Physics M.C.Q (1 Marks)

$=\frac{7}{100} \times 2 \times 3.14 \times \frac{4.5}{100}$

$=197.82 \times 10^{-4}$

$=19.8 \times 10^{-3} \mathrm{~N}$

$=19.8 \mathrm{mN}$

$=2(0.03)(4)(3.14)\left(2 \times 10^{-2}\right)^2$

$=3.01 \times 10^{-4}\,J$

$\Rightarrow R$ increases and $P$ decreases

Mass of water in the firsr tube,

$m=\pi r^{2} h \rho=\pi r^{2} \times\left(\frac{2 T \cos \theta}{r \rho g}\right) \times \rho$

$=\frac{2 \pi r T \cos \theta}{g}$

$m \propto r$

$\frac{ m _{2}}{ m _{1}}=\frac{ r _{2}}{ r _{1}}$

$\frac{ m _{2}}{5}=\frac{2 r }{ r }$

$m _{2}=10 g$

Also, $\mathrm{P}=\mathrm{P}_{0}+\frac{4 \mathrm{T}}{\mathrm{R}}\ldots$ (ii)

From $(i) \;and\; (ii)$

$\rho g Z_{0}=\frac{4 \mathrm{T}}{\mathrm{R}}$

$\mathrm{z}_{0}=\frac{4 \mathrm{T}}{\mathrm{pgR}}$

$=\frac{4 \times 2.5 \times 10^{-2}}{10^{3} \times 10 \times 10^{-3}}=10^{-2} \mathrm{m}=1 \;\mathrm{cm}$

$Change\,in\,area\,of\,the\,film$

$or,\,\,W = T \times \Delta A$

$Here,{A_1} = 4\,cm \times 2\,cm = 8\,c{m^2}$

${A_2} = 5\,cm \times 4\,cm = 20\,c{m^2}$

$\Delta A = 2\,\left( {{A_2} - {A_1}} \right) = 24\,c{m^2} = 24 \times {10^{ - 4}}{m^2}$

$W = 3 \times {10^{ - 4}}J,T = ?$

$\therefore \,\,T = \frac{W}{{\Delta A}} = \frac{{3 \times {{10}^{ - 4}}}}{{24 \times {{10}^{ - 4}}}} = \frac{1}{8} = 0.125\,N\,{m^{ - 1}}$

$For\,given\,value\,of\,T\,and\,r,h \propto \frac{{\cos \theta }}{\rho }$

$Also,{h_1} = {h_2} = {h_3}\,\,or\,\,\frac{{\cos {\theta _1}}}{{{\rho _1}}} = \frac{{\cos {\theta _2}}}{{{\rho _2}}} = \frac{{\cos {\theta _3}}}{{{\rho _3}}}$

$Since\,{\rho _1} > {\rho _2} > {\rho _{3'}}\,so\,\cos {\theta _1} > \cos {\theta _2} > \cos {\theta _3}$

$For\,0 \le \theta  < \frac{\pi }{2},{\theta _1} < {\theta _2} < {\theta _3}$

$Hence,\,0 \le {\theta _1} < {\theta _2} < {\theta _3} < \frac{\pi }{2}$

$\therefore $ Force on the frame = $ 4(2\,TL) = 8\,TL$

Force to pull $ = 2\pi RT = 2 \times \pi \times 5 \times 75 = 750\pi \, dynes$

$T = 72.8\;dyne/cm$

$F = \frac{{2TA}}{t} = \frac{{2 \times 70 \times {{10}^{ - 3}} \times {{10}^{ - 2}}}}{{0.05 \times {{10}^{ - 3}}}} = 28N$

$ = (0.07 - 0.06)l = 0.01 \times 2 = 0.02N$

$mg = 2Tl$==> $m = \frac{{2Tl}}{g}$

$ = \frac{{2 \times 3 \times {{10}^{ - 2}} \times 10 \times {{10}^{ - 2}}}}{{9.8}}kg = 0.6gm$

$2 \mathrm{T}^{\prime} \theta=\mathrm{T}(2 \theta . \mathrm{R})$

Tension $\left.\mathrm{T}^{\prime}=\mathrm{RT}\right]$

$F = 7.5 \times 2 \times 1.5$

$F = 22.5\, N$

$\left( {{P_0} + \rho g\frac{h}{2}} \right)\left( {2Rh} \right) - T2R$

$ \Rightarrow \left| {2\,{P_0}Rh + R\rho g{h^2} - 2RT} \right|$

Length of the slider, $l=30 cm =0.3 m$

A soap film has two free surfaces.

$\therefore$ Total length $=2 l=2 \times 0.3=0.6 m$

Surface tension, $S=\frac{\text { Force o-Weight }}{2 l}$ $=\frac{1.5 \times 10^{-2}}{0.6}$$=2.5 \times 10^{-2} N / m$

Therefore, the surface tension of the film is $2.5 \times 10^{-2}\; N m ^{-1}$

By force balancing in vertical direction

$S_F=2 T \sin \theta$

$S_F=2 T \theta$   $\left\{\begin{array}{l}\because \theta \text { is small } \\ \sin \theta=\theta\end{array}\right.$

$S \times 2 r \times 2 \theta=2 \times T \times \theta$

$S \times 2 r=$ Tension   

$S \times d=$ Tension    $\left\{\begin{array}{l}r \text {-radius } \\ d \text {-diameter }\end{array}\right.$

$\because  S=T$

So, Tension $=T d$   $\left\{\begin{array}{l}\text { Where, } \\ S_F=\text { Force due to surface tension } \\ T=\text { Tension in string } \\ \theta=\text { Small angle } \\ S=\text { Surface tension }\end{array}\right.$

= (surface energy of $n$ small drops) -(surface energy of
one big drop)

$ = n4\pi {r^2}T - 4\pi {R^2}T = 4\pi T(n{r^2} - {R^2})$

= $4 \times 3.14 \times {(1.4 \times {10^{ - 1}})^2} \times 75({125^{1/3}} - 1)$$ = 74\;erg$

$W = 2\pi ({D^2} - {d^2})T = 2\pi [{(2D)^2} - {(D)^2}]\;T = 6\pi {D^2}T$

$W \propto {R^2}$ ($T$ is constant)

If radius becomes double then work done will become four times.

$=T \times\left(4 \pi r^{2}\right) \times 2=8 \pi T r^{2}$

$ = 4 \times 3.14 \times {10^{ - 4}} \times 35 \times {10^{ - 1}}({10^{6/3}} - 1)$ $ = 4.4 \times {10^{ - 3}}\,J$

$n\frac{4}{3}\pi {r^3} = \frac{4}{3}\pi {R^3}$ $⇒$ $1000{r^3} = {R^3}$ $⇒$ $R = 10r$

$\therefore \frac{r}{R} = \frac{1}{{10}}$

$\frac{{{\rm{surface \,energy \,of \,one\, small\, drop}}}}{{{\rm{surface\, energy \,of\, one \,big\, drop}}}} = \frac{{4\pi {r^2}T}}{{4\pi {R^2}T}} = \frac{1}{{100}}$

$ = \pi {D^2}\sigma ({27^{1/3}} - 1)$$ = 2\pi {D^2}\sigma $

$\frac{{{\rm{surface\, energy\, of\, one \,big \,drop}}}}{{{\rm{surface \,energy \,of \,}}n{\rm{ drop}}}} = \frac{{4\pi {R^2}T}}{{n \times 4\pi {r^2}T}}$

$\frac{{{R^2}}}{{n{r^2}}} = \frac{{{n^{2/3}}{r^2}}}{{n{r^2}}}$ = $\frac{1}{{{n^{1/3}}}} = \frac{1}{{{{(1000)}^{1/3}}}} = \frac{1}{{10}}$

$\frac{4}{3} \pi R^{3}=n \frac{4}{3} \pi r^{3}$

$r=n^{-1 / 3} R$

workdone $=$ $final energy - initial energy$

$=\left[n 4 \pi n^{-2 / 3} R^{2}-4 \pi R^{2}\right] \sigma$

$=4 \pi R^{2}\left(n^{1 / 3}-1\right) \sigma$

${R^3} = 1000{r^3}$ $⇒$ $R = 10r \Rightarrow r = \frac{R}{{10}}$

$\therefore R = 20r$

$\frac{{{\rm{Surface \,energy\, of \,one \,big \,drop}}}}{{{\rm{Surface \,energy \,of \,8000 \,small\, drop}}}} = \frac{{4\pi {R^2}T}}{{8000\;4\pi {r^2}T}}$

$ = \frac{{{R^2}}}{{8000{r^2}}} = \frac{{{{\left( {20r} \right)}^2}}}{{8000{r^2}}} = \frac{1}{{20}}$

$\frac{{{\rm{Energy\, of \,big \,drop}}}}{{{\rm{Energy \,of\, small\, drop}}}} = \frac{{4\pi {R^2}T}}{{4\pi {r^2}T}} = \frac{{{R^2}}}{{{r^2}}}$$ = {(8)^{2/3}} = 4$

$ = 2 \times [(10 \times 0.6) - (10 \times 0.5)] \times {10^{ - 4}} = 2 \times {10^{ - 4}}{m^2}$

Work done = $T \times \Delta A$

$ = 7.2 \times {10^{ - 2}} \times 2 \times {10^{ - 4}} = 1.44 \times {10^{ - 5}}J$

$ \Rightarrow W = 4\pi  \times {(2 \times {10^{ - 3}})^2} \times 0.465({8^{1/3}} - 1) = 23.4\;\mu \,J$

Therefore, $W=S . T=2 \pi r . T$

$W = 8\pi {R^2}T$(As $V \propto {R^3}$ $\therefore R \propto {V^{1/3}}$)

$\therefore W \propto {V^{2/3}}$

$\frac{{{W_2}}}{{{W_1}}} = {\left( {\frac{{{V_2}}}{{{V_1}}}} \right)^{2/3}} = {(2)^{2/3}}$==> ${W_2} = {(4)^{1/3}}W$

${W_1} = 8\pi {R^2}{T_1}$

Work done to form a bubble of radius $2R$

${W_2} = 8\pi {(2R)^2}{T_2}$$ = 32\pi {R^2}{T_2}$ ?$\frac{{{W_1}}}{{{W_2}}} = \frac{{{T_1}}}{{4{T_2}}}$

If surface tension of soap solution is same then

${W_2} = 4{W_1}$

But in the problem temperature of solution is increased so its surface tension decreases.

$\therefore {W_2} < 4{W_1}$

$\Rightarrow r=\frac{2 \mathrm{T}}{\rho \mathrm{L}}$

$=4 \times 3.14 \times\left(1.4 \times 10^{-1}\right)^{2} \times 75\left[(125)^{\frac{1}{3}}-1\right]$

$=73.85$ erg $=74$ erg

$W\, = 2\pi (D^2-d^2)T\, = 2\pi [(2D)^2-(D)^2]T$

    $= 6\pi D^2T$

$\mathrm{w}=60\left(4 \pi \mathrm{r}_{2}^{2}-4 \pi \mathrm{r}_{1}^{2}\right)$

$w=60 \times 4 \pi(16-4)$

$w=60 \times 4 \pi \times 12$

$\mathrm{w}=9.043 \times 10^{3} \mathrm{erg}$

$=8 \pi \mathrm{T}\left[\frac{\mathrm{D}^{2}}{4}-\frac{\mathrm{d}^{2}}{4}\right]$

$=2 \pi \mathrm{T}\left[\mathrm{D}^{2}-\mathrm{d}^{2}\right]$

             $= 4\pi \frac {D^2}{4}T [(27)^{1/3}-1] = 2\pi D^2T$

$W = 20\left[ {2 \times \left( {12 \times 10 - 6 \times 10} \right)} \right]$

$W = 2400\, erg$

Surface energy of soap bubble :

$U=T \times 4 \pi r^2$

$U_1=T \times 4 \pi(\frac{r}{2})^2=T \times 4 \pi (\frac{r^2}{4}) =\pi T r^2$

$U_2=T \times 4 \pi(2 r)^2=16 \pi T r^2$

The work done in increasing the radius of soap bubble is equal to the increase in surface energy.

$\therefore W =\left( U _2- U _1\right)$

$=16 \pi Tr ^2-\pi Tr ^2$

$=15 \pi Tr ^2$

Thus, the work done in increasing the radius of soap bubble will be $=15 \pi Tr ^2$

$\because$ Volume $=$ constant

i.e., $\frac{4}{3} \pi R^3=n \times \frac{4}{3} \pi r^3$

Radius of each new droplet $=\frac{R}{n^{1 / 3}}$

Work done to break into ' $n$ ' drops $=S\left[n \times 4 \pi r^2-4 \pi R^3\right]$

$=S \times 4 \pi R^2\left[n^{1 / 3}-1\right]$

$ = 70\;dyne/cm$$ \equiv 68.66\;dyne/cm$

i.e. $\Delta P \propto \frac{1}{r}$ 

This means that bubbles $A$ and $C$ posses greater pressure inside it than $B$. So the air will move from $A$ and $C$ towards $B$.

$\therefore \;\frac{{{r_2}}}{{{r_1}}} = 4$ and $\frac{{{V_1}}}{{{V_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^3} = \frac{1}{{64}}$

$ \Rightarrow {P_{in}} - {P_{out}} = \frac{{2T}}{r} = \frac{{2 \times 70 \times {{10}^{ - 3}}}}{{0.1 \times {{10}^{ - 3}}}} = 1400\,Pa$

$⇒$  ${P_{in}} = 1400 + 1.013 \times {10^5}$$ = 0.014 \times {10^5} + 1.013 \times {10^5} = 1.027 \times {10^5}\,Pa$

$T = \frac{{Rhdg}}{4}$

$T  = \frac{{{{10}^{ - 2}} \times 2 \times {{10}^{ - 3}} \times 0.8 \times {{10}^3} \times 9.8}}{4}$

$ = 3.9 \times {10^{ - 2}}\,N/m$

At constant temperature $V \propto \frac{1}{P}$ (Boyle’s law)

Since as the bubble rises upward, pressure decreases, then from above law volume of bubble will increase i.e. its size increases.

$\Delta P \propto \frac{1}{r}$

As radius of soap bubble increases with time

$\Delta P \propto \frac{1}{t}$

$\Rightarrow \frac{2 \times 75 \times 10^{-3}}{0.05 \times 10^{-3}}=10^{3} \times 10 \times h$

$\Rightarrow 3000=10^{3} \times 10 \times h$

$\Rightarrow h=0.3 m=30 \mathrm{cm}$

$\therefore P_{1}>P_{2}$

hence air moves from smaller bubble to bigger bubble.

diameter of the hole $=\mathrm{d}=2 \mathrm{r}=\frac{4 \mathrm{T}}{\mathrm{hP}_{\mathrm{w}} \mathrm{g}}$

$d=\frac{4 \times 0.07}{0.4 \times 10^{3} \times 9.8}=0.07 \times 10^{-3} \mathrm{m}=0.07 \mathrm{mm}$

$\frac{4}{3}\pi r_1^3 = n \times \frac{4}{3}\pi r_2^3 = n \times \frac{4}{3}\pi {\left( {2{r_1}} \right)^3}$

$or\,\,n = \frac{1}{8} = 0.125$

$P_{\text {excess }}=\frac{T}{R} \quad \quad \because\left(\begin{array}{l}{r_1=R} \\ {r_{2}=0}\end{array}\right)$

Excess pressure $\Delta P =\frac{4 T }{ R }$

$\frac{\Delta P _1}{\Delta P _2}=\frac{ R _2}{ R _1}=\frac{2}{5}$

$\frac{ S _1}{ S _2}=\left(\frac{ R _2}{ R _1}\right)^2=\frac{25}{4}$

$P_1-P_0=\frac{4 T}{R_1} ; P_2-P_0=\frac{4 T}{R_2}$

$P_1-P_2=4 T\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\frac{4 T}{R}$

$\frac{1}{R}=\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{2}-\frac{1}{4}$

$\frac{1}{R}=\frac{2-1}{4}=\frac{1}{4}$

$R=4\,Cm$

Excess pressure in soap bubble $=\frac{4 S}{R}$

$P=\frac{4 S}{R}$   $\left\{\begin{array}{l}S \text { - Surface tension } \\ R \text { - Radius }\end{array}\right.$

For second bubble,

$2 P=\frac{4 S}{x}$   $\left\{\begin{array}{l}S \text { - Surface tension } \\ R \text { - Radius }\end{array}\right.$

Substitute value of $P$

$2 \times \frac{4 S}{R}=\frac{4 S}{x}$

$\Rightarrow x=\frac{R}{2}$

Ratio of Radii $=\frac{R / 2}{R}=\frac{1}{2}$

So, Ratio of volume $=\left(\right.$ Ratio of Radii )$^3$

$=\left(\frac{1}{2}\right)^3=\frac{1}{8}$

Let take $g=10 \,m / s ^2$

For water to enter the sphere, pressure required is $=\rho g h$

$=1 \times 10 \times \frac{40}{100} \times 1000 \quad\left(\rho=1000 \,kg / m ^3\right)$

$=4000 \,\frac{ N }{ m ^2}=\text { excess pressure }$

Let the hole have radius $=R$

Excess pressure $=\frac{2 T}{R}$    [One surface air, one surface water]

$\Rightarrow 4000=\frac{2 \times 0.07}{R}$

$\Rightarrow 2 R=0.07 \times 10^{-3} \,m$

$\Rightarrow d=0.07 \,mm$

$P_2=P_1 -\frac{2 T}{R}=P_0- \frac{2 T}{R}$

$P_3=P_2+\rho gL=P_0- \frac{2 T}{R}+\rho gL$

here $r=R \cos \theta=R$

$P_4=P_3-\frac{2 T}{R}=P_0- \frac{4T}{R}+\rho gL$

$P_4=P_0$

$\Rightarrow L p g=\frac{4 T}{R}$

$\Rightarrow L=\frac{4 T}{R p g}$

$\sin \left(90^{\circ}-\theta\right)=\frac{R-h}{R}$

$\Rightarrow \cos \theta=\frac{R-h}{R}$

$\Rightarrow \theta=\cos ^{-1}\left(\frac{R-h}{R}\right)$

If $r$ is less then $h$ will be more.

$ \Rightarrow r = \frac{1}{2}mm$

$\therefore D = 2r = 1\;mm$

So, $l = \frac{h}{{\cos \theta }} = \frac{3}{{\cos {{60}^o}}} = 6\;cm$

$r=\frac{2 T \cos \theta}{h d g}$

$ \Rightarrow \frac{{{h_1}}}{{{h_2}}} = \frac{{{T_1}}}{{{T_2}}} \times \;\frac{{{d_2}}}{{{d_1}}}$          ($r,\;\theta $ and $g$ are constants) 

$ \Rightarrow \frac{{{h_1}}}{{{h_2}}} = \frac{{60}}{{50}} \times \frac{{0.6}}{{0.8}} = \frac{9}{{10}}$

$⇒$  $r = \frac{{2T}}{{hdg}} = \frac{{2 \times 7.2 \times {{10}^{ - 2}}}}{{3 \times {{10}^{ - 2}} \times {{10}^3} \times 10}} = 4.8 \times {10^{ - 4}}$

 $d = 2r = 9.6 \times {10^{ - 4}}\,m$

$M \propto {R^2} \times \left( {\frac{1}{R}} \right)$   (As H $\propto 1/R$)

 ${\rm M} \propto R$. If radius becomes double then mass will becomes twice.

$\frac{{{h_2}}}{{{h_1}}} = \frac{{140}}{{70}} \times \frac{{\cos 60^\circ }}{{\cos 0^\circ }} \times \frac{1}{2} \times 1 = \frac{1}{2} \Rightarrow {h_2} = \frac{{{h_1}}}{2} = 3cm.$

$\Rightarrow \frac{T_{w}}{T_{m}} \cong \frac{1}{6.5}$

$\mathrm{F}=$ force due to surface tension at cross-section

$\left(\mathrm{S}_{1} \text { and } \mathrm{S}_{2}\right)+$ weight of tube.

$ = 2\pi RT + mg$

$\Rightarrow 2 \pi r=\frac{75 \times 10^{-4}}{12 \times 10^{-2}}=6.25 \times 10^{-2} \mathrm{m}$

Angle of contact, $\theta=0^{\circ}, \rho=10^{3} \mathrm{kgm}^{-3}$

The maximum height to which the sap can rise by capillary action is,

$h=\frac{2 S \cos \theta}{r \rho g}=\frac{2 \times 7.28 \times 10^{-2} \times \cos 0^{\circ}}{2.5 \times 10^{-5} \times 10^{3} \times 9.8}=0.59 m$

$\Rightarrow \frac{\mathrm{h}_{1}}{\mathrm{h}_{2}}=\frac{\mathrm{T}_{1}}{\mathrm{T}_{2}} \times \frac{\mathrm{d}_{2}}{\mathrm{d}_{1}} \quad[\mathrm{r}, \theta \text { and } \mathrm{g} \text { are constants }]$

$=\frac{60}{50} \times \frac{0.6}{0.8}=\frac{9}{10}$

$r=\frac{2 \times 73 \times 10^{-3}}{10 \times 10^{-2} \times 10^{3} \times 9.8}=0.015 \mathrm{cm}$

$\ell=\frac{h}{\cos \alpha}$

$=\frac{4 \mathrm{cm}}{\cos 30^{\circ}}$

$=\frac{8}{\sqrt{3}} \mathrm{cm}$

Hence, the radius of the first bore, $r_{1}=\frac{d_{1}}{2}=1.5 \times 10^{-3} m$

Diameter of the second bore, $d_{2}=6.0 mm$

Hence, the radius of the second bore, $r_{2}=\frac{d_{2}}{2}=3 \times 10^{-3} m$

Surface tension of water, $s=7.3 \times 10^{-2} N m ^{-1}$

Angle of contact between the bore surface and water, $\theta=0$

Density of water, $\rho=1.0 \times 10^{3} kg / m ^{-3}$

Acceleration due to gravity, $g=9.8 m / s ^{2}$ Let $h_{1}$ and $h_{2}$ be the heights to which water rises in the first and second tubes respectively. These heights are given by the relations:

$h_{1}=\frac{2 s \cos \theta}{r_{1} \rho g}$

$h_{2}=\frac{2 s \cos \theta}{r_{2} \rho g}$

The difference between the levels of water in the two limbs of the tube can be calculated as

$=\frac{2 s \cos \theta}{r_{i} \rho g }-\frac{2 s \cos \theta}{r_{2} \rho g }$

$=\frac{2 s \cos \theta}{\rho g}\left[\frac{1}{r_{1}}-\frac{1}{r_{2}}\right]$

$=\frac{2 \times 7.3 \times 10^{-2} \times 1}{1 \times 10^{3} \times 9.8}\left[\frac{1}{1.5 \times 10^{-3}}-\frac{1}{3 \times 10^{-3}}\right]$

$=4.966 \times 10^{-3} m$

$=4.97 mm$

Hence, the difference between levels of water in the two bores is $4.97\; mm$

Height of liquid in capillary $=\frac{2 T}{r \rho g}=h$

Pressure we need to apply $=\rho g h+P_0$   $\left\{\begin{array}{l}\text { Where, } \\ T=\text { Surface tension } \\ r=\text { Radius of capillary } \\ \rho=\text { Density of liquid } \\ P_0=\text { Atmospheric pressure }\end{array}\right.$

Substitute value of $h$

$P=\rho g \times \frac{2 T}{r \rho g}+P_0=\frac{2 T}{r}+P_0=\frac{4 T}{d}+P_0$

$\Rightarrow P=\frac{4 \times 0.07}{\left(0.28 \times 10^{-3}\right)}+P_0=1000 \,Nm ^{-2}+10^5 \,Nm ^{-2}$    $\left\{\begin{array}{l}\text { Given, } \\ T=0.07 \,N / m \\ d=0.28 \,mm \end{array}\right.$

$\Rightarrow P=\left(10^3+10^5\right) \,Nm ^{-2}=101 \times 10^3 \,Nm ^{-2}$

Neglecting the small mass in the meniscus, for full rise,

$2 \pi r T=\pi r^2 h \rho g$

$\rho=\frac{2 T}{r \rho g}=\frac{2 \times 0.07}{0.25 \times 10^{-3} \times 1000 \times 9.8}=0.057 m =5.7 cm$

But there the tbe is only $2 cm$ above the water and so water will rise by $2 cm$ and meet the tube at an angle such that

$2 \pi r T \cos 0^{\circ}=\pi r^2 h^{\prime} \rho g$

$\Rightarrow 2 T \cos \theta=h^{\prime} r \rho g$

$\Rightarrow \cos \theta=\frac{h^{\prime} r \rho g}{2 T}$

$\therefore \cos \theta=\frac{2 \times 10^{-2} \times 0.25 \times 10^{-3} \times 1000 \times 9.8}{2 \times 0.07}$

$\Rightarrow \theta=70^{\circ}$

$\Delta P =\frac{8 \eta v }{ a ^2} h$

$\frac{ dp }{ dt }= T (2 \pi a )-\rho\left(\pi a ^2 h \right) g-\left[\frac{8 \eta v }{ a ^2} h \right]\left(\pi a ^2\right)$

$F -\pi a ^2 \rho g h = T (2 \pi a )-\rho\left(\pi a ^2 h \right) g-8 \pi \eta h\left(\frac{ dh }{ dt }\right)$

$F = T (2 \pi a )-8 \pi \eta h \left(\frac{ dh }{ dt }\right)$

$\Rightarrow p_{\text {dynamic }} \times \text { Area } \geq \text { Force of surface tension }$

$\Rightarrow \frac{1}{2} \rho v^{2} \times \pi R^{2} \geq 2(2 \pi R T)$

Note that factor $2$ on right hand side appears as there are two sides of film surface.

$\Rightarrow \quad v \geq \sqrt{\frac{8 T}{\rho R}}$

$\Rightarrow \quad v_{\min }=\sqrt{\frac{8 T}{\rho R}}$

$ \frac{4}{3} \pi \mathrm{R}^3=27 \times \frac{4}{3} \times \pi \mathrm{r}^3 $

$ \mathrm{R}^3=27 \mathrm{r}^3 $

$ \mathrm{R}=3 \mathrm{r} $

$ \mathrm{r}=\frac{\mathrm{R}}{3} $

$ \mathrm{r}^2=\frac{\mathrm{R}^2}{9}$

Work done $=\mathrm{T} \Delta \mathrm{A}$

$=27 \mathrm{~T}\left(4 \pi \mathrm{r}^2\right)-\mathrm{T} 4 \pi \mathrm{R}^2$

$=27 \mathrm{~T} 4 \pi \frac{\mathrm{R}^2}{9}-4 \pi \mathrm{R}^2 \mathrm{~T}$

$8 \pi \mathrm{R}^2 \mathrm{~T}$

$R=10 r$

$E_1=1000 \times 4 \pi r^2 \times S$

$E_2=4 \pi(10 r)^2 S$

$\frac{E_1}{E_2}=\frac{10}{1}, x=10$

$\frac{4}{3} \pi R^3=1000 \frac{4}{3} \pi r^3$

$R=10 r$

$U_i=1000\left(4 \pi r^2 S\right)$

$U_f=4 \pi R^2 S$

$=100\left(4 \pi r^2 S\right)$

$U_f=\frac{1}{10} U_i$

$36960$ $\mathrm{erg}=\frac{40 \text { dyne }}{\mathrm{cm}} 8 \pi\left[(\mathrm{r})^2-\left(\frac{7}{2}\right)^2\right] \mathrm{cm}^2$

$\mathrm{r}=7 \mathrm{~cm}$

$10 \mathrm{r}=\mathrm{R}$

$\mathrm{R}=10 \mathrm{r}$

$\frac{\text { S.E. of } 1000 \text { drops }}{\text { S.E. of Big drop }}=\frac{1000\left(4 \pi^2\right) T}{4 \pi R^2 T}$

$=\frac{1000 \times \mathrm{r}^2}{(10 \mathrm{r})^2}=10=\frac{10}{\mathrm{x}}$

$\therefore \quad \mathrm{x}=1$

$\Delta \mathrm{P}=2\left(\frac{2 \mathrm{~S}}{\mathrm{R}}\right)=\frac{4 \mathrm{~S}}{\mathrm{R}}$

$\Rightarrow R=\frac{4 \times 0.28}{8 \times 10^3 \times 10 \times 4 \times 10^{-4}}$

$\Rightarrow \frac{0.28}{8} \mathrm{~m}=\frac{28}{8} \mathrm{~cm}$

$\Rightarrow R=3.5 \mathrm{~cm}$

$\text { Diameter }=7 \mathrm{~cm}$

$\mathrm{P}_1-\mathrm{P}_0=3\left(\mathrm{P}_2-\mathrm{P}_0\right)$

$\frac{4 \mathrm{~T}}{\mathrm{r}_1}=3 \frac{4 \mathrm{~T}}{\mathrm{r}_2}$

$\mathrm{r}_2=3 \mathrm{r}_1$

$\frac{\mathrm{V}_1}{\mathrm{~V}_2}=\frac{\frac{4}{3} \pi \mathrm{r}_1^3}{\frac{4}{3} \pi \mathrm{r}_2^3}=\frac{1}{27}$

$\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}}$

Height of capillary rise will be smaller in hot water and larger in cold water.

Statement $II$ is incorrect because height of liquid is given by $h==\frac{2 T \cos \theta_C}{\rho g r}$ where $r$ is radius of

Tube (assuming length of capillary is sufficient) Hence option $(3)$ is correct.

$\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\rho \mathrm{gr}} \text {; }$

If $\theta=0^{\circ}$ then rise is non-zero

$\therefore$ Statement- $1$ is incorrect.

Option$(1)$ is correct

$R$ : Radius of bigger drop

$r :$ Radius of smaller drop

Volume will remain same

$\frac{4}{3} \pi R ^3=1000 \times \frac{4}{3} \pi r^3$

$R =10 r$

$u _{ i }= T \cdot 4 \pi R ^2$

$u _{ f }= T .4 \pi r ^2 \times 1000$

$\frac{ u _{ f }}{ u _{ i }}=\frac{1000 r ^2}{ R ^2}$

$\frac{ u _{ f }}{ u _{ i }}=\frac{10}{1}$

So, $x =1$

Work done $=$ change in surface energy $\times T_S$

$= T _{ S } \times 8 \pi \times\left( R _2^2- R _1^2\right)$

$=2 \times 10^{-2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{-4}$

$=18.48 \times 10^{-4}\,J$

$R =10\,mm$

$T \times 1000 \times 4 \pi\left(10^{-3}\right)^2- T \times 4 \pi\left(10 \times 10^{-3}\right)^2=\Delta E$

$\Delta E =4 \times \pi \times 7 \times 10^{-2}[1000-100] \times 10^{-6}$

$\Delta E =7.92 \times 10^{-4}\,J$

$\frac{4}{3} \pi\left(10^{-3}\right)^3=125 \times \frac{4 \pi}{3} R_{\text {new }}^3$

$10^{-3}=5 R_{\text {new }}$

$R_{\text {new }}=\frac{10^{-3}}{5}\,m$

So, final surface energy $=0.45 \times 125 \times 4 \pi\left(\frac{10^{-3}}{5}\right)^2$

Increase in energy $=0.45 \times 4 \pi \times\left(10^{-3}\right)^2\left[\frac{125}{25}-1\right]$

$=4 \times 0.45 \times 4 \pi \times 10^{-6}$

$=2.26 \times 10^{-5}\,J$

$W = T \left(8 \pi\left( r _2^2- r _1^2\right)\right)$

$W =264 \times 10^{-4}\,J$

$\frac{ h _1}{ h _2}=\frac{ S _1}{ S _2} \frac{\rho_2}{\rho_1}$

$\frac{5}{ h _2}=\left[\frac{1}{2}\right]\left[\frac{2}{1}\right] \Rightarrow h _2=5\,cm =0.05\,m$

Info about angle of contact not there so most appropriate is $3$

$P=P_0+h \rho g+\frac{2 T}{r}$

$P-P_0=h \rho g+\frac{2 T}{r}$

$=0.1 \times 1000 \times 10+\frac{2 \times .075}{10^{-3}}$

$=1000+(0.15)(1000)$

$=1150\,Pa$

Where $T$ is surface tension and $A$ is surface area

$U _{ i }=\left(\frac{75 \times 10^{-5}}{10^{-2}} \frac{ N }{ m }\right) \times\left[4 \pi\left(1 \times 10^{-2}\right)^{2}\right]$

$=75 \times 10^{-3} \times 4 \pi \times 10^{-4}=942 \times 10^{-7}\,J$

To get final radius of drops by volume conservation

$\frac{4}{3} \pi R^{3}=729\left(\frac{4}{3} \pi r^{3}\right)$

$R=\text { Initial radius }$

$r=\text { final radius }$

$r=\frac{R}{(729)^{1 / 3}}=\frac{R}{9}=\frac{1}{9}\,cm$

Final surface energy

$U _{ f }=729[ TA ]$

$=729\left[\frac{75 \times 10^{-5}}{10^{-2}} \frac{ N }{ m }\right] \times\left[4 \pi\left(\frac{1}{9} \times 10^{-2}\right)^{2}\right]$

$=729\left[75 \times 10^{-3} \times \frac{4 \pi \times 10^{-4}}{81}\right]$

$=9\left[942 \times 10^{-7} J \right]$

Gain in surface energy

$\Delta U =9 \times 942 \times 10^{-7}-942 \times 10^{-7}$

$=8 \times 942 \times 10^{-7} J =7536 \times 10^{-7}\,J$

$=7.5 \times 10^{-4}\,J$

$\sigma \frac{ V }{2} g+2 \pi R T=\rho Vg$

$2 \pi RT =\frac{(2 \rho-\sigma)}{2} \frac{4}{3} \pi R ^{3} g ; \quad\left[ V =\frac{4}{3} \pi R ^{3}\right]$

$R ^{3}=\frac{3 T }{(2 \rho-\sigma) g} \Rightarrow R =\sqrt{\frac{3 \times 7.5 \times 10^{-2} N - m ^{-1}}{(2 \rho-\sigma) \times 10}}$

$R=\frac{3}{20 \sqrt{(2 \rho-\sigma)}} m =\frac{15}{\sqrt{2 \rho-\sigma}}\,\,cm$

$\Delta SE = T \Delta A$

$=0.075\left( A _{ f }- A _{1}\right)$

$A _{ i }=4 \pi r ^{2}$

$A _{ t }=4 \pi r _{0}^{2} \times 64$

By volume conservation

$\frac{4}{3} \pi r ^{3}=64 \cdot \frac{4}{3} \pi r_{0}^{3}$

$I _{0}=\frac{ r }{4}$

$A _{ f }=4 \pi\left(\frac{ r }{4}\right)^{2} \cdot 64=16 \pi r ^{2}$

$\Delta SE =0.075\left(16 \pi r ^{2}-4 \pi r ^{2}\right)$

$=0.075\left(12 \pi(0.01)^{2}\right)$

$=2.8 \times 10^{-4}\,J$

$\Rightarrow P_{1}-P_{0}=\frac{4 T}{2}=2$

$\therefore n ^{1 / 3} r = R$

$\therefore$ Total change in surface energy

$=\left(n\left(4 \pi r^{2}\right)-4 \pi R^{2}\right) T$

$\Rightarrow 4 \pi T \left( nr ^{2}- R ^{2}\right)$

$\therefore$ Heat energy

$=\frac{4 \pi T \left( nr ^{2}- R ^{2}\right)}{ J \times \frac{4}{3} \pi R ^{3}}=\frac{3 T }{ J }\left(\frac{ nr ^{2}}{ R ^{3}}-\frac{1}{ R }\right)$

Put $nr ^{3}= R ^{3}$

$\therefore \frac{3 T }{ J }\left(\frac{1}{ r }-\frac{1}{ R }\right)$

$\Rightarrow P=P_{2}-P_{1}$

$\Rightarrow \frac{4 T}{r}=\frac{4 T}{r_{2}}-\frac{4 T}{r_{1}}$

$\Rightarrow \frac{1}{r}=\frac{1}{r_2}-\frac{1}{r_1}$

$\Rightarrow r=\frac{r_{1} r_{1}}{r_{1}-r_{2}}$

$\therefore \mathrm{P}_{\mathrm{atm}}-\frac{2 \mathrm{~T}}{\mathrm{r}_{1}}+\rho \mathrm{g}(\mathrm{x}+\Delta \mathrm{h})=\mathrm{P}_{2 \mathrm{tm}}-\frac{2 \mathrm{~T}}{\mathrm{r}_{2}}+\rho g \mathrm{x}$

$\therefore \rho g \Delta \mathrm{h}=2 \mathrm{~T}\left[\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{I}_{2}}\right]$

$=2 \times 7.3 \times 10^{-2}\left[\frac{1}{2.5 \times 10^{-3}}-\frac{1}{4 \times 10^{-3}}\right]$

$\therefore \Delta \mathrm{h}=\frac{2 \times 7.3 \times 10^{-2} \times 10^{3}}{10^{3} \times 10}\left[\frac{1}{2.5}-\frac{1}{4}\right]$

$=2.19 \times 10^{-3} \mathrm{~m}=2.19 \mathrm{~mm}$

$\Delta {P}=\frac{4 {S}}{{r}_{1}}+\frac{4 {S}}{{r}_{2}}\ldots \text { (i) }$

The excess pressure inside the equivalent soap bubble

$\Delta {P}=\frac{4 {S}}{{R}_{{eq}}} \ldots \text { (ii) }$

From $ (i) \,\& \,(ii)$

$\frac{4 {S}}{{R}_{{eq}}}=\frac{4 {S}}{{T}_{1}}+\frac{4 {S}}{{T}_{2}}$

$\frac{1}{{R}_{{eq}}}=\frac{1}{{T}_{1}}+\frac{1}{{I}_{2}}$

$=\frac{1}{6}+\frac{1}{3}$

${R}_{{eq}}=2 \,{cm}$

$\Delta P _{2}=0.02=4 T / R _{2} \quad \ldots . .(2)$

Equation $(1)+(2)$

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

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

$\frac{V_{1}}{V_{2}}=\frac{R_{1}^{3}}{R_{2}^{3}}=\frac{8 R_{2}^{3}}{R_{2}^{3}}=8$

$h =\frac{2 S \cos \theta}{\rho gr }$

$S=\frac{\rho grh}{2 \cos \theta}$

$=\frac{(900)(10)\left(15 \times 10^{-5}\right)\left(15 \times 10^{-2}\right)}{2}$

$S=1012.5 \times 10^{-4}$

$S=101.25 \times 10^{-3}=101.25 mN / m$

In question closest integer is asked

so closest integer $=101.00$

$\mathrm{B}=\left(\frac{2}{3} \pi \mathrm{R}^{3}\right) \rho \mathrm{g}$

$\mathrm{F}=\mathrm{T}(2 \pi \mathrm{R})$

$\mathrm{m}=\mathrm{d}\left(\frac{4}{3} \pi \mathrm{R}^{3}\right)$

$\left(\frac{2}{3} \pi \mathrm{R}^{3}\right) \rho \mathrm{g}+\mathrm{T}(2 \pi \mathrm{R})=\mathrm{d}\left(\frac{4}{3} \pi \mathrm{R}^{3}\right) \mathrm{g}$

$\mathrm{T}(2 \pi \mathrm{R})=\left(\frac{2}{3} \pi \mathrm{R}^{3}\right) \mathrm{g}[2 \mathrm{d}-\rho]$

$\mathrm{R}=\sqrt{\frac{3 \mathrm{T}}{(2 \mathrm{d}-\rho) \mathrm{g}}}$

$R \rightarrow$ Radius of meniscus.

From figure, $\frac{r}{R}=\cos 30^{\circ}$

$R =\frac{2 r }{\sqrt{3}}=\frac{2 \times 0.15 \times 10^{-3}}{\sqrt{3}}$

$=\frac{0.3}{\sqrt{3}} \times 10^{-3} m$

Height of capillary

$h =\frac{2 T }{\rho gR }=2 \sqrt{3} T$

$h =\frac{2 \times 0.05}{667 \times 10 \times\left(\frac{0.3 \times 10^{-3}}{\sqrt{3}}\right)}$

$h=0.087 m$

$R = {\left( {\frac{{3k}}{{4\pi }}t} \right)^{1/3}}$

${P_{in}} = {P_{atm}} + \frac{{4T}}{R}$

$ \Rightarrow h \propto \frac{1}{r}$

When radius becomes double height become half 

$\therefore h' = \frac{h}{2}$

$Now,\,M = \pi {r^2}h \times \rho \,and\,M'$

$ = \pi {\left( {2r} \right)^2}\left( {h/2} \right) \times \rho  = 2M.$

$h = \frac{{2{s_2}\cos {\theta _2}}}{{{r_2}{\rho _2}g}}$

$ \Rightarrow \frac{{{r_1}}}{{{r_2}}} = \frac{2}{5}$

$\left( {{P_1}} \right)\frac{4}{3}\pi {r^3} = \left( {{P_2}} \right)\frac{4}{3}\pi \frac{{125{r^3}}}{{64}}$

$\frac{{\rho g\left( {10} \right) + \rho gh}}{{\rho g\left( {10} \right)}} = \frac{{125}}{{64}}$

$640 + 64\,h = 1250$

$On\,solving\,we\,get\,h = 9.5\,m$

${P_2} = {P_0} + \frac{{4T}}{6} + \frac{{4t}}{4}$

$or,\,\,\,{P_2} - {P_0} = \frac{{4T}}{6} + \frac{{4T}}{4}$              $...\left( i \right)$

Let $r$ be the radius of bubble with pressure difference

${P_2} - {P_0}\,so,$

${P_2} - {P_0} = \frac{{4T}}{r}$                                           $...(ii)$

$From\,e{q^n}\,\left( i \right)\,and\,\left( {ii} \right),$

$\frac{{4T}}{r} = \frac{{4T}}{6} + \frac{{4T}}{4}$

$\frac{1}{r} = \frac{1}{6} + \frac{1}{4} \Rightarrow r = 2.4\,cm$

Pressure difference across a double curvature film $\Delta P=2 T\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right)$

$\because \quad R_{1}=R$ and $R_{2}=\infty$

$\Delta P=2 T\left(\frac{1}{R}+\frac{1}{\infty}\right)$

$\Delta P=\frac{2T}{R}$

$r = 0.1\,cm = {10^{ - 3}}m$

Surface tension of liquid,

$S = 0.06\,N/m = 6 \times {10^{ - 2}}\,N/m$

Density of liquid, $\rho  = {10^3}\,kg/{m^3}$

Excess pressure inside the bubble,

${\rho _{exe}} = 1100N{m^{ - 2}}$

Depth of bubble below the liquid surface,

$h=?$

As we know,

${\rho _{Excess}} = h\rho g + \frac{{2s}}{r}$

$ \Rightarrow 1100 = h \times {10^3} \times 9.8 + \frac{{2 \times 6 \times {{10}^{ - 2}}}}{{{{10}^{ - 3}}}}$

$ \Rightarrow 1100 - 9800h + 120$

$ \Rightarrow 9800h = 1100 - 120$

$ \Rightarrow h = \frac{{980}}{{9800}} = 0.1\,m$

$g' = g\left( {1 - \frac{d}{R}} \right)$

$pressure,\,P = \rho gh$

$Hence\,ratio,\frac{x}{y}is\left( {1 - \frac{d}{R}} \right)$

Buoyant force = force due to surface tension

$\int T \times d l \sin \theta=\frac{4}{3} \pi R^{3} \rho_{w} g$

$\Rightarrow T \times 2 \pi r \times \frac{r}{R}=\frac{4}{3} \pi R^{3} \rho_{w} g$

$\Rightarrow r^{2}=\frac{2 R^{4} \rho_{w g} g}{3}$

$\Rightarrow r=R^{2} \sqrt{\frac{2 \rho_{w D g}}{3 T}}$

$\therefore {P_1} = P + \frac{{4T}}{{{R_1}}}.\,\,\,{P_2} = P + \frac{{4T}}{{{R_2}}}\,\,and\,\,{P_3} = P + \frac{{4T}}{{{R_3}}}$

$Also\,{P_1}{V_1} + {P_2}{V_2} = {P_3}{V_3}$

$\therefore \left( {P + \frac{{4T}}{{{R_1}}}} \right)\frac{{4\pi }}{3}R_1^3 + \left( {P + \frac{{4T}}{{{R_2}}}} \right)\frac{{4\pi }}{3}R_2^3$

$ = \left( {p + \frac{{4T}}{{{R_3}}}} \right)\frac{{4\pi }}{3}R_3^3$

$P\left( {\frac{{4\pi }}{3}R_1^3 + \frac{{4\pi }}{3}R_2^3 - \frac{{4\pi }}{3}R_3^2} \right)$

$ + \frac{{4T}}{3}\left( {4\pi R_1^2 + 4\pi R_2^2 - 4\pi R_3^2} \right) = 0$

$P\left( {{V_1} + {V_2} - {V_3}} \right) + \frac{{4T}}{3}\left( {{S_1} - {S_2} - {S_3}} \right) = 0$

$PV + \frac{{4T}}{3}S = 0\,\,\,\,\,\,\, \Rightarrow \,\,\,\,3PV + 4ST = 0$

$4\pi {R^2}\Delta R\rho L = 4\pi T\left[ {{R^2} - {{\left( {R - \Delta R} \right)}^2}} \right]$

$ \Rightarrow \rho {R^2}\Delta RL = T\left[ {{R^2} - {R^2} + 2R\Delta R - \Delta {R^2}} \right]$

$ \Rightarrow \rho {R^2}\Delta RL = T2R\Delta R\,\,\left[ {\Delta R\,is\,very\,small} \right]$

$ \Rightarrow R = \frac{{2T}}{{\rho L}}$

$Capillary\,rise\,h = \frac{{2T\cos \theta }}{{rdg}}$

As surface tension $T$ decreases with rise in temperature hence capillary rise also decreases.

$\cos \theta  = \frac{{{T_{SA}} - {T_{SL}}}}{{{T_{LA}}}}$

when water is on a waxy or oily surface

${T_{SA}} < {T_{SL}}\cos \theta \,is\,negative\,i.e.,$

${90^ \circ } < \theta  < {180^ \circ }$

$i.e.,\,angle\,of\,contact\,\theta \,increases$

$And\,for\,\theta  > {90^ \circ }\,liquid\,level\,in\,capillary\,tube$

$fall.\,i.e.,\,h\,decreases$

$PV ^\gamma=\text { constant [adiabatic process] }$

$\left( Pa _1+\frac{4 S }{ r _1}\right)\left(\frac{4}{3} \pi r _1^3\right)^{5 / 3}=\left( P _{2,}+\frac{4 S }{ r _2}\right)\left(\frac{4}{3} \pi r _2^3\right)^{5 / 3}$

$\frac{r_1^3}{r_2^3}=\left(\frac{P_{2_2}+\frac{4 S}{r_2}}{P_{a_1}+\frac{4 S}{r_1}}\right)$

$P ^{1-5} T ^y=\text { constant }$

$\left( P _{2,}+\frac{4 S }{ r _2}\right)^{1-5 / 3} T _2^{5 / 3}=\left( P _{ a _1}+\frac{4 S }{ r _1}\right)^{1-5 / 3} T _1^{5 / 3}$

$\left(\frac{T_2}{T_1}\right)^{s / 3}=\left(\frac{P_{2_1}+\frac{4 S}{I_1}}{P_{2_2}+\frac{4 S}{I_2}}\right)^{-2 / 3}$

$\left(\frac{T_2}{T_1}\right)^{s / 2}=\left(\frac{P_{a_2}+\frac{4 S}{r_2}}{P_{a_1}+\frac{4 S}{r_1}}\right)$

$(D)$ is correct

$\rho g h= T \left(\frac{1}{ R _1}+\frac{1}{ R _2}\right)$

$R _1 \gg R _2$

$\text { So } \frac{1}{ R _1} \ll \frac{1}{ R _2} \text { and } R _2= h / 2$

$\therefore \rho h = T \left[\frac{1}{ R _1}+\frac{1}{ R _2}\right]= T \left[0+\frac{1}{ h / 2}\right]$

$h ^2=\frac{2 T }{\rho g }$

$h =\sqrt{\frac{2 T }{\rho g}}=\sqrt{\frac{2 \times 0.07}{10^3 \times 10}}=\sqrt{\frac{14 \times 100}{10^4 \times 100}}$

$h =\sqrt{14} mm =3.741$

$\Rightarrow h _1=75 mm$ (in $T1$) [If we assume entire tube of $T 1$ ]

$\Rightarrow h _2=\frac{2 \times 0.075 \times \cos 60^{\circ}}{1000 \times 10 \times 0.2 \times 10^{-3}}=37.5 mm$ (in $T2$) [If we assume entire tube of $T2$]

Option $(1)$ : Since contact angles are different so correction in the height of water column raised in the tube will be different in both the cases, so option $(1)$ is correct

Option $(2)$ : If joint is $5 cm$ is above water surface, then lets say water crosses the joint by height $h$, then:

$\Rightarrow  P_0-\frac{2 T}{r}+\rho g h+\rho g \times 5 \times 10^{-2}$

$=P_0$

$\Rightarrow  \cos \theta=\frac{R}{r}, r=\frac{R}{\cos \theta}$

$\Rightarrow \rho g\left( h +5 \times 10^{-2}\right)=\frac{2 T \cos \theta}{ R }$

(image)

$\Rightarrow h =\frac{2 \times 0.075 \times \cos 60}{0.2 \times 10^{-3} \times 1000 \times 10}-5 \times 10^{-2}$

$\Rightarrow h =-$ ve, not possible, so liquid will not cross the interface, but angle of contact at the interface will change, to balance the pressure,

So option $(2)$ is wrong.

Option $(3)$ : If interface is $8 cm$ above water then water will not even reach the interface, and water will rise till $7.5 cm$ only in $T 1$, so option $(3)$ is right.

Option $(4)$ : If interface is $5 cm$ above the water in vessel, then water in capillary will not even reach the interface. Water will reach only till $3.75 cm$, so option $(4)$ is right.

where $\mathrm{R}$ is the radius of meniscus.

$h=\frac{2 \sigma}{\rho g R}$

$\mathrm{R}=\frac{\mathrm{r}}{\cos \theta}$, where $\mathrm{r}$ is the radius of capillary and $\theta$ is the angle of contact.

$h=\frac{2 \sigma \cos \theta}{\rho g r}$

$(A)$ For a given material, $\theta$ is constant.

Therefore, $\mathrm{h} \propto \frac{1}{\mathrm{r}}$

$(B)$ $\mathrm{h}$ depends on $\sigma$ as $\mathrm{h} \propto \sigma$.

$(C)$ If lift is going up with constant acceleration a,

$\mathrm{g}_{\text {eff }}=\mathrm{g}+\mathrm{a}$

$\mathrm{h}=\frac{2 \sigma \cos \theta}{\rho(\mathrm{g}+\mathrm{a}) \mathrm{r}} ; \mathrm{h} \text { decreases }$

$(D)$ $\mathrm{h}$ is proportional to $\cos \theta$.

$=S \times 4 \pi R^2\left[\mathrm{~K}^{1 / 3}-1\right]$

$\Rightarrow \mathrm{K}^{1 / 3}=\frac{\Delta \mathrm{U}}{4 \pi \mathrm{R}^2 \mathrm{~S}}+1=101$

$\Rightarrow \mathrm{K} \approx 10^6$

Using Pressure method : $P _0-\frac{2 S}{R_c}+h \rho g=P_0$

$2.$ $\frac{2 \pi \mathrm{r}^2 \mathrm{~T}}{\mathrm{R}}=\mathrm{mg}=\frac{4}{3} \pi \mathrm{R}^3 \rho g$

$3.$ Surface energy $=\mathrm{T}\left(4 \pi \mathrm{R}^2\right)=2.7 \times 10^{-6} \mathrm{~J}$

$P_A=16 N / m ^2$

$P_a=P_0+\frac{4 T}{r_B}=8+\frac{4 \times 0.04}{0.04}$

$P_B=12 N / m ^2$

for bubble $A, P V=n R T$

$(16)$ $\frac{4}{3} \pi(0.02)^3=n_A R T \ldots$ $(1)$

for bubble $B$

$(12)$ $\left(\frac{4}{3} \pi(0.04)^3\right)=n_B R T \ldots \ldots$

dividing $e q^n$ ( $i$ ) and $(2)$

$\frac{n_A}{n_6}=\frac{1}{6}, \frac{n_B}{n_A}=6$

Hence, option $\mathrm{B}$ is correct.

For example mercury surface in glass capillary is convex.