{3}\pi r^{3}= V \Rightarrow r =\left ( \dfrac{3V}{4 \pi} \right )^{1/3}$
So

$\dfrac{4}{3} \pi r_{1}^{3}= 2V$

$\Rightarrow r_{1}= 2^{1/3}\left ( \dfrac{3V}{4 \pi} \right )^{1/3}= 2^{1/3}r$
Now,
New Terminal velocity =

$\dfrac{2 2^{2/3} r^{2}(T-\sigma )g}{9 \eta}$

$= 2^{2/3} \times 10 cm/s$

Two hail stones with radii in the ratio of

$1:2$ fall from a great height through the atmosphere. Then the ratio of their momenta after they have attained terminal velocity is

0%
$1:1$
0%
$1:4$
0%
$1:16$
0%
$1:32$

Explanation

As the density of hailstones are same,

Mass of first =

$d \times \dfrac{4}{3} \pi r^{3}= M$

Velocity of first =

$\dfrac{2 r^{2}(T_{h}-T_{a})g}{9 \eta}= V$

For

$2^{nd}$ hail with radius

$= 2r$ ,

Mass of

$2^{nd}$ =

$d \times \dfrac{4}{3} \pi r^{3} 2^{3}= 8M$

Velocit of

$2^{nd}$ =

$\dfrac{2 g r^{2}(T_{h}-T_{a})4}{9 \eta}=4V$

Ratio of Momenta =

$MV : 8M \times 4V$

$MV : 32MV = 1:32$

A vessel is filled with water and kerosene oil. The vessel has a small hole in the bottom. Neglecting viscosity if the thickness of water layer is

$h_{1}$ and kerosene layer is

$h_{2}$ then the velocity

$v$ of flow of water will be (Given : density of water is

$\rho _{1}$ g/ cc and that of kerosene is

$\rho _{2}$ g/cc, neglecting viscosity):

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

Explanation

Berneoulli's Theorm :

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

Pressure at A and B,

$P_A=P_B=P_o$ and velocity at B be

$v$ and at A is zero.

Applying Berneoulli's Theorem at points A and B,

$P_o + \rho_1 g h_1 +\rho_2 g h_2 +0= P_o + 0 +\dfrac{1}{2}\rho_1 v^2$

$\implies \dfrac{1}{2}\rho_1 v^2= \rho_1 g h_1+ \rho_2 g h_2$

$v=\sqrt{\bigg[2g(h_1+h_2\dfrac{\rho_2}{\rho_1})\bigg]}$

A container having a hole at the bottom is free to move on a horizontal
surface. As the liquid comes out, the container moves in a backward direction
with an acceleration

$\alpha$ and finally acquires a velocity v (when all the liquid has drained out). Neglect the mass of the container. The correct option
out of the following is :

0%
Only v depends on h
0%
Only $\alpha$ depends on h
0%
Both v and $\alpha$ depend on h
0%
Neither v nor $\alpha$ depends on h .

Explanation

The velocity through the hole of the body will be

$v=\sqrt{(2\times g\times h)}$ as per the laws of fluid mechanics.

Since the container is initially at rest, its initial velocity=u=0.

Due to momentum conservation, the container will have some final velocity V.

This means that it has accelerated from 0m/s to V m/s.

Therefore, the acceleration must also depend on the height h. Hence, the answer is option c.

A ball of mass m and radius r is released in the viscous liquid. The value of its terminal velocity is proportional to :

0%
$(\dfrac{1}{ r})$ only
0%
$\dfrac{m}{ r}$
0%
$(\dfrac{m}{ r})^{1/2}$
0%
$m$ only.

The area of cross-section of the two arms of a hydraulic press are 1 cm

$^{2}$ and 10 cm

$^{2}$ respectively(fig.). A force of 5 N is applied on the water in the thinner arm. What force should be applied on the water in the thicker arm so that the water may remain in equilibrium?

0%
50 N
0%
60 N
0%
70 N
0%
80N

Explanation

For the water to remain in equilibrium, the pressure at same height i.e.,A and B should be same according to Pascals law,

$\dfrac{5N}{1cm^{2}}=\dfrac{xN}{10cm^{2}}$

$x=50N$

The cross-sectional area of a large tank is 0.5m

$^{2}$ . It has an opening near the bottom having cross-sectional area 1 cm

$^{2}$ . A load of 20 kg is applied on the water at the top. The velocity of water coming out of the opening, at the time when the height of water level is 50 cm above the bottom,is nearly (Take g

$=$ 10 ms

$^{-2}$ )

0%
$1.3 ms^{-1}$
0%
$3.3 ms^{-1}$
0%
$5.3 ms^{-1}$
0%
$7.3 ms^{-1}$

Explanation

Applying Bernoullis therom across point (1) and (2),

$P_{0}+\dfrac{20 \times 10}{0.5}+ 1000 \times 10\times 0.5=P_{0}+\dfrac{1}{2}\times 1000 \times V^{2}$
[

$A_{1}V_{1}=a_{1}v_{1}$ from continuity equation]

$\Rightarrow 400 + 5000 = 500 \times V^{2}$

$\Rightarrow \frac{5400}{500}= v^{2}\Rightarrow V^{2}= 10.8 m/s$

$\Rightarrow V \approx 3.3 m/s$

A non-viscous liquid of constant density

$500 kg/m^{3}$ flows in a variable cross-sectional tube. The area of cross-section of the tube at two points P and Q at heights of

$3 m$ and

$6 m$ are

$2\times 10^{-3} m^{3}$ and

$4\times 10^{-3}m^{3}$ respectively. The work done per unit volume by the forces of gravity as the fluid flows from point P to Q, is:

0%
$29.4 J/m^{3}$
0%
$-1.47 \times 10^{4} J/m^{3}$
0%
$-2.94 \times 10^{4} J/m^{3}$
0%
None of these

Explanation

Difference in heigh of P and Q,

$=(6-3)=3m$
work done by gravity per unit volume of the fluid

$work\ done=\rho gh$

$=-3\times g\times 500$

$=-14700J/m^{3}$ =

$-1.47\times 10^{4}J/^{3}$

The cross-sectional area of a large tank is

$0.5m^{2}$ . It has an opening near the bottom having cross-sectional area

$1 cm^{2}$ . A load of

$20 kg$ is applied on the water at the top. The velocity of water coming out of the opening, at the time when the height of water level is

$50 cm$ above the bottom is nearly:(Take

$g = 10 ms^{-2}$ )

0%
$1.15$
0%
$3.15$
0%
$5.15$
0%
$715$

Explanation

Applying Bernoullis theorem across the two points.

$\displaystyle \dfrac{2000}{0.5}+1000\times 10\times \dfrac{50}{100}=\dfrac{1}{2}\times 1000\times v^{2}$

$\Rightarrow 400+5000=500v^{2}$

$\Rightarrow \dfrac{54}{5}=v^{2}\Rightarrow v=3.286m/s$

$\Rightarrow v\approx 3.3m/s$ .....[ Neglecting the velocity of water at the top of beaker ]

Figure shows four containers of olive oil. The pressure at depth h is

0%
Greatest in A
0%
Greatest in D
0%
Least in B and C both
0%
Equal in all the containers

A solid sphere falls with a terminal velocity of 10 m/s in air. If it is allowed to fall in vacuum:

0%
terminal velocity will be more than 10 m/s
0%
terminal velocity will be less than 10 m/s
0%
terminal velocity will be 10 m/s
0%
there will be no terminal velocity

In the arrangement shown above, a block of mass 2700 kg is in equilibrium on applying a force F. The value of force F ? (

$d_{liquid}=0.75g cm^{-3} is$ )

0%
$147 N$
0%
$1755.18 N$
0%
$153 N$
0%
$918 N$

Explanation

Total pressure acting on side of

$A_1=\dfrac {F}{A_1}+h\rho g$

$=\dfrac {F}{6\times 10^{-4}}+(2\times 0.75\times 9.8\times 10^3)pa$
pressure acting on side

$A_2=\dfrac {2700\times 9.8}{90\times 10^{-4}}pa$

According to Pascal's law
Pressure acting on side of

$A_1=$ pressure acting on side

$A_2$

$\dfrac {F}{6\times 10^{-4}}+(2\times 0.75\times 9.8\times 10^3)$

$=\dfrac {2700\times 9.8}{90\times 10^{-4}}$

$\dfrac {F}{6\times 10^{-4}}=294\times 10^4-14.7\times 10^3$

$F=6\times 10^{-4}(292.53\times 10^4)$

$F=1755.18 N$

With the increase in the area of contact of an object, the pressure:
(Note : Thrust remains same)

0%
Increases
0%
Decreases
0%
Is not affected
0%
None of these

Explanation

Since,

$P = \frac{F}{A}$
Thus pressure is inversely proportional to area,

$i.e. \,\,P \propto \frac{1}{A}$

With the increase in the area of contact of an object the pressure decreases because pressure is inversely proportional to area of contact.

A diver releases bubbles of gas from the bottom of a lake. The bubbles increase to 10 times of their original volume when they reach the surface. Assuming that the pressure exerted by a column of water of 5m height is double the atmospheric pressure, the depth of the lake is...

0%
45m
0%
90m
0%
80m
0%
22.5m

The force exerted by water on the base of a tank, of base area

$1.5 m^2$ when filled with water up to a height of 1 m is (Density of water is

$1000 kg m^{-3}$ and

$g=10 ms^{-2})$ :

0%
1500 N
0%
15000 N
0%
3000 N
0%
30000 N

Explanation

$P = \rho g h \\ P = 10^{3} * 10 * 1 \\ P = 10^{4} N/m^2$

$F = PA \\ F = 10^{4} * 1.5 \\ F = 1.5 * 10^{4} = 15000N$

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

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

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

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

Explanation

The four pistons are initially in equilibrium. As additional force

$F$ is applied to each piston, the pressure in the fluid at each point must be increased by

$F.A$ so that each piston retains state of equilibrium.
Thus, the increment in the pressure at each point is

$\Delta P=\dfrac {F}{A}$ (by Pascal's law)
So, the pressure increment at the centre is

$4\dfrac {F}{A}$

The velocity of the liquid coming out of a small hole of a vessel containing two different liquids of densities

$2\rho$ and

$\rho$ as shown in the figure is

0%
$\sqrt {6gh}$
0%
$2\sqrt {gh}$
0%
$2\sqrt {2gh}$
0%
$\sqrt {gh}$

Explanation

Pressure at (1),

$P_1=P_{atm}+\rho g(2h)+2\rho gh$
Applying Bernoulli's theorem between points (1) and (2),

$(P_{atm}+2\rho gh)+(2\rho) gh=P_{atm}+\dfrac {1}{2}(2\rho )v^2$

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

A piston of cross-sectional area

$100 cm^2$ is used in a hydraulic press to exert a force of

$10^7$ dyne on the water. The cross-sectional area of the other piston which supports an object having a mass of 2000 kg is :

0%
$100 cm^2$
0%
$10^9 cm^2$
0%
$2\times 10^4 cm^2$
0%
$2\times 10^{10} cm^2$

Explanation

As we know , 1 dyne =

$10 ^{-5}$ N so

$10 ^{7}$ dyne = 100 N

hence , for Hydraulic press we have

$\dfrac{100N}{100 cm^{2}}$ =

$\dfrac{2000g}{A}$

$\Rightarrow$ Area =

$2\times 10 ^{4} cm^{2}$

So option C is correct.

A uniformly tapering vessel shown in figure, is filled with liquid of density

$900 kg/m^3$ . The force that acts on the base of the vessel due to liquid is :

0%
3.6 N
0%
7.2 N
0%
9.0 N
0%
12.6 N

Explanation

This cylinder so formed has volume exactly half the total volume,

$\therefore$ force due to liquid

$= 2\times A \times 8gh$

$= 2\times 10^{-3} \times 9.8\times 900\times 0.4$

$= 7.056\ N = 7.2\ N$ (Taking

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

A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. The pressure exerted by the box will be:

0%
maximum when length and breadth form the base
0%
maximum when breadth and width form the base
0%
maximum when width and length form the base
0%
the same in all the above three cases

Explanation

The force exerted by the box on base=Weight of object=

$W$

The pressure that it exerts on the base

$=\dfrac{W}{A}$

where

$A$ is the area of the base.

Hence the pressure is maximum for smallest area of the base, which is when formed by breadth and width.

Equal volumes of two immiscible liquids of densities

$\rho$ and

$2\rho$ are filled in a vessel as shown in figure. Two small holes are punched at depth h/2 and 3h/2 from the surface of lighter liquid. If

$V_{1}$ and

$V_{2}$ are the velocities of a flux at these two holes, then

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

0%
$\displaystyle \frac{1}{2\sqrt2}$
0%
$\displaystyle \frac{1}{2}$
0%
$\displaystyle \frac{1}{4}$
0%
$\displaystyle \frac{1}{\sqrt2}$

Explanation

Here, two small holes are punches at depth

$\dfrac{h}{2}$ and

$\dfrac{3h}{2}$ from the surface of lighter liquid. Hence, the height of first hole from the surface of lighter liquid is

$h_1 = \dfrac{h}{2}$
and the height of second hole from the first hole is(i.e. from surface of remaining liquid)

$h_2 = \dfrac{3h}{2} - \frac{h}{2} = h$
Now, we have the velocity of flux from first hole is

$V_1 = \sqrt 2h_1g$

$V_1 = \sqrt 2(\dfrac{h}{2})g$

$V_1 = \sqrt hg$
Also, the velocity of flux from second hole is

$V_2 = \sqrt 2h_2g$

$V_2 = \sqrt 2hg$
Therefore,

$\dfrac{V_1}{V_2} = \dfrac{\sqrt hg}{\sqrt 2hg}$

$\dfrac{V_1}{V_2} = \dfrac{1}{\sqrt 2}$

Equal volumes of two immiscible liquids of densities

$\rho$ and

$2\rho$ are filled in a vessel as shown in figure. Two small holes are punched at depth

$\frac{h}{2}$ and

$\dfrac{3h}{2}$ from the surface of lighter liquid. If

$v_1$ and

$v_2$ are the velocities of a flux at these two holes, then

$v_1/v_2$ is:

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

Explanation

Here, two small holes are punchesd at depth

$\frac{h}{2}$ and

$\frac{3h}{2}$ from the surface of lighter liquid. Hence, the height of first hole from the surface of lighter liquid is

$h_1 = \dfrac{h}{2}$
and the height of second hole from the first hole is(i.e. from surface of remaining liquid)

$h_2 = \dfrac{3h}{2} - \dfrac{h}{2} = h$
Now, we have the velocity of flux from first hole is

$v_1 = \sqrt 2h_1g$

$v_1 = \sqrt 2(\frac{h}{2})g$

$v_1 = \sqrt hg$
Also, the velocity of flux from second hole is

$v_2 = \sqrt 2h_2g$

$v_2 = \sqrt 2hg$
Therefore,

$\dfrac{v_1}{v_2} = \dfrac{\sqrt hg}{\sqrt 2hg}$

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

The area of cross-section of the wider tube shown in figure is

$800 cm^2$ . If a mass of 12 kg is placed on the massless piston, the difference in heights h in the level of water in the two tubes is :

0%
10 cm
0%
6 cm
0%
15 cm
0%
2 cm

Explanation

Given,

$m=12kg=12000g$

$g=980g/cm^2$

$A=800cm^2$

$\rho =1g/cm^3$

Pressure,

$P=\rho g h=\dfrac{mg}{A}$

$1\times 980\times h=\dfrac{12\times 1000\times 980}{800}$

$h=\dfrac{12\times 10}{8}=15cm$

The correct option is C.

1447124_119859_ans_fbbb1323bece485492478f45bd28407e.png

A Cylindrical vessel of cross-sectional area

$1000cm^{2}$ , is fitted with a frictionless piston of mass 10 kg, and filled with water completely. A small hole of cross-sectional area

$10mm^{2}$ is opened at a point 50 cm deep from the lower surface of the piston. The velocity of effluent from the hole will be

0%
10.5 m/s
0%
3.4 m/s
0%
0.8 m/s
0%
0.2 m/s

Explanation

Given that

$A = 1000 cm^3$

$m = 10 Kg =10000 g$

$A_h = 10 mm^2 = 0.1 m^2$

$(H-h) = 50 cm$
Here, the force on piston is

$F = mg$
Hence, Increase in pressure on the liquid in the wider tube is

$P = \frac{F}{A} = \frac{mg}{A}$
Let, H is the level of water to which piston will move, thus

$P = H\rho g$

$H = \frac{P}{\rho g}$

$H = \frac{mg}{A\rho g} = \frac{m}{A\rho}$

$H = \frac{10000}{1000}$ .......(

$\because \rho = 1$ )

$\therefore H = 10 cm$ from surface of tube.
Since, (H-h) = 50cm, h = 60cm = 0.6 m
The velocity of effluent from the hole is

$v = \sqrt h\rho g$

$v = (\sqrt 2gh)$

$v = \sqrt (2)(9.8)(0.6) = 3.4 ms^{-1}$

Some liquid is filled in a cylindrical vessel of radius

$R$ . Let

${ F }_{ 1 }$ be the force applied by the liquid on the bottom of the cylinder. Now the same liquid is poured into a vessel of uniform square cross-section of side

$R$ . Let

${ F }_{ 2 }$ be the force applied by the liquid on the bottom of this new vessel. (Neglect atmosphere pressure). Then

0%
${ F }_{ 1 }=\pi { F }_{ 2 }$
0%
${ F }_{ 1 }=\cfrac { { F }_{ 2 } }{ \pi }$
0%
${ F }_{ 1 }=\sqrt { \pi { F }_{ 2 }}$
0%
${ F }_{ 1 }={ F }_{ 2 }$

Explanation

Force applied on the base is equal to the weight of the liquid, since in both cases same liquid has been poured in the container, it will exert same force on the base of the container.

$F_1 = F_2$

The area of two holes A and B are

$2a$ and

$a$ respectively. The holes are at height

$\dfrac{H}{3}$ and

$\dfrac{2H}{3}$ from the surface of water. Find the correct option(s)

0%
the velocity of efflux at hole B is 2 times the velocity of efflux at hole A
0%
the velocity of efflux at hole B is $\sqrt{2}$ time the velocity of efflux at hole A
0%
the discharge is same through both the holes
0%
the discharge through hole A is $\sqrt{2}$ time the discharge through hole B

Explanation

$V_A=\left (2g\dfrac {H}{3}\right )^{\dfrac {1}{2}}$

$V_B=\left (\dfrac {4gH}{3}\right )=\sqrt 2\left (\dfrac {2gH}{3}\right )^{\frac {1}{2}}=\sqrt 2V_A$

$Q_A=V_A(2a)$
and

$Q_B=V_B(a)=V_A\sqrt 2(a)\Rightarrow \dfrac {Q_A}{Q_B}=\sqrt 2$

Match column I with column II :

List I	List II
Magnus energy	Pascal's Law
Loss of Energy	Archimedes' principle
Pressure is same at same level in a liquid	Viscous force
Hydraulic Machines	Lifting of asbestos roofs

0%
B $\rightarrow$ 4, C $\rightarrow$ 2, D $\rightarrow$ 1, A $\rightarrow$ 1
0%
A $\rightarrow$ 4, B $\rightarrow$ 3, C $\rightarrow$ 1, D $\rightarrow$ 1
0%
C $\rightarrow$ 4, A $\rightarrow$ 2, B $\rightarrow$ 1, D $\rightarrow$ 1
0%
D $\rightarrow$ 4, A $\rightarrow$ 2, B $\rightarrow$ 1, C $\rightarrow$ 1

Explanation

Magnus effect talks of difference in pressure caused due to difference in velocity to lift a structure. Lifting of asbestos roofs during cyclonic storms in based on this phenomenon. So,

$A\rightarrow 4$
Viscous force is wet friction and brings loss in internal energy. So,

$B\rightarrow 3$
At the same level in the same liquid in a container, the pressure is the same according to Pascal's law. So,

$C\rightarrow 1$
Hydraulic machines use the idea of Pascal's law to transfer equal pressure in all directions from the free surface where it is applied. So,

$D\rightarrow 1$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 3 - 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 3 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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