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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\therefore \;\frac{{{r_2}}}{{{r_1}}} = 4$ 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$

$\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$

$T = 72.8\;dyne/cm$

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

= (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.

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

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

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

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

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