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

A square box of water has a small hole located in one of the bottom corners. When the box is full and sitting on a level surface, complete opening of the hole results in a flow of water with a speed

$v_0$ , as shown in fig. When the box is still half empty, it is tilted by

$45^o$ so that the hole is at the lowest point. Now the water will flow out with a speed of:

0%
$v_0$
0%
$v_0\cdot 2$
0%
$\dfrac {v_0}{\sqrt 2}$
0%
$\dfrac {v_0}{\sqrt [4]{2}}$

Explanation

Using Bernoulli's principle we know that the velocity at the bottom is given as

$\sqrt {2gh}$ where h is the height of the water level.

$v_{o} = \sqrt{2gh}$
Now when square is tilted, effective height =

$\displaystyle\frac{h}{\sqrt{2}}$
Thus substituting this as as the height we get the answer as D.

$v_1 = \displaystyle\sqrt{2g\frac{h}{\sqrt{2}}} = \displaystyle\frac{v_{o}}{\sqrt[4]{2}}$

In a cylindrical vessel containing liquid of density

$\rho$ , there are two holes in the side walls at heights of

$h_1$ and

$h_2$ respectively such that the range of efflux at the bottom of the vessel is same. Find the height of a hole, for which the range of efflux would be maximum.

0%
$\dfrac {h_2+h_1}{2}$
0%
$\dfrac {h_2-h_1}{2}$
0%
$\dfrac {h_2h_1}{2}$
0%
$\dfrac {h_2}{h_1}$

Explanation

Hint: The range of efflux is $x=2\sqrt{(H-h)h}$

Correct Answer: Option A

Explanation of correct option:

$\textbf{Step1: Find height of cylinder}$

Two holes are present in the side walls at heights

$h_1$ and

$h_2$ in such a way that the range of efflux at the bottom of the vessel is same,

$\therefore x=2\sqrt{(H-h_1)h_1}=2\sqrt{(H-h_2)h_2}$

$\therefore (H-h_1)h_1=(H-h_2)h_2$

$H(h_1-h_2)=(h_1)^2-(h_2)^2$

$\therefore H=\dfrac{{h_1}^2-{h_2}^2}{h_1-h_2}$

Multiplying and dividing by

$(h_1+h_2)$

$H=\dfrac{(h_1)^2-(h_2)^2}{h_1-h_2} \dfrac{(h_1+h_2)}{(h_1+h_2)}$

Since,

$(h_1+h_2)(h_1-h_2)=(h_1)^2-(h_2)^2$

$\therefore H=h_1+h_2$

$\textbf{Step2: Find height of hole}$

Also, maximum height for which the range of efflux is maximum is half of the heights of the holes,

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

$\therefore h=\dfrac{h_1+h_2}{2}$

A piston of mass

$M= 3 kg$ and radius

$R=4 cm$ has a hole into which a thin pipe of radius

$r= 1 cm$ is inserted. The piston can enter a cylinder tightly and without friction, and initially it is at the bottom of the cylinder.

$750gm$ of water is now poured into the pipe so that the piston and pipe are lifted up as show. Then,

0%
the height $H$ of water in the cylinder is $\cfrac { 2 }{ \pi }m$
0%
the height $H$ of water in the cylinder is $\cfrac { 11 }{ 32\pi } m$
0%
the height $h$ of water in the pipe is $\cfrac { 2 }{ \pi }m$
0%
the height $h$ of water in the pipe is $\cfrac { 11 }{ 32\pi } m$

Explanation

Consider the equilibrium of piston,
Atmospheric pressure and weight should be balanced by pressure due to water on piston.

$\implies$

$(P^{}_{0})(A) + Mg = P(A)$

$P = P^{}_{0}$ +

$\rho gh$

$\implies$

$(P^{}_{0})(A) + Mg$ = (

$P^{}_{0}$ +

$\rho$ gh)(A)

$\implies$ h =

$\dfrac{M}{\rho A}$

$A = \pi (R^2 - r^2)$
=

$\pi (16 - 1)$

$cm^2$

$\implies h = \dfrac{3}{10^3 \times 15 \times \pi \times 10^{-4}}$
Therefore height 'h' of water in the pipe is

$\dfrac{2}{\pi}$ m

Mass of water pored is equal to mass of water in (cylinder+pipe)
Mass of water in cylinder =

$\pi \times R^2 \times H \times \rho$
Mass of water in cylinder =

$\pi \times r^2 \times h \times \rho$

$\implies 750\times 10^{-3}$ =

$\pi \times \rho$

$[4^{2} \times 10^{-4} \times H + 1^{2} \times 10^{-4} \times h]$
substituting

$h$ we get,
height H of water in the cylinder =

$\dfrac{11}{32 \pi}m$

The velocity of efflux is:

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

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 (must outside the hole) in the same horizontal line as shown in the figure in question.

Pressure at A:

$P_A= P_0 + h\rho_2g + (h-y)\rho_1g$

Pressure at A:

$P_B= P_0$

Applying Bernoulli's theorem at points A and B

$P_A + \dfrac{1}{2}\rho_2v_2^2 + (\rho_2 gh + \rho_1 g(h-y))= P_B + \dfrac{1}{2}\rho_1v_1^2 + (\rho_2 gh + \rho_1 g(h-y) )$ .....(1)

Substituting

$P_A$ and

$P_B$ in eqn(1) we get,

$P_0 + h\rho_2g + (h-y)\rho_1g + \dfrac{1}{2}\rho_2v_2^2 + (\rho_2 gh + \rho_1 g(h-y))= P_0 + \dfrac{1}{2}\rho_1v_1^2 + (\rho_2 gh + \rho_1 g(h-y) )$ ....(2)

Given,

$A=0.5\: m^2$

$a =5 cm^2 = 25 \times 10^{-4}\: m^2$

From continuity equation,

$Av_2 = av_1$

$\therefore \dfrac{v_2}{v_1} = \dfrac{a}{A}= \dfrac{25 \times 10^{-4}}{0.5}=50 \times 10^{-4}$

Given

$h= 60\: cm=0.6\: m$

$y=20 \: cm = 0.2 \:m$

$\rho_1= 900\: kg\:m^{-3}$

$\rho_2= 600\: kg\:m^{-3}$

$P_0 + h\rho_2g + (h-y)\rho_1g + \dfrac{1}{2}\rho_2v_2^2 = P_0 + \dfrac{1}{2}\rho_1v_1^2$

As given in the problem, we ignore

$v_2$ since a is very small.

$\therefore P_0 + h\rho_2g + (h-y)\rho_1g = P_0 + \dfrac{1}{2}\rho_1v_1^2$

$\therefore h\rho_2g + (h-y)\rho_1g = \dfrac{1}{2}\rho_1v_1^2$

Substituting the values,

$0.6 \times 600 \times 10 + (0.6-0.2) \times 900 \times 10 = \dfrac{1}{2} \times 900 \times v_1^2$

$\Rightarrow v_1 = 4\: ms^{-1}$

A cylindrical vessel of cross-sectional area

$1000 cm^2$ , is fitted with a frictionless piston of mass

$10 kg$ , and filled with water completely. A small hole of cross-sectional area

$10 mm^2$ is opened at a point

$50 cm$ deep from the lower surface of the piston. The velocity of efflux 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

$P = \dfrac{F}{A} = \dfrac{mg}{A}$
Let,

$H$ is the level of water to which piston will move, thus

$P = H\rho g$

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

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

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

$\because \rho = 1$ )

$\therefore H = 10 cm$ from surface of tube.

Since,

$(H-h) = 50cm, H = 60cm = 0.6 m$
The velocity of effluent from the hole is

$v = \sqrt {h\rho g}$

$v = (\sqrt {2gh})$

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

A non viscous liquid of constant density

$1000 kg/{ m }^{ 3 }$ flows in a streamline motion along a tube of variable cross section. The tube is kept inclined in the vertical plane as shown in the figure. The area of cross section of the tube at two points

$P$ and

$Q$ at heights of

$2 m$ and

$5 m$ are respectively

$4\times { 10 }^{ -3 }{ m }^{ 2 }$ and

$8\times { 10 }^{ -3 }{ m }^{ 3 }$ . The velocity of the liquid at point

$P$ is

$1 m/s$ . Then,

0%
work done by pressure per unit volume is $-29625 J/{ m }^{ 3 }$
0%
work done by pressure per unit volume is $+29625 J/{ m }^{ 3 }$
0%
work done by gravity is $-30000 J/{ m }^{ 3 }$
0%
work done by gravity is $+30000 J/{ m }^{ 3 }$

Explanation

From the equation of continuity at P and Q, we have

$a_av_1 = a_2v_2$

$\therefore v_2 =\dfrac{a_1}{a_2} \times v_1= \dfrac{4 \times 10^{-3}}{8 \times 10^{-3}} \times 1= \dfrac{1}{2}\:ms^{-1}$

From Bernoulli's theorem at points P and Q,

$p_1 + \dfrac{1}{2} \rho v_1^2 + \rho gh_1 = p_2 + \dfrac{1}{2}\rho v_2^2 + \rho gh_2$

Each term in the above eqn represents energy per unit volume.
Work done by the pressure per unit volume is:

$p_1 - p_2 = \dfrac {1}{2}\rho(v_2^2-v_1^2) + \rho g(h_2-h_1)$

$=\dfrac {1}{2}\times 1000( 0.5^2-1^2) + 1000 \times 10( 5-2)$

$=\dfrac {1}{2} \times 1000 \times (-0.75) + 1000 \times 10 \times 3$

$=-0.375 \times 10^3 + 30 \times 10^3$

$=29.625 \times 10^3\: Jm^{-3}$

Work done by gravity:

$W= \rho gh_1- \rho gh_2= -1000 \times 10 \times (5 -2)= -3 \times 10^4\: Jm^{-3}$

Find the height of the water in the long tube above the top when the water stops coming out of the hole.

0%
$-2h_0$
0%
$h_0$
0%
$h_2$
0%
$-h_1$

Explanation

Equating the pressure due air and due to water we get

$2p_0=(h_2+h_0)\rho g+p_0$ (since liquids at the same level have the same pressure)

$p_0=h_2\rho g+h_0\rho g$

$h_2\rho g=p_0-h_0\rho g$

$h_2=\displaystyle\dfrac {p_0}{\rho g}-\dfrac {h_0\rho g}{\rho g}=\dfrac {p_0}{\rho g}-h_0$
KE of the water

$=$ Pressure energy of the water at that layer

$\displaystyle\frac {1}{2}mv^2=m\times \dfrac {P}{\rho}$

$v^2=\displaystyle\dfrac {2P}{\rho}=\dfrac {2}{\rho}[p_0+\rho g(h_1-h_0)]$

$v=\left [\displaystyle\dfrac {2}{\rho}\left \{p_0+\rho g(h_1-h_0)\right \}\right ]^{\dfrac {1}{2}}$

Now we know:

$2P_0+\rho g(h_1-h_0)=p_0+\rho gX$

$\Rightarrow X=\displaystyle\frac {p_0}{\rho g}+(h_1-h_0)=h_2+h_1$
i.e, X is

$h_1$ metre below the top or X is

$-h_1$ above the top.

Find the height

$h_2$ of the water in the long tube above the top initially.

0%
$\dfrac {3p_0}{\rho g}-\dfrac {h_0}{3}$
0%
$\dfrac {2p_0}{\rho g}-\dfrac {h_0}{2}$
0%
$\dfrac {p_0}{\rho g}-h_0$
0%
$\dfrac {p_0}{2\rho g}-2h_0$

Explanation

Equating the pressure due air and due to water we get

$2p_0=(h_2+h_0)\rho g+p_0$ (since liquids at the same level have the same pressure)

$p_0=h_2\rho g+h_0\rho g$

$h_2\rho g=p_0-h_0\rho g$

$h_2=\displaystyle\dfrac {p_0}{\rho g}-\dfrac {h_0\rho g}{\rho g}=\dfrac {p_0}{\rho g}-h_0$

Find the speed with which water comes out of the hole.

0%
$\dfrac {1}{\rho}[p_0-\rho g(h_1-2h_0)]^{\dfrac {1}{2}}$
0%
$\left [\dfrac {2}{\rho}[p_0+\rho g(h_1-h_0)]\right ]^{\dfrac {1}{2}}$
0%
$\left [\dfrac {3}{\rho}[p_0+\rho g(h_1-h_0)]\right ]^{\dfrac {1}{2}}$
0%
$\left [\dfrac {4}{\rho}[p_0+\rho g(h_1-h_0)]\right ]^{\dfrac {1}{2}}$

Explanation

Equating the pressure due air and due to water we get

$2p_0=(h_2+h_0)\rho g+p_0$ (since liquids at the same level have the same pressure)

$p_0=h_2\rho g+h_0\rho g$

$h_2\rho g=p_0-h_0\rho g$

$h_2=\displaystyle\frac {p_0}{\rho g}-\frac {h_0\rho g}{\rho g}=\frac {p_0}{\rho g}-h_0$
KE of the water

$=$ Pressure energy of the water at that layer

$\displaystyle\frac {1}{2}mv^2=m\times \frac {P}{\rho}$

$v^2=\displaystyle\frac {2P}{\rho}=\frac {2}{\rho}[p_0+\rho g(h_1-h_0)]$

$v=\left [\displaystyle\frac {2}{\rho}\left \{p_0+\rho g(h_1-h_0)\right \}\right ]^{\dfrac {1}{2}}$

A tank is filled upto height

$L$ with a liquid and is placed on a platform of height

$h$ from the ground. To get maximum range

${ x }_{ m }$ a small hole is punched at a distance of

$y$ from the free surface of the liquid. Then

0%
${ x }_{ m }=2h$
0%
${ x }_{ m }=1.5h$
0%
$y=h$
0%
$y=0.75h$

Explanation

Velocity of efflux at y

$V_x = \sqrt{2gy}$
At y, vertical component is zero
The distance to ground is

$h = (L - y)$ if L is the distance from ground to top of vessel
Hence, time required to reach ground

$t = \sqrt{2(L-y)/g}$
Horizontal Range is

$V_x\times t$
As we have to maximize range, we differentiate

$V_xt$ with

$y$
We get range maximum for

$y = L/2 =h$
Accordingly, if we put

$y = h$ in range equation, we get range

$=2h$

Figure shows a capillary tube of radius

$r$ dipped into water. If the atmospheric pressure is

$P_0$ , the pressure at point

$A$ ( just below the meniscus ) is

0%
$P_0$
0%
$P_0 + \displaystyle \frac {2S}{r}$
0%
$P_0 - \displaystyle \frac {2S}{r}$
0%
$P_0 - \displaystyle \frac {4S}{r}$

Explanation

The pressure just above the meniscus is atmospheric pressure and is greater than the pressure just below the meniscus since the pressure on the concave side is greater than that on the convex side.
$P_{concave} - P_{convex} = S(\dfrac{1}{R_1} + \dfrac{1}{R_2})$
For a spherical surface $R_1=R_2=r$
Hence, $P_{concave} - P_{convex} =\dfrac{2S}{r}$
In the given scenario, $P_{concave}= P_0$ where $P_0$ is the atmospheric pressure.
$P_{convex} = P_0 -\dfrac{2S}{r}$

A tank, which is open at the top, contains a liquid up to a height

$H$ . A small hole is made in the side of a tank at a distance y below the liquid surface. The liquid emerging from the hole lands at a distance

$x$ from the tank.

0%
If $y$ is increased from zero to $H$ , $x$ will first increase and then decrease.
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x is maximum for $y = H/2$ .
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The maximum value of $x$ is $H$ .
0%
The maximum value of $x$ will depend on the density of the density of the liquid.

Explanation

The velocity of efflux

$=v=\sqrt{2gy}$
The emerging liquid moves as a projectile and reaches the ground in time

$t$ , So we have

$H-y=\dfrac { 1 }{ 2 } gt^{ 2 }\quad \\ \Rightarrow t=\sqrt { \dfrac { 2(H-y) }{ g } } \\ \Rightarrow x=vt=\sqrt { 2gy } \times \sqrt { \dfrac { 2(H-y) }{ g } } =2\sqrt { (H-y)y }$

$\Rightarrow x=2\sqrt { (H-y)y }$
Differentiating

$x$ w.r.t. to

$y$ , we have

$\dfrac { dx }{ dy } =2\times \dfrac { 1 }{ 2\sqrt { (H-y)y } } \times (H-2y)\\ \Rightarrow \dfrac { dx }{ dy } =\dfrac { H-2y }{ \sqrt { (H-y)y } }$
clearly

$\dfrac { dx }{ dy } >0$ for

$y<\dfrac { H }{ 2 }$
and

$\dfrac { dx }{ dy } <0$ for

$y>\dfrac { H }{ 2 }$
which implies that

$x$ first increases and then decreases, as

$y$ is increased from

$0$ to

$H$

$\Rightarrow$ option A is correct
for maxima or minima, we have

$\dfrac { dx }{ dy } =0$

$\Rightarrow y=\dfrac{H}{2}$
Since the

$x$ is first increasing which implies that

$y=\dfrac{H}{2}$ is a point of maxima.

$\Rightarrow$ option B is correct

${ x }_{ max }=x\left( y=\dfrac { H }{ 2 } \right)$

$\Rightarrow { x }_{ max }=2\sqrt { \left( H-\dfrac { H }{ 2 } \right) \dfrac { H }{ 2 } } =H$

$\Rightarrow$ option C is correct.
It is clear from the above equation that

${ x }_{ max }$ is independent of density of the liquid..

$\Rightarrow$ option D is incorrect.

What work should be done in order to squeeze all water from a horizontally located cylinder (figure shown above) during the time

$t$ by means of a constant force acting on the piston? The volume of water in the cylinder is equal to

$V$ , the cross-sectional area of the orifice to

$s$ , with

$s$ being considerably less than the piston area. The friction and viscosity are negligibly small.

0%
$\displaystyle A=\frac{1}{2}\rho\frac{V^3}{{(St)}^2}$
0%
$\displaystyle A=\frac{3}{2}\rho\frac{V^3}{{(St)}^2}$
0%
$\displaystyle A=\frac{5}{2}\rho\frac{3V^3}{{(St)}^2}$
0%
None of these

Explanation

Volume is

$V$ , Orifice area is

$S$ .

Let the discharge velocity be

$v$ .

So,

$v\times St=V\Rightarrow v=V/St\ -(1)$

Total mass is

$M=\rho\times V\ -(2)$ .

So, work done is equal to kinetic energy which is

$Mv^2/2$ .

From

$(1)\ and\ (2)$ ,

kinetic energy is

$KE=\dfrac{1}{2}\rho\dfrac{V^3}{(St)^2}$

Air is streaming past a horizontal airplane wing such that its speed is

$90 ms^{-1}$ at the lower surface and

$120 ms^{-1}$ over the upper surface. If the wing is 10 m long and has an average width of 2m, the difference of pressure on the two sides and the gross lift on the wing respectively, are (density of air

$=1.3 kg m^{-3})$

0%
5 Pa, 900 N
0%
95 Pa, 900 N
0%
4095 Pa, 900 N
0%
4095 Pa, 81900 N

Explanation

Applying Bernoulli's principle, we have

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

$\Rightarrow P_2-P_1=\frac{1}{2} \rho \left ( v_1^2 - v_2^2 \right ) \\\Rightarrow \Delta P=\frac{1}{2} \times 1.3 \times \left ( 120^2 - 90^2 \right ) \\\Rightarrow \Delta P=0.65 \times (120+90) \times (120-90) \\\Rightarrow \Delta P=0.65 \times 210 \times 30=4095 Pa$
The difference in pressure provides the lift to the aeroplane. So
lift on the aeroplane=pressure difference

$\times$ area of wings

$= \Delta P \times A = 4095 \times (10 \times 2)=81900N$

A massless conical flask filled with a liquid is kept on a table in a vacuum. The force exerted by the liquid on the base of the flask is

$W_1$ .The force exerted by the flask on the table is

$W_2$ .

0%
$W_1 = W_2$
0%
$W_1 > W_2$
0%
$W_1 < W_2$
0%
The force exerted by the liquid on the walls of the flask is $\left (W_1 - W_2 \right)$ .

Explanation

Pressure acts equally in all directions, and a liquid exerts forces on these walls of its container normal to the walls. Hence, the liquid exerting forces on the sloping walls of the conical flask will have a net upward component. As the force exerted by the liquid on the base of the flask is

$W_1$ ,we get the force exerted by the liquid on the walls of the flask as

$W_1-W_2$

The initial speed of efflux without cylinder is:

0%
$\displaystyle v=\sqrt{\frac{g}{3}[3H+4h]}$
0%
$\displaystyle v=\sqrt{\frac{g}{2}[4H+3h]}$
0%
$\displaystyle v=\sqrt{\frac{g}{2}[3H-4h]}$
0%
None of these

Explanation

Applying Bernoulli's equation just inside and just outside the hole,

$\displaystyle P_{0}+\left ( \frac{H}{2} \right )\left ( d \right )g+\left ( \frac{H}{2}-h \right )\left ( 2d \right )\left ( g \right )=\frac{1}{2}\left ( 2d \right )v^{2}+P_{0}$

$\therefore$

$\displaystyle v=\sqrt{\frac{g}{2}\left ( 3H-4h \right )}$

A cylindrical vessel open at the top is 20 cm high and 10 cm in diameter. A circular hole of cross sectional area

$1 cm^{2}$ is cut at the centre of the bottom of the vessel. Water flows from a tube above it in to the vessel at the rate of

$10^{2} cm^{3}/s$ . The height of water in the vessel under steady state is
(Take

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

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

Explanation

In steady state, Volume flow rate entering the vessel=volume flow rate leaving the vessel

$\therefore$

$Q=av=a\sqrt{2gh}$ or

$\displaystyle h=\frac{Q^{2}}{2ga^{2}}$

$\displaystyle =\frac{\left ( 10^{3} \right )^{2}}{\left ( 2\times 1000 \right )\left ( 1 \right )^{2}}=5$ cm

The density

$D$ of the material of the floating cylinder is:

0%
$5d/4$
0%
$3d/4$
0%
$4d/5$
0%
$4d/3$

Explanation

Weight=upthrust

$\therefore$

$\displaystyle \left ( L\frac{A}{5}Dg \right )=\frac{L}{4}\times \frac{A}{5}\times 2d\times g+\frac{3L}{4}\times \frac{A}{5}\times d\times g$

$\therefore$

$\displaystyle D=\frac{5d}{4}$

A container of large uniform cross-sectional area

$A$ resting on a horizontal surface, holds, two immiscible, non-viscous and incompressible liquids of densities

$d$ and

$2d$ each of height

$H/2$ as shown in the figure. The lower density liquid is open to the atmosphere having pressure

$P_{0}. A$ homogeneous solid cylinder of length

$L(L< H /2)$ , cross-sectional area

$A/5$ is immersed such that it floats with its axis vertical at the liquid-liquid interface with length

$L/4$ in the denser liquid.
The cylinder is then removed and the original arrangement is restored. A tiny hole of area

$s(s < < A)$ is punched on the vertical side of the container at a height

$h(h< H/2)$ . As a result of this, liquid starts flowing out of the hole with a range

$x$ on the horizontal surface.

The horizontal distance traveled by the liquid, initially, is :

0%
$\sqrt{(3H+4h) h}$
0%
$\sqrt{(3h+4H) h}$
0%
$\sqrt{(3H-4h) h}$
0%
$\sqrt{(3H-3h) h}$

Explanation

Applying Bernoulli's equation just inside and just outside the hole,

$\displaystyle P_{0}+\left ( \frac{H}{2} \right )\left ( d \right )g+\left ( \frac{H}{2}-h \right )\left ( 2d \right )\left ( g \right )=\frac{1}{2}\left ( 2d \right )v^{2}+P_{0}$

$\therefore$

$\displaystyle v=\sqrt{\frac{g}{2}\left ( 3H-4h \right )}$

$\displaystyle t=\sqrt{\frac{2h}{g}}$

$\therefore$

$x=vt=\sqrt{\left ( 3H-4h \right )h}$

An open vessel full of water is falling freely under gravity. There is a small hole in one filee of the vessel as shown in the figure. The water which comes out from the hole at the instant when hole is at height H above the ground, strikes the ground at a distance of

$x$ from

$P$ . Which of the following is correct for the situation described?

0%
The value of x is $\displaystyle 2\sqrt{\frac{2hH}{3}}$
0%
The value of x is $\displaystyle \sqrt{\frac{4hH}{3}}$
0%
The value of $x$ can't be computed from information provided
0%
No water comes out from the hole

Explanation

Since the whole vessel is falling under gravity, apparent pressure at the hole inside the vessel=

$P_0+\rho(g-a)h_1=P_0+\rho(g-g)h_1=P_0$ .

The pressure at hole outside the vessel is

$P_0$ .

Since there is no pressure difference, no water comes out of the hole.

A spherical ball is dropped in a long column of viscous liquid Which of the following graphs represent the variation of
(i) gravitational force with time
(ii) viscous force with time
(iii) net force acting on the bass with time

0%
Q, R, P
0%
R, Q, P
0%
P, Q, R
0%
R, P, Q

A small steel ball falls through a syrup at a constant speed of 10 m/s. If the steel ball is pulled upwards with a force equal to twice its effective weight, how fast will it move upwards?

0%
10 cm/s
0%
20 cm/s
0%
5 cm/s
0%
-5 cm/s

Equal volumes of a liquid is poured into containers A and B such that the area of cross-section of container A is double the area of cross-section of container B. If

$P_A$ and

$P_B$ are the pressures exerted at the bottom of the containers then

$P_A : P_B =$ __________

0%
1 : 2
0%
1 : 1
0%
3 : 2
0%
2 : 1

A metallic shpere of radius

$\displaystyle 1.0\times 10^{-3}m$ and density

$\displaystyle 1.0\times 10^{4}kg/m^{3}$ enters a tank of water after a free fall, falling through a distance of h in the earth's gravitational field. If its velocity remains unchanged after entering water, determine the value of h.

[Given : coefficient of viscosity of water =

$\displaystyle 1.0\times 10^{-3}N-s/m^{2},g=10m/s^{2}$ and density of water =

$\displaystyle 1.0\times 10^{3}kg/m^{3}$ ]

0%
20 m
0%
40 m
0%
80 m
0%
10 m

Explanation

The velocity attained by the sphere in falling freely from a height h is

$\displaystyle v=\sqrt{2gh}$ ............(i)
This is the terminal velocity of the sphere in water Hence by Stoke's law we have

$\displaystyle v=\frac{2}{9}\frac{r^{2}(\rho -\sigma )}{\eta }$
Where r is the radius of the sphere

$\displaystyle \rho$ is the density of the material of the sphere

$\displaystyle \sigma (=1.0\times 10^{3}kg/m^{3})$ is the density of water and

$\displaystyle \eta$ is coefficient of viscosity of water

$\therefore$

$\displaystyle v=\frac{2\times (1.0\times 10^{-3})^{2}(1.0\times 10^{4}-1.0\times 10)}{9\times 1.0\times 10^{-3}}=20m/s$
from equation (i) we have

$\displaystyle h=\frac{v^{2}}{2g}=\frac{20\times 20}{2\times 10}=20m$

A cylinder of height h filled with water and is kept on lock of height h/The level of water in the cylinder is kept constant. Four holes numbered 1, 2, 3 and 4 are at the side of the cylinder and at height 0, h/4, h/2 and 3h/4, respectively. When all four holes are opened together, the hole from which water will reach farthest distance on the plane PQ is the hole number.

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

Explanation

We know from Torricelli's theorem, that the range of the liquid falling from a certain height is given by:

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

where

$H$ is the total height of the container and

$h$ is the height where the hole is.

For

$R = R_{max}$ ;

$\dfrac{dR}{dh} = 0$

$\dfrac{dR}{dh} = 2 \times \Big( \dfrac{1}{2 \sqrt{h}} \sqrt{H - h} + \sqrt{h} \dfrac{-1}{2 \sqrt{H - h}} \Big)$

$\dfrac{dR}{dh} = \Big( \dfrac{ \sqrt{H - h}}{ \sqrt{h}} + \dfrac{- \sqrt{h}}{ \sqrt{H - h}} \Big)$

$\dfrac{dR}{dh} = \dfrac{H - h -h}{\sqrt{h(H - h)}} = 0$

$\Rightarrow H - h -h = 0$

$\Rightarrow H = 2h$

For

$R = R_{max}$

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

Taking PQ as the reference,

$H = h + \dfrac{h}{2} = \dfrac{3h}{2}$

So hole must be at height

$\dfrac{H}{2} = \dfrac{\Big( \dfrac{3h}{2} \Big)}{2} = \dfrac{3h}{4}$

For hole 1,

$h_1 = \dfrac{h}{2} + 0 = \dfrac{h}{2}$

For hole 2,

$h_2 = \dfrac{h}{2} + \dfrac{h}{4} = \dfrac{3h}{4}$

For hole 3,

$h_3 = \dfrac{h}{2} + \dfrac{h}{2} = h$

For hole 4,

$h_4 = \dfrac{h}{2} + \dfrac{3h}{4} = \dfrac{5h}{4}$

Since, hole 2 is at the

$\dfrac{H}{2}$ height required for longest range.

$\therefore$ Hole 2 is from which water will reach farthest distance on the plane PQ.

Hence, the correct answer is OPTION B.

When pressure is applied through a piston at the top of a closed tube containing water, the pressure is transmitted to:

0%
Only the bottom of container
0%
All directions
0%
Only the side faces and the bottom of the container
0%
None of these

A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is

$\displaystyle 1.8\times 10^{-5}kg/m \ s$ , what will be the terminal velocity of the drop? (density of water =

$\displaystyle 1.0\times 10^{3}kg/m^{2}$ and g = 9.8 N/kg). The density of air can be neglected

0%
$\displaystyle 2.72\times 10^{-4}m/s$
0%
$\displaystyle 3.72\times 10^{-4}m/s$
0%
$\displaystyle 0.72\times 10^{-4}m/s$
0%
$\displaystyle 12.72\times 10^{-4}m/s$

Explanation

By Stoke's law the terminal velocity of a water drop of radius r is given by

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

$\displaystyle \rho$ is the density of water

$\displaystyle \sigma$ is the density of air and

$\displaystyle \eta$ the coefficient of viscosity of air Here

$\displaystyle \sigma$ is negligible and r = 0.0015 mm = 1.5 x

$\displaystyle 10^{-3}$ = 1.5 x

$\displaystyle 10^{-6}$ m Substituting the values

$\displaystyle v=\frac{2}{9}\times \frac{(1.5\times 10^{-6})^{2}\times (1.0\times 10^{3})\times 9.8}{1.8\times 10^{-5}}=2.72\times 10^{-4}m/s$

A given shaped glass tube having uniform cross section is filled with water and is mounted on a rotatable shaft as shown in figure. If the tube is rotated with a constant angular velocity

$\displaystyle \omega$ , then

0%
Water levels in both sections A and B go up
0%
Water level in section A goes up and that in B comes down
0%
Water level in section A comes down and that in B it goes up
0%
Water levels remain same in both sections

For a fluid which is flowing steadily in a horizontal tube as shown in the figures, the level in the vertical tubes is best represented by.

A liquid flows along a horizontal pipe AB of uniform cross-section. The difference between the level of the liquid in tubes P and Q is

$10\ cm$ . The diameter of the tubes P and Q are the same. Then:
(

$\displaystyle \\ g=9.8{ ms }^{ -2 }$ )

0%
level in P is greater than that of Q and velocity of flow is $1.4 \ m/s$
0%
level in Q is greater than that of P and velocity of flow is $1.4\ m/s$
0%
level in P is greater than that of Q and velocity of flow is $0.7\ m/s$
0%
level in Q is greater than that of P and velocity of flow is $0.7\ m/s$

Explanation

Assuming laminar flow, velocity of liquid inside P is zero as it is perpendicular to the direction of flow of liquid. Hence, liquid level in P is same as the height of the pipe AB.

For pipe Q, it is inclined against the direction of water and there is extra pressure on the bottom of Q. This causes rise in level in Q.

Applying Bernoulli theorem for Q,

$P_o+\rho g h=P_o+\dfrac{1}{2}\rho v^2$

$v=\sqrt {2gh}=\sqrt {2 \times 9.8 \times 10}=1.4 m/s$

In a hydraulic lift, the small piston has an area of

$2 cm^2$ and the large piston has an area of

$80 cm^2$ . What is the mechanical advantage of the hydraulic lift?

0%
$40$
0%
$42$
0%
$10$
0%
$4$

Explanation

Area of small piston is

$2 cm^{2}$ and the area of large piston is

$80 cm^{2}$

We know that

$P=\frac{F}{A}$

$\Rightarrow \frac{P_{in}}{P_{out}}=\frac{F_{in}}{F_{out}} \times 40$

Therefore the mechanical advantage is

$40$

Therefore correct option is

$A$

A horizontally oriented tube AB of length

$l$ rotates with a constant angular velocity

$\omega$ about a stationary vertical axis OO' passing through the end A (fig). The tube is filled with an ideal fluid. The end A of the tube is open, the closed end B has a very small orifice. Find the velocity of the fluid relative to the tube as a function of the column 'height'

$h$ .

0%
$v=\omega\sqrt{h(l-h)}$
0%
$v=\omega \sqrt{h(2l-h)}$
0%
$v=\omega h$
0%
$v=\omega \sqrt{2h(l-h)}$

Explanation

As the tube rotates, the centrifugal force (as seen in the rotating frame of the tube) creates a pressure head causing the water to be jetted out of the orifice at the end of tube

Consider, an infinitesimally thin slice of water of width $dx$ at a distance $x$ from the one end of the tube. Let the area of cross-section of the tube be $s$ . The mass of water contained in this infinitesimally thin slice of water is given by $dm=\rho sdx$ , here $\rho$ is the density of water. the centrifugal force by this slice of water on the next is given by,

$df=(dm)w^2x=\rho s \omega^2xdx$ ...............(1)

the net force acting at the end of the tube just before the orifice can be obtained by integrating over all such thin slice as,

$\int^f_0 df = \int^l _ {l-h} \rho s \omega^2xdx$ or,

$f=\dfrac{1}{2}\rho s \omega^2h(2l-h)$ or,

$P=f/s = \dfrac{1}{2}\rho \omega^2h(2l-h)$ .................(2)

In (2), $P$ is the pressure head generated by the centrifugal force acting on the water. the net pressure at the end of the tube just before the orifice is given by $P + P_o$ , when $P_o$ is the atmosphereic pressure. We assume that at this point the water is still i.e its velocity is zero. Right outside the tube, the water jets into an area of pressure $P_i$ (atmospheric pressure), say at a velocity $v$ . Using Bernoulli's equation at point just before and just outside the orifice we have,

$P+P_o+\dfrac{1}{2}\rho (0)^2=P_0+\dfrac{1}{2}\rho v^2$ or,

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

or, $v=\omega \sqrt{h(2l-h)}$

The area of cross-section of the pump plunger and the press plunger of hydraulic press are

$0.03{m}^{2}$ and

$9{m}^{2}$ respectively. How much is the force acting on the pump plunger of the hydraulic press overcomes a load of

$900kgf$ ?

0%
$13kgf$
0%
$3kgf$
0%
$9kgf$
0%
$36kgf$

The flow of blood in a large artery of an anesthetised dog is diverted through a venturi meter. The wider part of the meter has a cross-sectional area equal to that of the artery

$A=8 \ mm^2$ .The narrower part has an area

$a=4 \ mm^2$ .The pressure drop in the artery is 24 Pa. What is the speed of blood in the artery?Density of blood is

$1.06\times 10^3 kg/m^3$ .

0%
$13.5\times 10^{-2}\ ms^{-1}$
0%
$12.5\times 10^{-2}\ ms^{-1}$
0%
$26.5\times 10^{-3}\ ms^{-1}$
0%
$32.5\times 10^{-}\ ms^{-1}$

Explanation

Given :

$A_a = 8mm^2$

$A_n = 4mm^2$

$P_a-P_n = 24$ Pa

Let the speed of blood in artery and narrow part be

$v_a$ and

$v_n$ respectively.

From continuity equation :

$A_av_a = A_nv_n$

$\therefore$

$8v_a = 4v_n$

$\implies v_n = 2v_a$

Using Bernoulli's equation :

$P_a+\rho gh_a +\dfrac{1}{2}\rho v_a^2=$

$P_n+\rho gh_n +\dfrac{1}{2}\rho v_n^2$

where

$h_n = h_n$

$\therefore$

$P_a - P_n = \dfrac{1}{2}\rho (v_n^2-v_a^2)$

$\therefore$

$24 = \dfrac{1}{2}\times 1.06\times 10^3 (4v_a^2 - v_a^2)$

$\implies$

$v_a = 12.5\times 10^{-2}$

$m/s$

A liquid flows through two capillary tubes connected in series. Their lengths are

$L$ and

$2L$ and radii

$r$ and

$2r$ respectively. Then the pressure differences across the first and the second tube are in the ratio

0%
$8:1$
0%
$4:1$
0%
$2:1$
0%
$1:8$

Explanation

given :-

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

Since flow rate across 1 = flow rate across 2

[as flow rate remains constant]

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

$\dfrac { { P }_{ 1 } }{ { P }_{ 2 } } =\dfrac { { l }_{ 1 } }{ { l }_{ 2 } } \times { \left( \dfrac { { r }_{ 2 } }{ { r }_{ 1 } } \right) }^{ 4 }=\dfrac { 1 }{ 2 } \times { 2 }^{ 4 }=8$

A capillary tube of radius

$r$ is immersed in water and water rises to a height of

$h$ . Mass of water in the capillary tube is

$5\times 10^{-3}kg$ . The same capillary tube is now immersed in a liquid whose surface tension is

$\sqrt{2}$ times the surface tension of water. The angle of contact between the capillary tube and this liquid is

$45^o$ . The mass of liquid which rises into the capillary tube now is (in kg):

0%
$5\times 10^{-3}$
0%
$2.5\times 10^{-3}$
0%
$5\sqrt{2}\times 10^{-3}$
0%
$3.5\times 10^{-3}$

Explanation

Since the force on liquid is proportional to

$Scos(\theta )$ , in the latter case, force will be

$\sqrt { 2 } cos(\dfrac { \pi }{ 4 } )=1$ times the force in first case. Hence, since the same force is applied, mass of water rising in the tube will be same.

What is a pressure on the smaller piston if both the pistons are at same horizontal level? (Take

$g=10m$

${s}^{-2}$ )

0%
$3Pa$
0%
$4\times {10}^{3}Pa$
0%
$6\times {10}^{5}Pa$
0%
$1000Pa$

Explanation

By Pascal's principle, the pressure on small piston

$=$ pressure on big piston.

Here pressure on big piston

$=$ the pressure due to car lift

$=F/A=\dfrac{3000g}{5\times 10^{-2}}=6\times 10^5 Pa$ as

$(g=10 m/s^2)$

There are two round tables in the physics classroom: one with the radius of

$50\ cm$ , the other with a radius of

$150\ cm$ . What is the relationship between the two forces applied on the tabletops by the atmospheric pressure? then value of

$\dfrac{F_1 }{F_2}$ is:

0%
$\dfrac{1}{9}$
0%
$\dfrac{1}{3}$
0%
$3$
0%
None of the above

Explanation

We know that

$P=\cfrac{F}{A}$

For the first table ,

$P_{1}=P_{0}$ and

$A_{1}=\pi \times (50)^{2}$ . Let the force be

$F_{1}$ .

For the second table,

$P_{2}=P_{0}$ and

$A_{2}=\pi \times (150)^{2}$ . Let the force be

$F_{2}$ .

So we have

$F_{1}=p_{1}A_{1}=(\pi \times 50^{2})P_{0}$ and

$F_{2}=p_{2}A_{2}=(\pi \times 150^{2})P_{0}$

By dividing above two equations , we get

$\cfrac{F_{1}}{F_{2}} = \cfrac{1}{9}$

Therefore option

$A$ is correct.

A capillary tube is attached horizontally to a constant pressure head arrangement. If the radius of the capillary tube is increased by

$10$ %, then the rate of flow of the liquid shall change nearly by

0%
$+10$ %
0%
$46$ %
0%
$-10$ %
0%
$-40$ %

Explanation

'v 'is directly proportional to the fourth power of 'r.'

'V' is directly proportional to the fourth power of (r+r/10) i.e. (11r/10)=1.1r

V/v=

${1.1r}^4$ /

${r}^4$ =1.4641

(V/v -1)100=0.4641x100=46.41

Hence (B) is correct.

Three capillaries of internal radii

$2r$ ,

$3r$ and

$4r$ , all of the same length, are joined end to end. A liquid passes through the combination and the pressure difference across this combination is

$20.2 cm$ of mercury. The pressure difference across the capillary of internal radius

$2r$ is

0%
$2 cm$ of $Hg$
0%
$4 cm$ of $Hg$
0%
$8 cm$ of $Hg$
0%
$16 cm$ of $Hg$

A small hole is made at a height of

$h = \dfrac{1}{2} m$ from the bottom of a cylindrical water tank and at a depth of

$h = 2 m$ from the upper level of water in the tank. The distance, where the water emerging from the hole strikes the ground, is:

0%
$22m$
0%
$1 m$
0%
$2 m$
0%
None of these.

Explanation

Range of water coming out = 2

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

given h =

$\dfrac { 1 }{ 2 }$ m & H-h = 2m.

So Range = 2

$\sqrt { \frac { 1 }{ 2 } \times 2 }$ = 2m

A tank is filled upto a height

$h$ with a liquid and is placed on a platform of height

$h$ from the ground. To get maximum range,

${x}_{m}$ a small hole is punched at a distance of

$y$ from the free surface of the liquid. Then

0%
${ x }_{ m }=2h$
0%
${ x }_{ m }=1.5h$
0%
$v=h$
0%
$y=0.75h$

A wooden cylinder of diameter

$4r$ , height

$h$ and density

$\rho /3$ is kept on a hole of diameter

$2r$ of a tank filled with water of density

$\rho$ as shown in the figure. The height of the base of cylinder from the base of tank is

$H$ . Now level of the liquid starts decreasing slowly. When the level of the liquid is at a height

$h_1$ , above the cylinder the block just starts to lift. At what value of

${h}_{1}$ , will the block rise.

0%
$\cfrac { 4h }{ 9 }$
0%
$\cfrac { 5h }{ 9 }$
0%
$\cfrac { 5h }{ 3 }$
0%
None of these

The block is maintained at the position by external means and the level of liquid is lowered. The height

${h}_{2}$ when the external force reduces to zero is

0%
$\cfrac { 4h }{ 9 }$
0%
$\cfrac { 5h }{ 9 }$
0%
$\cfrac { 2h }{ 3 }$
0%
None of these

Two fully blown balloons are suspended as shown in the diagram. A stream of air is passed in between the balloons. What will happen to the balloons?

0%
They will remain as they are
0%
They will go apart
0%
They will come closer
0%
They will go up

In the adjoining figure, the cross-sectional area of the smaller tube is a and the larger tube is 2a. A block of mass 2 m is kept in the smaller tube have the same base area an as that of the tube. The difference between water levels of the two tubes are (density of water is

$\rho$ and in

${ \rho }_{ 0 }$ is atmospheric pressure)

0%
$\dfrac { { \rho }_{ 0 } }{ pg } +\dfrac { m }{ ap }$
0%
$\dfrac { { \rho }_{ 0 } }{ pg } +\dfrac { m }{ 2ap }$
0%
$\dfrac { 2m }{ ap }$
0%
$\dfrac { m }{ ap }$

The figure shows the commonly observed decrease in diameter of a water stream as it falls from a tap. The tap has internal diameter

${ D }_{ 0 }$ and is connected to a large tank of water. The surface of the water is at a height b above the end of the tap. By considering the dynamics of a thin "cylinder' of water in the stream answer the following: (ignore any resistance to the flow and any effects of surface tension, given

$\rho _{ w }$ density of water)
1. Which of the following equation expresses the fact that the flow rate at the tap is the same as at the stream point with diameter D and velocity v (i.e. D in terms of

${ D }_{ 0 }$ ,

${ v }_{ 0 }$ and v will be):

0%
$D=\dfrac { { D }_{ 0 }{ v }_{ 0 } }{ v }$
0%
$D=\dfrac { { D }_{ 0 }{ { v }_{ 0 } }^{ 2 } }{ { v }^{ 2 } }$
0%
$D=\dfrac { { D }_{ 0 }v }{ { v }_{ 0 } }$
0%
$D={ D }_{ 0 }\sqrt { \dfrac { { v }_{ 0 } }{ v } }$

Explanation

$1$ Flow rate will be same

i.e

$A_1V_1=A_2V_2\\ \cfrac{\pi D^2V^2}{4}=\cfrac{\pi D_0^2V_0^2}{4}\\ \Rightarrow D^2V^2=D_0^2V_0^2\\ \Rightarrow DV=D_0V_0\\\Rightarrow D=\cfrac{D_0V_0}{V}$

A plane is in level flight at constant speed and each of its two wings has an area of 25 m

$^2$ . If the speed of the air on the upper and lower surfaces of the wing are 270 km h

$^{-1}$ and 234 km h

$^{-1}$ respectively, then the mass of the plane is (Take the density of the air = 1 kg m

$^{-3}$ )

0%
1550 kg
0%
1750 kg
0%
3500 kg
0%
3200 kg

Explanation

Let

$v_1$ ,

$v_2$ are the speed of air on the lower and upper surfaces of the wings of the plane

$P_1$ and

$P_2$ are the pressure there.

According to Bernoulli's theorem

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

$\displaystyle P_1 - P_2 = \frac{1}{2} \rho (v_2^2 - v_1^2)$

Here,

$v_1$ = 234 km h

$^{-1}$ = 234 x 58 m s

$^{-1}$ =65 m s

$^{-1}$

$v_2$ = 270 km h

${-1}$ = 270 x

$\displaystyle \frac{5}{18}$ = 75 m s

$^{-1}$

Area of wings = 2 x 25 m

$^2$ = 50 m

$^2$

$\therefore \,P_1 - P_2=\frac {1}{2} \times 1 (75^2 -65^2)$ Upward force on the plane =

$(P_1 - P_2) A$

$=\dfrac {1}{2} \times 1 \times (75^2 -65^2) \times 50m$

As the plane is in level flight, therefore upward force balances the weight of the plane.

$\therefore mg = (P_1 - P_2) A$

Mass of the plane,

$\therefore \displaystyle m = \dfrac{(P_1 - P_2)}{g} \times = \dfrac{1}{2} \times \dfrac{1 \times (75^2 - 65^2)}{10} \times 50$

$\displaystyle = \dfrac{(75+65)(75-65) \times 50}{2 \times 10} = 3500 kg$

A narrow tube completely filled with a liquid is lying on a series of cylinders as shown in the figure. Assuming no sliding between any surfaces, the value of acceleration of the cylinders for which liquid will not come out of the tube from anywhere is given by

0%
$\frac{gH}{2L}$
0%
$\frac{gH}{L}$
0%
$\frac{2gH}{L}$
0%
$\frac{gH}{\sqrt{2L}}$

A liquid flows through a horizontal tube as shown in figure. The velocities of the liquid in the two sections, which have areas of cross-section

$A_1$ and

$A_2$ , are

$v_1$ and

$v_2$ , respectively The difference in the levels of the liquid in the two vertical tubes is h. then

0%
$v_2^2 - v_1^2 = 2gh$
0%
$v_2^2 + v_1^2 = 2gh$
0%
$v_2^2 - v_1^2 = gh$
0%
$v_2^2 + v_1^2 = gh$

Explanation

Hint: Use Bernoulli’s equation to find the difference of pressure at both point.
Step – 1: From Bernoulli’s equation, find the difference of pressure at both point

Bernoulli’s equation is given by,

$\dfrac{P_1}{\rho} + \dfrac{{V_1}^2}{2} = \dfrac{P_2}{\rho} + \dfrac{{V_2}^2}{2}$

$\dfrac{P_1}{\rho} -\ \dfrac{P_2}{\rho} = \dfrac{{V_2}^2}{2} - \dfrac{{V_1}^2}{2}$

$P_1 - P_2 = \dfrac{\rho}{2} ({V_2}^2 - {V_1}^2)$

Step – 2: Find the difference between the pressure at two different pressure

There is difference h in height of 2 points. So, difference in pressure will be

$\rho gh$ .

Step – 3: Put the value in the equation

$\rho gh = \dfrac{\rho}{2} ({V_2}^2 - {V_1}^2)$

$({V_2}^2 - {V_1}^2) = 2gh$

$Answer:$

Hence, option A is the correct answer.

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

0:0:1

Practice Class 11 Engineering Physics Quiz Questions and Answers

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