Q: The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at the instant the first touches the ground. How far above the ground is the second drop at that instant?

Height of tap = 5m and (g) = \$10 m/sec^{2}\$For the first drop, \$5=ut+gt^{2}\$\$=(0\times t)+\times 10t^{2v}=5t^{2}\$ or \$t^{2}=1\$ or t=1.It mean that the third drop leaves after one second of the first drop. Or, each drop leaves after every 0.5 sec.Distance covered by the second drop in 0.5 sec.\$=(0.5)^{2}=1.25m\$Therefore, distance of the second drop above the ground\$=5-1.25=3.75 m\$

Q: What is the pressure on a swimmer 10 m below the surface of a lake?

Let atmospheric pressure \$=P_o=1.05\times 10^5\$ PaPressure \$10\$m below surface of lake,\$P=P_o+\displaystyle\rho gh\$\$=1.05\times 10^5+1000\times 9.8\times 10\$\$-1.99\times 10^5Pa\$\$=2\$ atm (\$\rho\rightarrow\$ Density of water \$=1000\$ kg\$/m^3\$, \$g=9.8m/s^2\$ \$h\rightarrow\$ Depth\$=10\$m).

Q: When a large bubble rises from the bottom of a lake to the surface, its radius doubles. If atmospheric pressure is equal to that of column of water height H, then the depth of lake is :-

It is given that the atmospheric pressure is,\$ P_o = \rho g H \$, where \$ \rho \$ is the density of water.Let the depth of the lake be d, the pressure at this depth would be,P = \$ P_o + \rho g d \$Hence, P = \$ \rho g (d+H) \$Since, the surrounding temperature is constant, we can assume that the process takes place isothermally. Therefore, we can apply the Boyle's Law\$ P_1 V_1 = P_2 V_2 \$Here, 1 denotes the water at depth d and 2 denotes the surface of water.Hence, \$ P_1 = \rho g (d+H) \$\$ P_2 = \rho g H \$\$ V_2 = 8V_1 \$ (since, radius is doubled, volume becomes 8 times)Substituting the values, in Boyle's Law, d = 8H - H = 7H

Q: A small sphere of radius \$r,\$ falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to

Let's consider,\$r=\$ radius of a small sphere\$V_T=\$Terminal velocity\$Power =\$ rate of production of heat\$P=FV_T\$. . . . .. .. .(1)From Stoke's formula,\$F=6\pi \eta V_T r\$. . . . . . . . . .(2)substitute equation (2) in equation (1), we get\$P=6\pi \eta rV_T^2\$ . . . . . . . . .(3)we know that the, the terminal velocity is given by\$V_T=\dfrac{2r^2(\rho-\sigma)}{9\eta}\$\$V_T\propto r^2\$. . . . . . .(4)from equation (3) and equation (4), we get\$P\propto r^5\$The correct option is A.

Q: A force of \$100N\$ is applied on an area of \$4m^2\$. the pressure being applied on the area is

Given, \$froce=100N\$and \$area=4m^2\$\$\therefore\$\$pressure=\dfrac{force}{area}=\dfrac{100N}{4m^{2}} =25 Pa\$

Q: The pressure on a swimmer 10 m below the surface lake is:(Atmospheric pressure=\$1.01\times 10^5\$ Pa,Density of water\$=1000kg/m^3\$ )

Given : \$h = 10 m\$, \$\rho = 1000 kg/m^3\$, \$g = 10 m/s^2\$Pressure on swimmer \$=\$ pressure of atmosphere + pressure of water \$ = 1 atm + \rho gh \times 10^{-5} atm\$ \$ = 1 atm + 1000\times 10 \times 10\times 10^{-5} atm\$ \$ = 2atm\$

Question 1

State and prove Bernoulli's theorem.

Accepted Answer

Bernoulli's Theorem:According to Bernoulli's theorem, the sum of the energies possessed by a flowing ideal liquid at a point is constant provided that the liquid is incompressible and non-viseous and flow in streamline.Potential energy + Kinetic energy + Pressure energy = Constant$P + \frac{1} {2}pv^{2} + pgh =  Constant$$gh + \frac{1} {2}v^{2} + \frac{P} {p} = C$ ............(11.11)Where C is a constant.This relation is called Bernoulli's theorem.Dividing eqn. (11.11) by g, we get$h + \frac{P} {pg} + \frac{1} {2}  \frac{v^{2}} {g} = C'$ ............(11.12)Where C is another constant.For horizontal flow, h remains same throughout.So, $ \frac{P} {pg} + \frac{v^{2}} {2g}  = Constant$ or;  $ P + \frac{1} {2} pv^{2}  = Constant$ P is static pressure of the liquid and $\frac{1} {2} pv^{2}$ is its dynamic and velocity pressure.Thus, for horizontal motion, the sum of static and dynamic pressure is constant. If $p_1v_1$ and $p_2v_2$ represent pressure and velocities at two points. Then$ P_1 + \frac{1} {2} pv^{2}_1  =  P_2 + \frac{1} {2} pv^{2}_2 $ Concepts : 1. In Bernoulli's theorem $ P + pgh + \frac{1} {2} ev^{2}$= constant. The term $\left(p + pgh \right)$ is called static pressure and the term $\frac{1} {2} pv^{2} $ is the dynamic pressure of the fluid.2. Bernoulli theorem is fundamental principle of the energy.3. The equation $\frac{P} {pg} + \frac{1} {2} \frac{v^{2}} {g} + h = constant$the term $\frac{P} {pg}$ = pressure headthe term $\frac{v^{2}} {2g}$ = velocity headh = gravitational head.Derivation of Bernoulli's Theorem :The energies possessed by a flowing liquid are mutually convertible. When one type of energy increases, the other type of energy decreases and vice-versa. Now, we will derive the Bernoulli's theorem using the work-energy theorem.Consider the flow of liquid. Let at any time, the liquid lies between two areas of flowing liquid $A_1$ and $A_2$. In time interval $\Delta t$, the liquid displaces from $A_1$ by $\Delta x_1 = v_1\Delta t$ and displaces from $A_2$ by $\Delta x_2 = v_2\Delta t$. Here $v_1$ and $v_2$ are the velocities of the liquid at $A_1$ and $A_2$.The work done on the liquid is $P_1 A_1 \Delta x_1$ by the force and $P_2 A_2 \Delta x_2$ against the force respectively.Net work done,$W = P_1 A_1 \Delta x_1 - P_2 A_2 \Delta x_2$$\Rightarrow  W = P_1 A_1 v_1 \Delta t - P_2 A_2 v_2 \Delta t$$= \left (P_1 - P_2 \right) \Delta V$ .........(11.13)Here, $\Delta V \rightarrow $ the volume of liquid that flows through a cross-section is same (from equation of continuity).But, the work done is equal to net change in energy (K.E. + P.E.) of the liquid, and $\Delta K =  \frac{1} {2} p \Delta V \left (v^{2}_1 - v^{2}_2 \right)$  ........ (11.14)and $\Delta U =   p \Delta V \left (h_2 - h_1 \right)$  ........ (11.15)$\therefore \left (P_1 - P_2 \right) \Delta V = \frac{1} {2} p \Delta V \left (v^{2}_1 - v^{2}_2 \right) + pg \Delta V \left (h_2 - h_1 \right)$ $P_1 + \frac{1} {2} v^{2}_1 + pgh_1 = P_2 + \frac{1} {2} v^{2}_2 + pgh_2$or $ P + \frac{1} {2} pv^{2} + pgh = constant$ .......... (11.16)This is the required relation for Bernoulli's theorem.$\therefore A_1 v_1 = A_2 v_2$So, more the cross-sectional area, lesser is the velocity and vice-versa.So, Bernoulli's theorem,$ P_1 + \frac{1} {2} p_1 v^{2}_1 = P_2 + \frac{1} {2} p_2 v^{2}_2$

Question 2

The water drops fall at regular intervals from a tap 5 m above the ground. The third drop is leaving the tap at the instant the first touches the ground. How far above the ground is the second drop at that instant?

Accepted Answer

Height of tap = 5m and (g) = $10 m/sec^{2}$For the first drop, $5=ut+gt^{2}$$=(0\times t)+\times 10t^{2v}=5t^{2}$ or $t^{2}=1$ or t=1.It mean that the third drop leaves after one second of the first drop. Or, each drop leaves after every 0.5 sec.Distance covered by the second drop in 0.5 sec.$=(0.5)^{2}=1.25m$Therefore, distance of the second drop above the ground$=5-1.25=3.75 m$

Question 3

Derive an expression for excess pressure inside a drop of liquid.

Accepted Answer

The net upward force on the top hemisphere of the liquid is just the pressure difference times the area of the equatorial circle :${ F }_{ upward }=\left( { P }_{ 1 }^{ \prime  }-{ P }_{ 0 } \right) \pi { r }^{ 2 }$ where ${ F }_{ upward }=$ upward force on the liquid due to air-currents; ${ P }_{ 1 }^{ \prime  }=$ Final pressure; ${ P }_{ 0 }=$ Initial pressure; $r=$ radius of the drop, assuming it to be spherical.The surface tension force downward around circle is equal to the surface tension of the circumference, since only one surface contributed to the force :Thus, ${ F }_{ downward }=T\left( 2\pi r \right) $ where, $T=$ surface tension.This gives, ${ P }_{ 1 }-{ P }_{ 0 }=\dfrac { 2T }{ r } $ which represents the excess pressure inside the liquid drop.

Question 4

What is the pressure on a swimmer 10 m below the surface of a lake?

Accepted Answer

Let atmospheric pressure $=P_o=1.05\times 10^5$ PaPressure $10$m below surface of lake,$P=P_o+\displaystyle\rho gh$$=1.05\times 10^5+1000\times 9.8\times 10$$-1.99\times 10^5Pa$$=2$ atm ($\rho\rightarrow$ Density of water $=1000$ kg$/m^3$, $g=9.8m/s^2$ $h\rightarrow$ Depth$=10$m).

Question 5

Give the expression for the excess pressure in a soap bubble.

Accepted Answer

Excess pressure inside a layer of liquid with radius of curvature $R$ is $\dfrac{2T}{R}$where $T$ is the surface tension of the liquid.Since a soap bubble has two spherical surfaces, the excess pressure inside it is$2\times\dfrac{2T}{R}$$=\dfrac{4T}{R}$

Question 6

Derive a relation between surface tension and surface energy.

Accepted Answer

Consider a rectangular frame (see fig. 11.20), having a sliding wire on one of its arms. Dip the frame in a soap solution and take it out. A soap film is formed on the frame and have two surfaces. Both the surfaces are in contact with the sliding wire. So, surface tension acts on the wire due to both the surface.Let T be the surface tension of the soap solution and L be the length of the wire.The force exerted by each surface on the wire = $T \times L$So, total force on the wire = $2TL$  ......(11.23)Let the surfaces contract by $\Delta x$.Work done on the film, $W = F \times \Delta x$$= 2T L \Delta x$ ......(11.24)Here $2L \times \Delta x \rightarrow$ total increase in the area of both of the surfaces of film.Let $2L \times \Delta x = \Delta A$$\therefore   W = T \Delta A$or  $T = \frac{W} {\Delta A}$  .........(11.25)So, surface tension of a liquid is equal to the work done in increasing the surface area of its free surface by one unit. In other words, surface tension is equal to surface energy per unit area.$E_s = T \Delta A$The unit of T is $Jm^{-2}$ and its dimensional formula is $\left[M^{1} L^{0} T^{-2} \right]$

Question 7

When a large bubble rises from the bottom of a lake to the surface, its radius doubles. If atmospheric pressure is equal to that of column of water height H, then the depth of lake is :-

Accepted Answer

It is given that the atmospheric pressure is,$ P_o = \rho g H $, where $ \rho $ is the density of water.Let the depth of the lake be d, the pressure at this depth would be,P = $ P_o + \rho g d $Hence, P = $ \rho g (d+H) $Since, the surrounding temperature is constant, we can assume that the process takes place isothermally. Therefore, we can apply the Boyle's Law$ P_1 V_1 = P_2 V_2 $Here, 1 denotes the water at depth d and 2 denotes the surface of water.Hence, $ P_1 = \rho g (d+H) $$ P_2 = \rho g H $$ V_2 = 8V_1 $ (since, radius is doubled, volume becomes 8 times)Substituting the values, in Boyle's Law, d = 8H - H = 7H

Question 8

A small sphere of radius $r,$ falls from rest in a viscous liquid. As a result, heat is produced due to viscous force. The rate of production of heat when the sphere attains its terminal velocity, is proportional to

Accepted Answer

Let's consider,$r=$ radius of a small sphere$V_T=$Terminal velocity$Power =$ rate of production of heat$P=FV_T$. . . . ..  ..  .(1)From Stoke's formula,$F=6\pi \eta V_T r$. . . . . . . . . .(2)substitute equation (2) in equation (1), we get$P=6\pi \eta rV_T^2$ . . . . . . . . .(3)we know that the, the terminal velocity is given by$V_T=\dfrac{2r^2(\rho-\sigma)}{9\eta}$$V_T\propto r^2$. . . . . . .(4)from equation (3) and equation (4), we get$P\propto r^5$The correct option is A.

Question 9

A force of $100N$ is applied on an area of $4m^2$. the pressure being applied on the area is

Accepted Answer

Given, $froce=100N$and $area=4m^2$$\therefore$$pressure=\dfrac{force}{area}=\dfrac{100N}{4m^{2}} =25 Pa$

Question 10

The pressure on a swimmer 10 m below the surface lake is:(Atmospheric pressure=$1.01\times 10^5$ Pa,Density of water$=1000kg/m^3$ )

Accepted Answer

Given :  $h = 10 m$,  $\rho = 1000 kg/m^3$,  $g = 10 m/s^2$Pressure on swimmer  $=$ pressure of atmosphere + pressure of water                                      $ = 1  atm  + \rho gh \times 10^{-5} atm$                                      $ = 1 atm + 1000\times 10 \times 10\times 10^{-5} atm$                                       $ = 2atm$