Let P₁ and P₂ be the pressure inside the two spherical soap bubbles, then
\(P_{1}-P=\frac{4T}{r_{1} } \Rightarrow P_{1}=P+\frac{4T}{r_{1}}\)
Similarly,
\(P_{2}-P=\frac{4T}{r_{2} } \Rightarrow P_{2}=P+\frac{4T}{r_{2}}\)
When bubbles coalesce
P₁V₁ + P₂V₂ = PV   ---- (1)
∴The pressure inside the new bubble,
\(P=P+\frac{4T}{r}\)
Substituting for P, P₁ and P₂ in equation (1),
P₁V₁ + P₂V₂= PV
\((P+\frac{4T}{r_{1}})V_{1}+(P+\frac{4T}{r_{2}})V_{2}=(P+\frac{4T}{r})V\)
\((P+\frac{4T}{r_{1}})\frac{4}{3}\pi r^{3}_{1} +(P+\frac{4T}{r_{2}})\frac{4}{3}\pi r^{3}_{2}=(P+\frac{4T}{r})\frac{4}{3}\pi r^{3}\)
[∵ Volume of the sphere V = \(\frac{4}{3} \pi r^{3}\)]
or \(\frac{4}{3} \pi P(r^{3}_{1}+r^{3}_{2}-r^{3})+\frac{16\pi T}{3}(\frac{r^{3}_{1}}{r_{1}}+\frac{r^{3}_{2}}{r_{2}}-\frac{r^{3}}{r})=0\)
\(\frac{4}{3} \pi P(r^{3}_{1}+r^{3}_{2}-r^{3})+\frac{16\pi T}{3}(r^{2}_{1}+r^{2}_{2}-r^{2})=0\)   --- (2)
Given change in Volume,
\(V=\frac{4}{3}\pi r^{3}_{1}+\frac{4}{3}\pi r^{3}_{2}-\frac{4}{3}\pi r^{3}\)    --- (3)
Change in Area,
A = 4ㅠr²₁ + 4ㅠr²₂ - 4ㅠr²   --- (4)
Using equations (3) and (4) in equation (2) we get,
\(P(\frac{4}{3} \pi r_{1}^{3}+\frac{4}{3} \pi r_{2}^{3}-\frac{4}{3} \pi r^{3})+\frac{4}{3}(4\pi r^{2}_{1}+4\pi r^{2}_{2}-4\pi r^{2})=0\)
\(P(V)+\frac{4T}{3}(A)=0\)
\(PV+\frac{4T}{3}A=0 \Rightarrow 3 PV+4TA=0\)
∴ 3PV + 4TA = 0.

The maximum constant velocity acquired by a body while falling through a viscous fluid is called its terminal velocity.
Terminal velocity, \(V=\frac{2}{9}. r^{2} \frac{(\rho-\sigma)g}{\eta}\)
\(⇒ \frac{V}{r{2}}=\frac{2}{9}[\frac{(\rho-\sigma)g}{\eta}]\)   
\(\frac{V}{r^{2}}=\frac{2g}{9\eta}(\rho-\sigma)\)    --- (1)
Similarly, the bigger drip,
\(\frac{V^{'}}{r^{2}}=\frac{2g}{9\eta}(\rho-\sigma)\)   --- (2)
Dividing equ (1) by (2),
\(\frac{V}{V^{'}}=\frac{r^{2}}{R^{2}}\Rightarrow V^{'}=V(\frac{R}{r})^{2}\)  -- (3)
IfN drops, then
Volume of one big drop =Volume of N droplets.
\(\frac{4}{3} \pi R^{3}=N(\frac{4}{3}\pi r^{3})\)      [∵ Volume of the sphere \(V=\frac{4}{3} \pi r^{3}\) ]
R³ = N(r³)
R³ = N^1/3(r)
​​​​​​∴ Terminal velocity of bigger drop,
=\((\frac{R}{r})^{2} \times V\) from equation --- (1)
\(= (\frac{N^{\frac{1}{3}}}{\frac{r}{r}})^{2} \times V=(N^{1/3})^{2} \times V\)  
= N^2/3 \(\times\)V from equation   --- (2)

Given L₁ = L₂ = L ; A₁ = A₂ = A
k₁ = 3 units, k₂ = 4 units
If R₁ & R₂ are the thermal resistances of two rods then
\(R_{1}=\frac{L_{1}}{k_{1}A_{1}}=\frac{L}{3\times A}\)
\(R_{2}=\frac{L_{2}}{k_{2}A_{2}}=\frac{L}{4\times A}\)
If R_s is the equivalent thermal resistance of 2 rods in series and k_s is the equivalent thermal conductivity
\(R_{s}=\frac{L_{1}+L_{2}}{k_{s}\times A}=\frac{2L}{k_{s}A}\)
as \(R_{s}=R_{1}+R_{2}\),
So, \(\frac{2L}{k_{s}A}=\frac{L}{3A}+\frac{L}{4A}=\frac{L}{12A}\)
\(k_{s}=\frac{12\times 2}{7}=3.428=3.43\)

From ideal gas equation
pV = nRT (or) \(r=\frac{nRT}{V}\)
As gas expands such that
ρT² = a constant = C
So \((\frac{nRT}{V})T^{2}\) = C (or) T³ α V
(or) T³ = kV
differentiating it w.r.t.T, we have
\(3T^{2}=k \frac{dV}{dT}\)
Dividing it by T³=kV
\(\frac{3}{T}=\frac{1}{V}.\frac{dV}{dT}\)
Coefficient of volume expansion r = \(\frac{1}{V}.\frac{dV}{dT}=\frac{3}{T}\)

Here ρ = 7.8 g/cm³ ; rw = 10 cm/s
ηw = 8.5\(\times\)10^-4 ρa.s
Pg=1.2g/cm³ ; ηg =13.2 ρa s; Vg=?
Terminal velocity V = \(\frac{2r^{2}(\rho-\rho_{0})g}{9\eta}\)
\(v \propto \frac{(\rho-\rho_{0})}{\eta}\)
When ball falls in water, then
\(V_{w} \propto \frac{(\rho-\rho_{w})}{\eta_{w}}\)
When ball falls in glycerine
\(V_{g} \propto \frac{(\rho-\rho_{g})}{\eta_{g}}\)
\(\frac{V_{g}}{V_{w}}=(\frac{\rho-\rho_{g}}{\rho-\rho_{w}})\frac{\eta_{w}}{\eta_{g}}\)
\(V_{g}={V_{w}}=(\frac{\rho-\rho_{g}}{\rho-\rho_{w}})\frac{\eta_{w}}{\eta_{g}}\)
= 10\((\frac{7.8-1.2}{7.8-1})\times \frac{8.5 \times 10^{-4}}{13.2}\)
= 6.25 \(\times\)10^-4 cm/s