$$\displaystyle \frac{1}{2} \sqrt{2gh}$$

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

Both Assertion and Reason are correct and Reason is 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 5

The area of pistons in a hydraulic machine are

$5 c{m}^{2}$ and

$625 c{m}^{2}$ . The force on the smaller piston so that a load of

$1250 N$ on the larger piston can be supported, is X N. Find

$\dfrac{X}{2}$ N.

0%
5
0%
10
0%
1250
0%
None of the above

Explanation

We know that,

$Pressure = \dfrac{Force}{Area}$

For Larger Piston,

$P_{l} = \dfrac{1250}{625} = 2N cm^{-2}$
So, Pressure on the smaller piston should be equal to

$2Ncm^{-2}$

So, Force on smaller piston

$X= Pressure \times Area= 2 \times 5 = 10N$

$\therefore \dfrac{X}{2}=\dfrac{10}{2}=5 N$

Two vessels

$A$ and

$B$ of cross-sections as shown in figure contain a liquid up to the same height. As the temperature rises, the liquid pressure at the bottom (neglecting expansion of the vessels) will:

0%
increase in $A$ , decrease in $B$
0%
increase in $B$ , decrease in $A$
0%
increase in both $A$ and $B$
0%
decrease in both $A$ and $B$

A tank has a small hole at its bottom of area of cross-section a. Liquid is being poured in the tank at the rate

$Vm^3/s$ , the maximum level of liquid in the container will be(Area of tank A):

0%
$\dfrac {V}{gaA}$
0%
$\dfrac {V^2}{2ga^2}$
0%
$\dfrac {V^2}{gAa}$
0%
$\dfrac {V^2}{2gaA}$

Explanation

At maximum level the discharge due to small hole will be equal pouring rate

$V=va$ , let the maximum height be H , assuming a<<A

$V=a\sqrt{2gH}$

$H=\dfrac{V^2}{2ga^2}$

Figure shows a capillary rise

$h$ . If air is blown through the horizontal tube in the direction as shown then rise in capillary tube will be

0%
$= h$
0%
$> h$
0%
$< h$
0%
$zero$

Explanation

If air is blown then by using Bernoulli equation we can conclude as the air is blown the pressure over capillary will reduce

and

$P_0=P+\rho gH$ , where P is the pressure in horizontal tube, now as air is blown , P will decrease H will increase because

$P_o$ is constant (atmospheric pressure)

In a hydraulic machine, a force of

$2 N$ is applied on the piston of area of cross section

$10 c{m}^{2}$ . What force is obtained on its piston of area of cross section

$100 c{m}^{2}$ ?

0%
20 N
0%
10 N
0%
5 N
0%
40 N

Explanation

We know that,

$Pressure = \dfrac{Force}{Area}$

$\Rightarrow Pressure = \dfrac{2}{10} = 0.2Ncm^{-2}$
Now,

$Pressure = 0.2$
Again,

$Force = Pressure \times Area$

$Force = 0.2 \times 100 = 20 N$

Two cylindrical vessels fitted with pistons A and B of area of cross section

$8 c{m}^{2}$ and

$320 c{m}^{2}$ respectively, are joined at their bottom by a tube and they are completely filled with water. Find : (i) the pressure on piston A, (ii) the pressure on piston B, and (iii) the thrust on piston B.(consider effort on piston A

$E=4kgf$ )

0%
(i) $5$ $kgf$ $c{m}^{-2}$

(ii) $5$ $kgf$ $c{m}^{-2}$

(iii) $160$ $kgf$
0%
(i) $5$ $kgf$ $c{m}^{-2}$

(ii) $5$ $kgf$ $c{m}^{-2}$

(iii) $16$ $kgf$
0%
(i) $10$ $kgf$ $c{m}^{-2}$

(ii) $10$ $kgf$ $c{m}^{-2}$

(iii) $160$ $kgf$
0%
(i) $0.5$ $kgf$ $c{m}^{-2}$

(ii) $0.5$ $kgf$ $c{m}^{-2}$

(iii) $160$ $kgf$

Explanation

Let

$E$ be the effort applied on the smaller piston

$A$ and

$L$ be the load required on the larger piston

$B$ .
Area of cross section of pistion

$A = 8 cm^{2}$
Area of cross section of pistion

$A = 320 cm^{2}$
(i) Pressure acting on piston

$A$ in the downward direction

$= Thrust/Area = E/8 = 4/8 = 0.5$

$\therefore$ Pressure acting on water

$= E/8 =0.5$ (according to Pascal's Law)
(ii)

$\therefore$ Pressure acting on piston

$B = E/8 =0.5$ (according to Pascal's Law)
(iii) Thrust acting on piston B in the upward direction

$= Pressure \ \ \times Area \ \ of B$

$= \dfrac{E \times 320}{8} = 160$

The diameter of neck and bottom of a bottle are

$2 cm$ and

$10 cm$ respectively. The bottle is completely filled with oil. If the cork in the neck is pressed in with a force of

$1.2 kgf$ , the amount of force applied on the bottom is

0%
F= $30$ $kgf$
0%
F= $15$ $kgf$
0%
F = $1.2 kgf$
0%
F= $60$ $kgf$

Explanation

Radius of the cork at the neck is

$1 cm$ .

Area of the cork,

$A_{c}= \pi \times 1^2 cm^2$

So, pressure applied through the cork,

$P_{c}=\dfrac{1.2}{\pi \times 1^2} kgf.{cm}^{-2}$

Same pressure is applied on the bottom of the bottle as per Pascal's law.

Radius of the bottom of bottle is

$5 cm$ .

Area of the bottom,

$A_{b}= \pi \times 5^2 cm^2$

So, force applied at the bottom =

$P_{c} \times A_{b}=\dfrac{1.2}{\pi \times 1^2} \times \pi \times 5^2=1.2 \times 25=30 kgf$

A water tank of height

$10m$ , completely filled with water is placed on a level ground. It has two holes one at

$3 m$ and the other at

$7 m$ from its base. The water ejecting from:

0%
both the holes will fall at the same spot
0%
upper hole will fall farther than that from the lower hole
0%
upper hole will fall closer than that from the lower hole
0%
more information is required

Explanation

Let

$h$ be the height of the hole.

From Bernoulli's Theorem,

$v=\sqrt{2g(10-h)}$

This is the velocity of water from a hole at height

$h$ from bottom.

Thus the range of the water from the hole=

$vt=\sqrt{2g(10-h)}\sqrt{\dfrac{2h}{g}}=2\sqrt{h(10-h)}$

Thus it is same for

$h=3m$ and

$h=7m.$

In a surface tension experiment with a capillary tube, water rises upto 0.1 m. If the same experiment is repeated in an artificial satellite revolving around the earth, then the water will rise in the capillary upto a height of

0%
0.1 m
0%
9.8 m
0%
0.98 m
0%
full length of capillary tube

A tank is filled up to a height

$2\ H$ 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%
$1.5$
0%
$2\ H$

Explanation

Pressure at a height

$x=\rho g(3H-x)$

Using Bernoulli's Equation inside and outside the hole,

$\dfrac{1}{2}\rho v^2=\rho g(3H-x)$

$\implies v=\sqrt{2(3H-x)g}$

Range of the liquid=

$vt=v\sqrt{\dfrac{2x}{g}}$

$=2\sqrt{x(3H-x)}$

This is maximum for

$x=1.5H$

The correct sketch is

0%
0%
0%
0%
None of these

Explanation

$x_{1}=2\sqrt{a\times 3a}=2\sqrt{3a}$

$x_{2}=2\sqrt{2a\times 2a}=2\sqrt{4a}$

$x_{1}=2\sqrt{3a}$

$x_{1}=x_{3}< x_{2}$

Speed of efflux is

0%
$\sqrt{3gh}$
0%
$\sqrt{2gh}$
0%
$\sqrt{gh}$
0%
$\displaystyle \frac{1}{2} \sqrt{2gh}$

Explanation

Applying Bernoulli's Theorem inside and outside the hole,

$P_{inside}+\dfrac{1}{2}\rho (0)^2+\rho gH=P_{outside}+\dfrac{1}{2}\rho v ^2+\rho gH$

Thus

$P_0+\rho gh=P_0+\dfrac{1}{2}\rho v^2$

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

A large open tank has two holes in the wall. One is a square hole of side

$L$ at a depth h from the top and the other is a circular hole of radius

$R$ at a depth

$4 h$ from the top, When the tank is completely filled with water, quantities of water flowing out per second from both holes are the same. Then

$R$ is equal to

0%
$\displaystyle \frac{L}{\sqrt{2 \pi}}$
0%
$2/\pi L$
0%
$L$
0%
$\displaystyle \frac{L}{2 \pi}$

Explanation

Since the water quantity from both the holes are equal, we have

$a_1v_1=a_2v_2$

$\Rightarrow L^2\sqrt{2gy}=\pi R^2\sqrt{2g(4y)} \because$ velocity of efflux at depth h is given by

$v =\sqrt{2gh}$

$\Rightarrow R=\frac{L}{\sqrt{2\pi}}$

Three points

$A, B$ and

$C$ on a steady flow of a non viscous and incompressible fluid are observed. The pressure, velocity and height of the points

$A, B$ and

$C$ are

$(2,3, 1). (1, 2, 2)$ and

$(4, 1, 2)$ respectively. Density of the fluid is

$1 kgm^{-3}$ and all otner parameters are given in SI units. Then which of the following is correct?

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

0%
points A and B lie on the same stream line
0%
point B and C lie on same stream line
0%
point C and A lie on same stream line
0%
None of the above

Explanation

Given :

$\rho = 1 kg/m^3$

Using Bernoulli's theorem,

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

For point A :

$2 + 1 (10) (1) + \dfrac{1}{2} \times (1) (3^2) =16.5$

For point B :

$1 + 1 (10) (2) + \dfrac{1}{2} \times (1) (2^2) =23$

For point C :

$4 + 1 (10) (2) + \dfrac{1}{2} \times (1) (1^2) = 24.5$

As the value of the constants are different in each case, thus all these points do not lie on same stream line.

In two figures

0%
$v_{1}/v_{2}= 1/2$
0%
$t_{1}/t_{2}=2/1$
0%
$R_{1}/R_{2}=1$
0%
$v_{1}/v_{2}=1/4$

Explanation

Velocity of efflux

$v =\sqrt{2gH}$

For figure 1,

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

For figure 2,

$v_2 = \sqrt{2g(4h)} = 2v_1$

$\implies$

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

A leakage begins in water tank at position

$P$ as shown in the figure. The initial gauge pressure pressure above that of the atmosphere) at

$P$ was

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

$1000 kg/m^{3}$ the initial velocity with which water gushes out is:

0%
$3.2 ms^{-1}$
0%
$32 ms^{-1}$
0%
$28 ms^{-1}$
0%
$2.8 ms^{-1}$

Explanation

$\displaystyle \Delta p=\dfrac{1}{2}\rho v^{2}$

$\therefore$

$\displaystyle v=\sqrt{\dfrac{2\Delta P}{\rho }}=\sqrt{\dfrac{2\times 5\times 10^{5}}{1000}}=31.5 m/s$

Distance

$x_{3}$ is given by

0%
$\sqrt{3} a$
0%
$\sqrt{2} a$
0%
$\displaystyle \frac{1}{2}\sqrt{3} a$
0%
$\displaystyle 2\sqrt{3} a$

Explanation

$x_{3}=2\sqrt{h_{Top}\times h_{bottom}}$

$=2\sqrt{3a\times a}$

$=2a\sqrt{3}$

Two holes

$1$ and

$2$ are made at depths

$h$ and

$16 h$ respectively. Both the holes are circular but radius of hole-

$1$ is two times

0%
Initially equal volumes of liquid will flow from both the holes in unit time
0%
Initially more volume of liquid will flow from hole-2 per unit time
0%
After some time more volume of liquid will flow from hole-1 per unit time
0%
After some time more volume of liquid will flow from hole-2 per unit time

Explanation

Initially

$\displaystyle \frac{dV_{1}}{dt}=v_{1}a_{1}=\left ( \sqrt{2gh} \right )\left ( \pi \right )\left ( 2R \right )^{2}$ (i)

$\displaystyle \frac{dV_{2}}{dt}=v_{2}a_{2}=\left ( \sqrt{2g\left ( 16h \right )} \right )\left ( \pi \right )\left ( R \right )^{2}$ (ii)
From Eqs. (i) and (ii) we can see that,

$\displaystyle \frac{dV_{1}}{dt}=\frac{dV_{2}}{dt}$
After some time

$v_{1}$ and

$v_{2}$ both will decrease, but decrease in the value of

$v_{1}$ is more dominating. So,

$a_{1}v_{1}$ or

$\displaystyle \frac{dV_{1}}{dt}< a_{2}v_{2}$ or

$\displaystyle \frac{dV_{2}}{dt}$

A viscous liquid flows through a horizontal pipe of varying cross-sectional area, Identify the option which correctly represent,, the variation of heighl of rise of liquid in each vertical tube:

0%
0%
0%
0%
None of these

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

Velocity & Range of the distance are directly proportional to the pressure above the liquid and it is pure mechanics and given by

$\boxed { v=\sqrt { 2gh } } \& \boxed { R=2\sqrt { (H-h)h. } }$

Eight spherical rain drops of the same mass and radius are falling down with a terminal speed of

$6cms^{-1}$ . If they coalesce to form one big drop, what will be its terminal speed? Neglect the buoyancy due to air:

0%
$1.5cms^{-1}$
0%
$6cms^{-1}$
0%
$24cms^{-1}$
0%
$32cms^{-1}$

Explanation

When eight equal drops coalesces, volume of water remains same and a bigger drop is formed. Lets say

$r$ is radius of drop before these combine and

$R$ is radius after these combine.
We have

$8 \times \frac{4}{3} \pi r^3 =\frac{4}{3} \pi R^3 \\\Rightarrow R=2r$
and we know that terminal velocity

$V_T \propto$ radius

$^2$ , so we have

$\frac{V_T'}{V_T}=\frac{R^2}{r^2} \\\Rightarrow V_T' =\left (\frac{R}{r} \right)^2 V_T = \left (2 \right)^2 \times 6 cms^{-1}=24 cms^{-1}$

A steel ball of mass

$m$ falls in a viscous liquid with terminal velocity

$v$ , then the steel ball of mass

$8\ m$ will fall in the same liquid with terminal velocity:

0%
$v$
0%
$4\ v$
0%
$8\ v$
0%
$16\sqrt{2}\ v$

Explanation

The terminal velocity $V$ of the body of radius $r$ , density $\rho$ falling through a medium of density $\rho_o$ is given by
$V=\displaystyle\frac{2r^2(\rho-\rho_o)g}{9 \eta}$ where $\eta$ is the coefficient of viscosity of medium.
clearly $V \propto r^2$
$\Rightarrow V \propto (volume)^{2/3}$
$\Rightarrow V \propto (m)^{2/3}$
$\Rightarrow \dfrac{V_2}{V_1} =\left (\frac{m_2}{m_1} \right)^{2/3}$
$\Rightarrow \dfrac{V_2}{V} =\left (\frac{8m}{m} \right)^{2/3}$
$\Rightarrow V_2=4V$

Uniform speed of

$2 cm$ diameter ball is

$20 cm/s$ in a viscous liquid. Then, the speed of

$1 cm$ diameter ball in the same liquid is:

0%
$5cms^{-1}$
0%
$10cms^{-1}$
0%
$40cms^{-1}$
0%
$80cms^{-1}$

Explanation

The terminal velocity

$V$ of the body of radius

$r$ , density

$\rho$ falling through a medium of density

$\rho_o$ is given by

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

$\eta$ is the coefficient of viscosity of medium.

clearly

$V \propto r^2$ , So we have

$\dfrac{V_2}{V_1}=\left(\dfrac{r_2}{r_1}\right)^2$

${V_2}=\left(\dfrac{1}{2}\right)^2 V_1=\dfrac{20}{4}m/s=5m/s$

There is a small hole at the bottom of tank filled with water. If total pressure at the bottom is 3 atm (1 atm

$= 10^{5} Nm^{-2})$ , then velocity of water flowing from hole is

0%
$\sqrt{400}ms^{-1}$
0%
$\sqrt{600}ms^{-1}$
0%
$\sqrt{60}ms^{-1}$
0%
None of these

Explanation

$\displaystyle \Delta p=\frac{1}{2}\rho v^{2}$

$\therefore$

$\displaystyle v=\sqrt{\frac{2\Delta p}{\rho }}$

$\displaystyle =\sqrt{\frac{2\left ( 3\: atm-1\: atm \right )}{\rho }}$

$\displaystyle =\sqrt{\frac{2\times 2\times 10^{5}}{10^{3}}}=\sqrt{400}$ m/s

A horizontal pipeline carries water in a streamline flow. At a point along the tube where the cross sectional area is

$10^{-2}m^{2}$ , the water velocity is 2m/s and the pressure is 8000 Pa. The pressure of water at another point wher cross sectional area is

$0.5 \times 10^{-2} m^{2}$ is

0%
4000Pa
0%
1000Pa
0%
2000Pa
0%
3000Pa

Explanation

Area at other point is half. So speed will be double.
Now,

$\displaystyle p_{1}+\frac{1}{2}\rho v_{1}^{2}=p_{2}+\frac{1}{2}\rho v_{2}^{2}$

$\therefore$

$\displaystyle p_{2}=p_{1}+\frac{1}{2}\rho \left ( v_{1}^{2}-v_{2}^{2} \right )$

$\displaystyle =\left ( 8000 \right )+\frac{1}{2}\times 1000\left ( 4-16 \right )$

$=2000$ Pa

For the determination of the coefficient of viscosity of a given liquid, a graph between square of the radius of the spherical steel balls and their terminal velocity is plotted. The slope of the graph is given by

0%
$\cfrac { { r }^{ 2 } }{ v }$
0%
$\cfrac { { v }^{ 2 } }{ r }$
0%
$\cfrac{v}{r}$
0%
$\cfrac{v}{{r}^{2}}$

Explanation

According to Stoke's law, terminal velocity

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

$\implies$

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

Thus the

$v$ Vs

$r^2$ graph is a straight line having slope

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

A cylindrical vessel is filled with water up to height

$H$ . A hole is bored in the wall at a depth

$h$ from the free surface of water. For maximum range,

$h$ is equal to :-

0%
$\displaystyle \dfrac{H}{4}$
0%
$\displaystyle \dfrac{H}{2}$
0%
$\displaystyle \dfrac{3H}{4}$
0%
$H$

Explanation

The cylinder is filled upto a height of

$H$ and a hole is drilled at a depth h from the surface of water.

This means that the velocity of water flowing out of the hole is

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

The time taken for the stream of water to reach ground,

$h = 0 + 0.5 g t^2$

This gives,

$t = \sqrt {h/5}$

The range of this stream would be,

$x = ut$

$x = \sqrt {2g h(H-h)/5}$

For maximum

$x$ ,

$dx/dh = 0$ , which gives

$h = H/2$

(a) A force of

$200\;N$ acts on an area of

$0.02\;m^2$ . Find the pressure in Pascal.
(b) What force will exert a pressure of

$50,000\ Pa$ on an area of

$0.5\;m^2$ ?
(c) Find out the area of a body which experiences a pressure of

$500\ Pa$ by a force of

$100\;N$ .

0%
(a) $\;10000\;Pa$ , (b) $\;25000\;N$ , (c) $\;0.2\;m^2$
0%
(a) $\;1000\;Pa$ , (b) $\;2500\;N$ , (c) $\;2\;m^2$
0%
(a) $\;100000\;Pa$ , (b) $\;250000\;N$ , (c) $\;0.2\;m^2$
0%
(a) $\;100\;Pa$ , (b) $\;250\;N$ , (c) $\;0.2\;m^2$

Explanation

We know

$Pressure(P)=\dfrac{Force(F)}{Area(A)}$

(a)

$F=200N$

$A=0.02m^2$

$P=\dfrac{200}{0.02}=10000Pa$

(b)

$P=50000Pa$

$A=0.5m^2$

$F=PA=50000\times 0.5=25000N$

(c)

$P=500Pa$

$F=100N$

$A=\dfrac{F}{P}=\dfrac{100}{500}=0.2m^2$

The diagram (fig.) shows a venturimeter, through which water is flowing. The speed of water at a is 2 cm/sec. The speed of water at Y (taking

$g=1000\:cm/sec^{2}$ ) is :-

0%
$23\:cm/sec$
0%
$32\:cm/sec$
0%
$101\:cm/sec$
0%
$1024\:cm/sec$

Explanation

The height difference between the points x and y is 5.1 mm

Therefore, the pressure difference between the 2 points will be,

$P = \rho g h$

$P = 1 \times 1000 \times 0.51$

P = 510

Applying Bernoulli's theorem,

$\delta P + 0.5 \rho \delta v^2 = 0$

$\delta P = 510$

$510 = 0.5 \times 1 \times (v_2^2 - 2^2)$

Hence,

$v_2^2 = 1024 \ or \ v_2 = 32 cm/sec$

The small piston of a hydraulic lift has an area of

$0.20m^2$ . A car weighing

$1.20\times 10^4N$ sits on a rack mounted on the large piston. The large piston has an area of

$0.90m^2$ . How large a force must be applied to the small piston to support the car?

0%
$2.67\times 10^3N$
0%
$2.67\times 10^4N$
0%
$2.67\times 10^5N$
0%
$2.67\times 10^6N$

Explanation

Hydraulic lift works on the principle of pascal's law

$P=\frac { { F }_{ 1 } }{ a_{ 1 } } =\frac { { F }_{ 2 } }{ { a }_{ 2 } }$
applying the given formula to the question.

$\frac { 1.2\times { 10 }^{ 4 } }{ 0.90 } =\frac { F }{ 0.2 } \\ F=\frac { 0.24\times { 10 }^{ 4 } }{ 0.90 } =2.67\times { 10 }^{ 3 }\quad N$

Water rises in a capillary upto a height h. If now this capillary is tilted by an angle of

$45^{\circ}$ , then the length of the water column in the capillary becomes

0%
2h
0%
$\displaystyle \frac{h}{2}$
0%
$\displaystyle \frac{h}{\sqrt{2}}$
0%
$h\sqrt{2}$

Explanation

The expression for the capillary rise in a tube is given by,

H =

$\dfrac {2T cos \theta}{\rho g r}$

for all other parameters kept constant, if I change the angle of inclination to

$45^o$

$cos \theta$ will go from 1 to

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

Therefore, the height of capillary tube rise will also change from

$H to \dfrac{H}{\sqrt2}$

The manometer shown below is used to measure the difference between water level of the two tanks. Calculate this difference for the conditions indicated.

0%
4 cm
0%
40 cm
0%
100 cm
0%
12 cm

Explanation

Answer is A.

$P_a\, +\, h_1 pg\, -\, 40p_1g\, +\, 40pg\, =\, P_a\, +\, h_2 pg$

$h_2pg\, -\, h_1 pg\, =\, 40 pg\, -\, 40 p_1g$
As p1 = 0.9p, the eation becomes

$h_2pg\, -\, h_1 pg\, =\, 40 pg\, -\, 36 pg$
Therefore,

$h_2\, -\, h_1 \, =$ 4 cm.
Hence, the difference for the conditions indicated is 4 cm.

In a hydraulic lift, used at a service station, the radius of the large and small piston is 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%
3.75 kg
0%
37.5 kg
0%
7.5 kg
0%
75 kg

Explanation

Using pascal's law

$\frac{F_1}{A_1}=\frac{F_2}{A_2}$

$\frac{1500}{\pi r_1^2}=\frac{W}{\pi r_2^2}=\frac{1500}{20^2}=\frac{W}{1^2}$

$W=3.75kg$

A tank is filled with water up to height

$H$ . Water is allowed to come out of a hole P in one of the walls at a depth

$D$ below the surface of water. The horizontal distance

$x$ in terms of

$H$ and

$D$ will be expressed as :

0%
$x\, =\, \sqrt {D(H - D)}$
0%
$x\, =\, \sqrt {\displaystyle \dfrac {D(H - D)}{2}}$
0%
$x\, =\, 2\sqrt {D(H - D)}$
0%
$x\, =\, 4\sqrt {D(H - D)}$

Explanation

Answer is C.

The velocity of water stream coming out of P =

$\sqrt { 2gD }$ .
The time taken by the water stream to fall through the height H-D =

$\sqrt { \dfrac { 2(H-D) }{ g } }$ .
Therefore,

$x=\sqrt { 2gD } \times \sqrt { \dfrac { 2(H-D) }{ g } } \quad =2\sqrt { D(H-D) }$ .
Hence, the horizontal distance x in terms of H and D is

$2\sqrt { D(H-D) }$ .

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

$cm^2$ and 10

$cm^2$ respectively (figure). A force of 50 N is applied on the water in the thicker arm. What force should be applied on the water in the thinner arm so that the water may remain in equilibrium?

0%
5 N
0%
10 N
0%
25 N
0%
50 N

Explanation

Answer is A.

In equilibrium, the pressures at the two surfaces should be equal as they lie in the same horizontal level. If the atmospheric pressure is P and a force F is applied to maintain the equilibrium, the pressures are

$P_0\, +\, \displaystyle \frac{50N}{10cm^2}\, and\, P_0\, +\, \displaystyle \frac{F}{1 cm^2}$
This givens F = 5 N.
Hence, force that should be applied on the water in the thinner arm so that the water may remain in equilibrium is 5 N.

Two parallel glass plates are dipped partly in a liquid of density

$'d'$ keeping them vertical. If the distance between the plates is

$'x'$ Surface tension for liquid is

$T$ & angle of contact is

$\displaystyle \theta$ then rise of liquid between the plates due to capillary will be

0%
$\displaystyle \dfrac{T\cos \theta }{xd}$
0%
$\displaystyle \dfrac{2T\cos \theta }{xdg}$
0%
$\displaystyle \dfrac{2T}{xdg\cos \theta}$
0%
$\displaystyle \dfrac{T\cos \theta }{xdg}$

Explanation

weight of liquid of height 'h' = (area of tube x h) x g x d= (3.14/4)hdg

${x}^2$

vertical component of surface tension force=(1/2)x(Txcircumference)x cosθ=3.14Txcosθ

therefore, (3.14/4)hgd

${x}^2$ =(1/2)Tx3.14xcosθ

h=(2Tcosθ)/(gdx)

θ
θθs=Tx3.14x

Water flows through a frictionless duct with a crosssection varying as shown in fig. Pressure p at points along the axis is represented by

Water flows in a horizontal tube as shown in figure. The pressure of water changes by 600

$N/m^2$ between x and y where the areas of cross - section are 3

$cm^2$ and 1.5

$cm^2$ respectively. Find the rate of flow of water through the tube.

0%
$189\, cm^3/s$
0%
$159\, cm^3/s$
0%
$189\, cm^2/s$
0%
$159\, cm^2/s$

Explanation

Let the velocity at x =

$v_x$ and that at y =

$v_y$ .
By the equation of continuity,

$\displaystyle \frac{v_y}{v_x}\, =\, \displaystyle \frac{3 cm^2}{1.5 cm^2}\, =\, 2$
By Bernoulli's equation,

$P_x\, +\, \displaystyle \frac{1}{2}\, p\, v_x^2\, =\, P_y\, +\, \displaystyle \frac{1}{2}\, pv_y^2$
or,

$P_x\, -\, O_y\, =\, \displaystyle \frac{1}{2}\, p(2v_y)^2\, -\, \displaystyle \frac{1}{2}\, pv_y^2\, =\, \displaystyle \frac{3}{2}\, pv_y^2$
or,

$600\, \displaystyle \frac{N}{m^2}\, =\, \displaystyle \frac{3}{2}\, \left (1000\, \displaystyle \frac{kg}{m^3} \right )\, v_x^2$
or,

$v_x\, =\, \sqrt{0.4\, m^2/ s^2}\, =\, 0.63\, m/s$
The rate of flow

$=\, (3\, cm^2)\, (0.63\, m/s)\, =\, 189\, cm^3/s$ .

For a fluid which is flowing steadily, the level in the vertical tubes is best represented by

Explanation

The tube is wider earlier and it becomes narrow as it progresses. Pressure in the thin tube will be more compared to that of the thick region of the tube. So, there would be enough pressure due to the thin region, so that liquid will move in that vertical part. But as the liquid is moving forward, its velocity will decrease. So, figure D is incorrect. And height in the second vertical part will be less compared to the height in the first vertical tube.

$Answer:$

Hence, option A is the correct answer.

The height of mercury which exerts the same pressure as 20 cm of water column is :( Given relative density of mercury=13.56)

0%
1.47 cm
0%
14.8 cm
0%
148 cm
0%
None of these

Explanation

Let pressure

$= P,$ density

$= \rho,$ mass

$= m$

$P_{\text{mercury}} = h_m\rho_mg \rightarrow$ Pressure exerted by mercury

$P_{\text{water}} = h_w\rho_wg \rightarrow$ Pressure exerted by water

Since given

$P_{\text{mercury}}$

$= P_{\text{water}}$ at

$h_w = 20 cm$

$\Rightarrow h_m\rho_m = h_w\rho_w$ where

$\rho_m = 13.56 \rho_w$ relative density

$\Rightarrow \dfrac{\rho_w}{\rho_m} = \dfrac{h_m}{h_w}$

$\Rightarrow \dfrac{1}{10.8}= \dfrac{h_m}{20}$

$\Rightarrow h_m = \dfrac{20}{10.8} = 1.47 cm$

Two identical cylindrical vessels with their bases at the same level; contain liquid of density

$\rho$ . The area of both is

$S$ , but the height of liquid in one vessel is

${h}_{1}$ and in other

${h}_{2}$ . The work done when both cylinders are connected, by gravity in equalising levels is:

0%
$\displaystyle\frac { 1 }{ 4 } g\rho S{ \left( { h }_{ 2 }-{ h }_{ 1 } \right) }^{ 2 }$
0%
$g\rho S{ \left( { h }_{ 2 }-{ h }_{ 1 } \right) }^{ 2 }$
0%
$g\rho S\left( { h }_{ 2 }-{ h }_{ 1 } \right)$
0%
$\displaystyle\frac { 1 }{ 4 } g\rho S\left( { h }_{ 2 }-{ h }_{ 1 } \right)$

Explanation

The water level will rise to a height of

$\displaystyle\frac{1}{2}\left({h}_{2} - {h}_{1}\right)$

$\therefore$ Increase in volume

$= \displaystyle\frac{1}{2}\left({h}_{2} - {h}_{1}\right)S$

$\therefore$ Work done

$= \displaystyle\frac{1}{2}\left({h}_{2} - {h}_{1}\right)S \times \rho \times g\displaystyle\frac{1}{2}\left({h}_{2} - {h}_{1}\right)$

$= \displaystyle\frac{1}{4}g \rho S{ \left( { h }_{ 2 }-{ h }_{ 1 } \right) }^{ 2 }$ .

The pressure of gas in a metal cylinder is 4 atmospheres at

$27 C$ , then the pressure at

$54 C$ : (in atmosphere)

0%
$4.36$
0%
$8$
0%
$3$
0%
${400}/{109}$

Explanation

According to the gas equation

$\therefore \displaystyle\frac{{P}_{1}{V}_{1}}{{T}_{1}} = \displaystyle\frac{{P}_{2}{V}_{2}}{{T}_{2}}$
Here

${P}_{1} = 4 atm V = V$ ,
and

${T}_{1} = 27 + 273 = 300 K$

${P}_{2} =$ ?,

${V}_{2} = V$
and

${T}_{2} = 54 + 273 = 327 K$

$\therefore \displaystyle\frac{4 \times V}{300} = \displaystyle\frac{{P}_{2} \times V}{327}$
or,

${P}_{2} = \displaystyle\frac{4 \times 327}{300}$

$= 4.36 atm$ .

A container is filled with water to a height of 10 m. The pressure exerted by the water at the bottom of the container is _____ Pa.

0%
$9.8 * 10^4$
0%
980
0%
$980 * 10^5$
0%
$19.6 * 10^4$

A fixed cylindrical vessel is filled with water up to height

$H$ . A hole is bored in the wall at a depth

$h$ from the free surface of water. For maximum horizontal range

$h$ is equal to :

0%
$H$
0%
$3H/4$
0%
$H/2$
0%
$H/4$

Explanation

Let velocity at hole be

$v$ , density be

$\rho$ . Applying Bernoulli theorem for top-most layer and hole,

$P_{atm}+ \rho gH+ 1/2\rho 0^{2} = P_{atm}+ \rho gh+1/2\rho V^{2}$

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

Time taken to reach ground

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

Horizontal range

$( Horizontal_{range})$

$=V \times t$

$= \sqrt {2gH-2gh} \times \sqrt{2h/g}$

Differentiating

$Horizontal_{range}$ and equating it to zero for maximum

$Horizontal_{range}$ we get maximum

$Horizontal_{range}$ when

$h=H/2.$

The height of mercury which exerts the same pressure as 20 cm of water column, is_____

0%
1.47 cm
0%
14.8 cm
0%
148 cm
0%
none of these

Explanation

$density\times area\times h=constant$

$Swater\times area\times hwater=Smercury\times area\times hmercury$

$1{ 9/cc }\times 20cm\quad =\quad 13.56\quad 9/cc\times h$

$h=\dfrac { 20 }{ 13.56 } cm=1.47cm$

$\boxed { h=1.47cm }$

The pressure exerted by a liquid column at the bottom of the container at a point inside a fluid

0%
does not depend on the area of cross-selection of container.
0%
dependent of the density of the fluid.
0%
equal in all directions.
0%
All the above are true

Water flows through a frictionless duct with a cross section varying as shown in fig. Pressure

$p$ at points along the axis is represented by

Two capillary tubes of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube is filled with water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is

$7.3 \times 10^{-2} N/m$ . Take the angle of contact to be zero and density of water to be

$10^3 kg/m^3(g = 9.8 m/s^2)$

0%
5 mm
0%
10 mm
0%
15 mm
0%
20 mm

Explanation

$\varrho gh=\dfrac { 2T }{ R }$

${ h }_{ 1 }=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times 1.5\times { 10 }^{ 3 } }$

${ h }_{ 2 }=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times 3\times { 10 }^{ -3 } }$

$\triangle h={ h }_{ 1 }-{ h }_{ 2 }$

$=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times { 10 }^{ -3 } } \left( \dfrac { 2 }{ 3 } -\dfrac { 1 }{ 3 } \right)$

$=\dfrac { 2\times 7.3\times { 10 }^{ -2 } }{ { 10 }^{ 3 }\times 9.8\times 3\times { 10 }^{ -3 } }$

$\boxed { \triangle h=5mm }$

A 20 cm long capillary tube is dipped vertically in water and the liquid rises upto 10 cm. If the entire system is kept in a freely falling platform, the length of the water column in the tube will be

0%
5 cm
0%
10 cm
0%
15 cm
0%
20 cm

Explanation

The height raised by liquid in capillary tube

$H=\dfrac{2l\,cos\theta}{\rho gh}$

Where H is rise in the capillary tube.
As in freely falling platform, a body experience weightlessness.
So, the liquid will rise upto to length of the capillary.
i.e height raised by the liquid will be 20 cm.

The work done in splitting a drop of water of

$1 mm$ radius into

$\displaystyle { 10 }^{ 6 }$ droplets is (surface tension of water

$\displaystyle 72\times { 10 }^{ -3 }{ N }/{ M }$ ) ;

0%
$\displaystyle 5.98\times { 10 }^{ -5 }J$
0%
$\displaystyle 10.98\times { 10 }^{ -5 }J$
0%
$\displaystyle 16.95\times { 10 }^{ -5 }J$
0%
$\displaystyle 8.95\times { 10 }^{ -5 }J$

Explanation

Radius of new droplet if be r then,

$\displaystyle { 10 }^{ 6 }\times \frac { 4 }{ 3 } \pi { r }^{ 3 }=\frac { 4 }{ 3 } \pi \times { \left( 0.001 \right) }^{ 3 }$

$\displaystyle { r }^{ 3 }={ 10 }^{ -15 }\Rightarrow r={ 10 }^{ -5 }$
Increase in surface area

$\displaystyle =\left[ 4\pi \times { \left( { 10 }^{ -5 } \right) }^{ 2 }\times { 10 }^{ 6 } \right] -\left[ 4\pi \times { \left( { 10 }^{ -3 } \right) }^{ 2 } \right]$

$\displaystyle =\left[ 4\pi \times { 10 }^{ -4 } \right] -\left[ 4\pi \times { 10 }^{ -6 } \right] =4\pi { 10 }^{ -6 }\left[ 100-1 \right]$

$\displaystyle =4\pi \times { 10 }^{ -6 }\times 99=4\pi \times { 10 }^{ -6 }\times 99$
Work done = surface tension x increase in surface area

$\displaystyle =72\times 4\pi \times 99\times { 10 }^{ -6 }\times { 10 }^{ -3 }=8.95\times { 10 }^{ -5 }J$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Fluids Quiz 5 - 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 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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