Surface Energy Questions in English

(A) Let $R$ be the radius of the large drop and $r$ be the radius of each small droplet.
Since the volume remains constant,we have $\frac{4}{3} \pi R^3 = 512 \times \frac{4}{3} \pi r^3$.
This simplifies to $R^3 = 512 r^3$,which gives $R = 8r$ or $r = \frac{R}{8}$.
The initial surface energy is $U = 4 \pi R^2 T$,where $T$ is the surface tension.
The new surface energy $U'$ is the sum of the surface energies of $512$ droplets: $U' = 512 \times (4 \pi r^2 T)$.
Substituting $r = \frac{R}{8}$ into the equation: $U' = 512 \times 4 \pi (\frac{R}{8})^2 T$.
$U' = 512 \times 4 \pi \frac{R^2}{64} T = 8 \times (4 \pi R^2 T) = 8 U$.

(D) The work done in blowing a soap bubble of radius $r$ is given by $W = T \times \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area. Since a soap bubble has two surfaces,$\Delta A = 2 \times (4 \pi r^2) = 8 \pi r^2$.
Thus,$W_1 = 8 \pi r^2 T$,where $T$ is the surface tension at room temperature.
For the second bubble of radius $2r$ formed from the heated solution,the work done is $W_2 = 8 \pi (2r)^2 T'$,where $T'$ is the surface tension at the higher temperature.
$W_2 = 8 \pi (4r^2) T' = 32 \pi r^2 T'$.
Since the soap solution is heated,its surface tension decreases,meaning $T' < T$.
Comparing $W_1$ and $W_2$: $W_2 = 4 W_1 \times (T'/T)$.
Since $T' < T$,it follows that $W_2 < 4 W_1$.

(A) The volume of a soap bubble is given by $V = \frac{4}{3} \pi r^3$,which implies $r = (\frac{3V}{4\pi})^{1/3}$.
The work done $W$ in blowing a soap bubble of radius $r$ is equal to the surface tension $T$ multiplied by the change in surface area. Since a soap bubble has two surfaces (inner and outer),the total surface area is $2 \times (4 \pi r^2) = 8 \pi r^2$.
Thus,$W = 8 \pi r^2 T$.
Substituting the expression for $r$,we get $W = 8 \pi (\frac{3V}{4\pi})^{2/3} T$.
This shows that $W \propto V^{2/3}$.
For a bubble of volume $2V$,the new work $W'$ is given by $\frac{W'}{W} = (\frac{2V}{V})^{2/3} = 2^{2/3}$.
Therefore,$W' = 2^{2/3} W$.

(A) soap bubble has two surfaces (inner and outer), so the work done $W$ in changing its radius from $r_1$ to $r_2$ is given by $W = T \times \Delta A \times 2$, where $\Delta A = 4\pi(r_2^2 - r_1^2)$.
Thus, $W = 8\pi T(r_2^2 - r_1^2)$.
Given: $T = 0.03 \ Nm^{-1}$, $r_1 = 3 \ cm = 0.03 \ m$, $r_2 = 5 \ cm = 0.05 \ m$.
Substituting the values:
$W = 8 \times \pi \times 0.03 \times ((0.05)^2 - (0.03)^2)$
$W = 8 \times \pi \times 0.03 \times (0.0025 - 0.0009)$
$W = 8 \times \pi \times 0.03 \times 0.0016$
$W = 0.24 \pi \times 0.0016 = 0.000384 \pi \ J$
$W = 0.384 \pi \ mJ \approx 0.4 \pi \ mJ$.

(B) The volume of a spherical soap bubble is given by $V = \frac{4}{3} \pi r^3$,which implies $V \propto r^3$ or $r \propto V^{1/3}$.
For a soap bubble,the work done $W$ in blowing it to a radius $r$ is $W = T \times \Delta A$,where $\Delta A = 2 \times (4 \pi r^2) = 8 \pi r^2$ (since a soap bubble has two surfaces).
Therefore,$W \propto r^2$.
Substituting $r \propto V^{1/3}$,we get $W \propto (V^{1/3})^2 = V^{2/3}$.
Let $W_1 = W$ for volume $V_1 = V$,and $W_2$ be the work for volume $V_2 = 2V$.
Then,$\frac{W_2}{W_1} = \left( \frac{V_2}{V_1} \right)^{2/3} = (2)^{2/3} = (2^2)^{1/3} = 4^{1/3}$.
Thus,$W_2 = 4^{1/3}W$.

(C) Let $r$ be the radius of each small drop and $R$ be the radius of the big drop.
Since the volume remains constant,the volume of $1000$ small drops equals the volume of the big drop: $1000 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$.
This simplifies to $R^3 = 1000 r^3$,so $R = 10r$.
The initial surface energy $U_i$ of $1000$ small drops is $U_i = 1000 \times (4 \pi r^2 T)$,where $T$ is the surface tension.
The final surface energy $U_f$ of the big drop is $U_f = 4 \pi R^2 T$.
Substituting $R = 10r$,we get $U_f = 4 \pi (10r)^2 T = 400 \pi r^2 T$.
The ratio of final surface energy to initial surface energy is $\frac{U_f}{U_i} = \frac{400 \pi r^2 T}{1000 \times 4 \pi r^2 T} = \frac{400}{4000} = \frac{1}{10}$.

(C) soap bubble has two surfaces (inner and outer). The work done $W$ in blowing a soap bubble of radius $R$ is given by the formula: $W = \text{Surface Tension} \times \text{Change in Surface Area} \times 2$.
$W = T \times (4 \pi R^2) \times 2 = 8 \pi R^2 T$.
For a bubble of radius $2R$, the work done $W'$ is:
$W' = T \times (4 \pi (2R)^2) \times 2 = 8 \pi (4R^2) T = 32 \pi R^2 T$.
Comparing $W'$ with $W$:
$W' = 4 \times (8 \pi R^2 T) = 4W$.

(C) The length of the wires is $l = 10 \text{ cm} = 0.1 \text{ m}$.
The increase in separation is $\Delta x = 1 \text{ mm} = 10^{-3} \text{ m}$.
$A$ water film has two surfaces,so the change in area $\Delta A$ is given by $2 \times (l \times \Delta x)$.
$\Delta A = 2 \times (0.1 \text{ m} \times 10^{-3} \text{ m}) = 2 \times 10^{-4} \text{ m}^2$.
The work done $W$ is given by $W = T \times \Delta A$,where $T$ is the surface tension.
$W = (7.2 \times 10^{-2} \text{ N/m}) \times (2 \times 10^{-4} \text{ m}^2) = 14.4 \times 10^{-6} \text{ J} = 1.44 \times 10^{-5} \text{ J}$.

(B) The surface tension $T = 3 \times 10^{-2} \,N/m$.
The area of the film $A = 20 \,cm \times 5 \,cm = 100 \,cm^2 = 100 \times 10^{-4} \,m^2 = 10^{-2} \,m^2$.
A soap film has two surfaces, so the total increase in surface area is $2A$.
The work done $W$ is given by $W = T \times (2A)$.
Substituting the values: $W = 3 \times 10^{-2} \times 2 \times 10^{-2} = 6 \times 10^{-4} \,J$.

(D) soap bubble has two surfaces (inner and outer). The work done $W$ in changing the surface area is given by $W = T \times \Delta A \times 2$.
Initial diameter $D_1 = D$, so initial radius $r_1 = D/2$. Initial surface area $A_1 = 4 \pi r_1^2 = 4 \pi (D/2)^2 = \pi D^2$.
Final diameter $D_2 = 2D$, so final radius $r_2 = D$. Final surface area $A_2 = 4 \pi r_2^2 = 4 \pi D^2$.
Change in area $\Delta A = A_2 - A_1 = 4 \pi D^2 - \pi D^2 = 3 \pi D^2$.
Since the bubble has two surfaces, the total change in area is $2 \times \Delta A = 2 \times 3 \pi D^2 = 6 \pi D^2$.
Therefore, the work done $W = T \times 6 \pi D^2 = 6 \pi TD^2$.

(B) soap bubble has two surfaces (inner and outer), so its total surface area is $2 \times 4 \pi R^2 = 8 \pi R^2$.
Initial surface energy $E_i = T \times (8 \pi R^2) = 8 \pi R^2 T$.
When the radius is doubled, the new radius becomes $R' = 2R$.
The new surface area is $2 \times 4 \pi (2R)^2 = 2 \times 4 \pi (4R^2) = 32 \pi R^2$.
Final surface energy $E_f = T \times (32 \pi R^2) = 32 \pi R^2 T$.
Work done $W = E_f - E_i$.
$W = 32 \pi R^2 T - 8 \pi R^2 T = 24 \pi R^2 T$.

(A) Molecules in the bulk of a liquid are surrounded by other molecules on all sides,resulting in a net attractive force of zero. However,molecules on the surface are attracted only by the molecules below them,as there are no liquid molecules above the surface. To bring a molecule from the bulk to the surface,work must be done against these inward attractive forces. This work is stored as potential energy. Therefore,molecules on the surface of a liquid in equilibrium possess maximum potential energy compared to those in the bulk.

(D) When the surface area of a liquid is increased,molecules from the interior of the liquid rise to the surface.
As these molecules reach the surface,work is done against the cohesive force.
This work is stored in the molecules in the form of potential energy.
Thus,the potential energy of the molecules lying on the surface is greater than that of the molecules in the interior of the liquid.

(C) The surface energy $(E)$ of a soap film is given by the product of surface tension $(T)$ and the total surface area. Since a soap film has two surfaces,the total area is $2A$.
$E = T \times 2A$
When the area of the frame is reduced by $50 \%$,the new area becomes $A' = A - 0.5A = 0.5A = A/2$.
The new surface energy $(E_1)$ is:
$E_1 = T \times 2(A/2) = T \times A$
The percentage change in surface energy is calculated as:
$\text{Percentage change} = \frac{E - E_1}{E} \times 100$
$\text{Percentage change} = \frac{2TA - TA}{2TA} \times 100 = \frac{TA}{2TA} \times 100 = 50 \%$
Thus,the energy of the soap film changes by $50 \%$.

(C) The work done to blow a soap bubble of radius $r$ is given by $W = T \times \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area. Since a soap bubble has two surfaces,$\Delta A = 2 \times (4 \pi r^2) = 8 \pi r^2$.
For the first bubble of radius $R$: $W_1 = 8 \pi R^2 T_1$.
For the second bubble of radius $2R$: $W_2 = 8 \pi (2R)^2 T_2 = 32 \pi R^2 T_2$.
If the temperature remained constant $(T_1 = T_2)$,the work $W_2$ would be exactly $4 W_1$.
However,heating the solution decreases the surface tension $(T_2 < T_1)$.
Therefore,$W_2 = 4 W_1 \times (T_2 / T_1)$. Since $T_2 < T_1$,it follows that $W_2 < 4 W_1$.

(D) The work done in blowing a soap bubble is given by $W = T \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area. Since a soap bubble has two surfaces,$\Delta A = 2 \times (4 \pi r^2) = 8 \pi r^2$.
For a spherical bubble,the volume is $V = \frac{4}{3} \pi r^3$,which implies $r^3 = \frac{3V}{4\pi}$,or $r = (\frac{3V}{4\pi})^{\frac{1}{3}}$.
Substituting $r$ into the area formula,we get $A = 8 \pi (\frac{3V}{4\pi})^{\frac{2}{3}}$,which shows that $A \propto V^{\frac{2}{3}}$.
Therefore,the work done $W$ is proportional to $V^{\frac{2}{3}}$,i.e.,$W \propto V^{\frac{2}{3}}$.
If the volume changes from $V$ to $2V$,the new work $W'$ is given by $\frac{W'}{W} = (\frac{2V}{V})^{\frac{2}{3}} = 2^{\frac{2}{3}} = (2^2)^{\frac{1}{3}} = 4^{\frac{1}{3}}$.
Thus,$W' = W(4)^{\frac{1}{3}}$.

(A) Let the radius of the large drop be $R = D/2$. Let the radius of each small drop be $r$.
Since the volume remains constant,the volume of the large drop equals the sum of the volumes of $n = 3375$ small drops:
$\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$
$R^3 = 3375 r^3$
Taking the cube root on both sides: $R = 3375^{1/3} r = 15r$,so $r = R/15$.
The initial surface area is $A_i = 4 \pi R^2$.
The final surface area is $A_f = n \times 4 \pi r^2 = 3375 \times 4 \pi (R/15)^2 = 3375 \times 4 \pi (R^2 / 225) = 15 \times 4 \pi R^2 = 60 \pi R^2$.
The change in surface area is $\Delta A = A_f - A_i = 60 \pi R^2 - 4 \pi R^2 = 56 \pi R^2$.
The change in surface energy is $\Delta U = S \times \Delta A = S \times 56 \pi R^2$.
Substituting $R = D/2$,we get $\Delta U = S \times 56 \pi (D/2)^2 = S \times 56 \pi (D^2 / 4) = 14 \pi D^2 S$.

(B) Let the radius of the large drop be $R = 1 \ cm = 10^{-2} \ m$. Let the number of small droplets be $n = 10^6$. Let the radius of each small droplet be $r$.
Since the volume remains constant,$\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$.
$r^3 = \frac{R^3}{n} = \frac{(10^{-2})^3}{10^6} = \frac{10^{-6}}{10^6} = 10^{-12} \ m^3$.
So,$r = 10^{-4} \ m$.
The change in surface energy $\Delta U$ is given by $\Delta U = T \times \Delta A$,where $\Delta A$ is the change in surface area.
$\Delta A = n(4 \pi r^2) - 4 \pi R^2 = 4 \pi (n r^2 - R^2)$.
$\Delta A = 4 \pi (10^6 \times (10^{-4})^2 - (10^{-2})^2) = 4 \pi (10^6 \times 10^{-8} - 10^{-4}) = 4 \pi (10^{-2} - 10^{-4}) = 4 \pi (0.01 - 0.0001) = 4 \pi (0.0099) \ m^2$.
$\Delta U = 35 \times 10^{-3} \times 4 \times 3.1416 \times 0.0099 \approx 4356 \times 10^{-6} \ J$.

(D) Let the radius of the bigger drop be $R$. Since the volume remains constant during the merger of two drops,we have:
$2 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$
$R^3 = 2r^3 \Rightarrow R = 2^{1/3} r$
The surface energy $E$ of the bigger drop is given by the product of surface tension $T$ and its surface area $A = 4 \pi R^2$:
$E = T \times 4 \pi R^2$
Substituting $R = 2^{1/3} r$ into the equation:
$E = T \times 4 \pi (2^{1/3} r)^2$
$E = T \times 4 \pi \times 2^{2/3} r^2$
$E = 4 \times 2^{2/3} \pi r^2 T$
Since $4 = 2^2$,we have $2^2 \times 2^{2/3} = 2^{2 + 2/3} = 2^{8/3}$.
Therefore,$E = 2^{8/3} \pi r^2 T$.

(D) Initial surface area $A_i = 4 \pi r^2 + 4 \pi (2r)^2 = 4 \pi r^2 + 16 \pi r^2 = 20 \pi r^2$.
Initial surface energy $E_i = A_i S = 20 \pi r^2 S$.
Volume conservation: $\frac{4}{3} \pi r^3 + \frac{4}{3} \pi (2r)^3 = \frac{4}{3} \pi R^3$.
$r^3 + 8r^3 = R^3 \implies R^3 = 9r^3 \implies R = 9^{1/3} r$.
Final surface area $A_f = 4 \pi R^2 = 4 \pi (9^{1/3} r)^2 = 4 \pi (9^{2/3}) r^2$.
Given $9^{2/3} = 4.326$,so $A_f = 4 \pi (4.326) r^2 = 17.304 \pi r^2$.
Final surface energy $E_f = 17.304 \pi r^2 S$.
Energy released $\Delta E = E_i - E_f = 20 \pi r^2 S - 17.304 \pi r^2 S = 2.696 \pi r^2 S \approx 2.7 \pi r^2 S$.

(A) The energy required to break a drop is equal to the increase in surface area multiplied by the surface tension $T$.
Surface area of the initial large drop = $4 \pi R^2$.
Total surface area of $n$ smaller drops = $n \times (4 \pi r^2) = 4 \pi r^2 n$.
Increase in surface area = (Final surface area) - (Initial surface area) = $4 \pi r^2 n - 4 \pi R^2$.
Therefore,the energy required $\Delta U = (4 \pi r^2 n - 4 \pi R^2) T$.

(D) Let the radius of the big drop be $R$ and the radius of each small drop be $r$.
Given the number of small drops,$n = 1000$.
Since the volume remains constant during the process,we have:
$V_{\text{final}} = V_{\text{initial}}$
$\frac{4}{3} \pi R^3 = n \left( \frac{4}{3} \pi r^3 \right)$
$R^3 = 1000 r^3 \Rightarrow R = 10r$ ...$(i)$
Initial surface energy,$SE_i = n \times (T \times 4 \pi r^2)$,where $T$ is the surface tension.
Final surface energy,$SE_f = T \times 4 \pi R^2$.
The ratio of final surface energy to total initial surface energy is:
$\frac{SE_f}{SE_i} = \frac{T \times 4 \pi R^2}{n \times T \times 4 \pi r^2} = \frac{R^2}{n r^2}$
Substituting $R = 10r$ and $n = 1000$:
$\frac{SE_f}{SE_i} = \frac{(10r)^2}{1000 r^2} = \frac{100 r^2}{1000 r^2} = \frac{1}{10}$.
Thus,the ratio is $1: 10$.

(B) The surface energy $E$ of a liquid drop is given by the formula $E = T \times A$,where $T$ is the surface tension and $A$ is the surface area of the drop.
The surface area $A$ of a spherical drop of radius $r$ is $4 \pi r^2$.
Thus,$E = T \times 4 \pi r^2$.
According to the problem,the total energy is $2 \pi$ times the surface tension,so $E = 2 \pi T$.
Equating the two expressions for $E$: $4 \pi r^2 T = 2 \pi T$.
Dividing both sides by $2 \pi T$ (assuming $T \neq 0$),we get $2 r^2 = 1$,which means $r^2 = 1/2$.
Therefore,$r = 1/\sqrt{2}$.
The diameter $d$ of the drop is $2r = 2 \times (1/\sqrt{2}) = \sqrt{2} \ m$.

(D) soap bubble has two surfaces (inner and outer).
Initially, the surface area of the soap bubble is $A_1 = 2 \times (4 \pi r^2) = 8 \pi r^2$.
Under isothermal conditions, the radius becomes $2r$.
The final surface area is $A_2 = 2 \times (4 \pi (2r)^2) = 2 \times (16 \pi r^2) = 32 \pi r^2$.
The increase in surface area is $\Delta A = A_2 - A_1 = 32 \pi r^2 - 8 \pi r^2 = 24 \pi r^2$.
The energy spent (work done) is given by $W = T \times \Delta A$.
Therefore, $W = T \times 24 \pi r^2 = 24 \pi T r^2$.

(B) soap bubble has two surfaces (inner and outer) in contact with air. Therefore,the change in surface area is $2 \times \Delta A$.
Given:
Surface tension $T = 0.03 \,N/m$
Surface area $A = 40 \,cm^2 = 40 \times 10^{-4} \,m^2$
The work done $W$ is given by the formula:
$W = T \times \Delta A_{total} = T \times 2 \times A$
Substituting the values:
$W = 0.03 \times 2 \times 40 \times 10^{-4}$
$W = 0.06 \times 40 \times 10^{-4}$
$W = 2.4 \times 10^{-4} \,J$

(A) Given: Radius of big drop $R = 1 \,cm = 10^{-2} \,m$, Number of small drops $n = 10^6$, Surface tension $T = 460 \times 10^{-3} \,N/m$.
Since the total volume remains constant:
$n \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3$
$10^6 \times r^3 = R^3 \implies r = \frac{R}{10^2} = 10^{-4} \,m$.
The energy expended is equal to the increase in surface area multiplied by surface tension:
$W = \Delta A \times T = (n \times 4 \pi r^2 - 4 \pi R^2) \times T$
$W = 4 \pi (n r^2 - R^2) T$
Substitute $r = R/100$:
$W = 4 \pi R^2 (n \times \frac{1}{10^4} - 1) T$
$W = 4 \times 3.14 \times (10^{-2})^2 \times (10^6 \times 10^{-4} - 1) \times 460 \times 10^{-3}$
$W = 4 \times 3.14 \times 10^{-4} \times (100 - 1) \times 0.46$
$W = 12.56 \times 10^{-4} \times 99 \times 0.46 \approx 0.057 \,J$ (Wait, re-calculating: $4 \times 3.14 \times 10^{-4} \times 99 \times 0.46 = 0.057 \,J$).

(D) The work done $W$ in blowing a soap bubble of radius $r$ is given by the formula $W = T \times \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area.
Since a soap bubble has two surfaces (inner and outer),the change in surface area is $\Delta A = 2 \times (4 \pi r^2) = 8 \pi r^2$.
Given: Diameter $d = 3 \ cm$,so radius $r = 1.5 \ cm = 1.5 \times 10^{-2} \ m$.
Surface tension $T = 0.035 \ N/m$.
Substituting the values:
$W = 0.035 \times 8 \times \pi \times (1.5 \times 10^{-2})^2$
$W = 0.035 \times 8 \times 3.14159 \times 2.25 \times 10^{-4}$
$W = 0.28 \times 3.14159 \times 2.25 \times 10^{-4}$
$W \approx 1.979 \times 10^{-4} \ J$
$W \approx 198 \times 10^{-6} \ J = 198 \ \mu J$.
Thus,the correct option is $D$.

(A) The work done in increasing the radius of a soap bubble from $r_1$ to $r_2$ is given by the formula $W = T \times \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area. Since a soap bubble has two surfaces,the change in area is $\Delta A = 2 \times (4\pi r_2^2 - 4\pi r_1^2) = 8\pi(r_2^2 - r_1^2)$.
For $W_1$ (from $r$ to $2r$): $W_1 = 8\pi T ((2r)^2 - r^2) = 8\pi T (4r^2 - r^2) = 8\pi T (3r^2) = 24\pi T r^2$.
For $W_2$ (from $2r$ to $3r$): $W_2 = 8\pi T ((3r)^2 - (2r)^2) = 8\pi T (9r^2 - 4r^2) = 8\pi T (5r^2) = 40\pi T r^2$.
Therefore,the ratio $W_1: W_2 = (24\pi T r^2) : (40\pi T r^2) = 24:40 = 3:5$.

(B) The work done in increasing the area of a film is given by $W = T \times \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area.
Since a thin film has two surfaces,the total change in area is $\Delta A = 2 \times l \times (d_2 - d_1)$.
Given: $l = 8 \ cm = 0.08 \ m$,$d_1 = 0.6 \ cm = 0.006 \ m$,$d_2 = 0.8 \ cm = 0.008 \ m$,and $T = 0.07 \ N/m$.
Change in distance $\Delta d = d_2 - d_1 = 0.8 \ cm - 0.6 \ cm = 0.2 \ cm = 0.002 \ m$.
$\Delta A = 2 \times 0.08 \ m \times 0.002 \ m = 0.00032 \ m^2$.
Work done $W = 0.07 \ N/m \times 0.00032 \ m^2 = 0.0000224 \ J$.
$W = 22.4 \times 10^{-6} \ J = 22.4 \ \mu J$.

(C) soap bubble has two surfaces (inner and outer),so the change in surface area is $\Delta A = 2 \times (4\pi r_2^2 - 4\pi r_1^2) = 8\pi(r_2^2 - r_1^2)$.
Given: Surface tension $T = 3.5 \times 10^{-2} \ N/m$,initial radius $r_1 = 1 \ cm = 1 \times 10^{-2} \ m$,final radius $r_2 = 2 \ cm = 2 \times 10^{-2} \ m$.
Work done $W = T \times \Delta A = T \times 8\pi(r_2^2 - r_1^2)$.
$W = 3.5 \times 10^{-2} \times 8 \times 3.14 \times [(2 \times 10^{-2})^2 - (1 \times 10^{-2})^2]$.
$W = 3.5 \times 10^{-2} \times 8 \times 3.14 \times (4 \times 10^{-4} - 1 \times 10^{-4})$.
$W = 3.5 \times 10^{-2} \times 8 \times 3.14 \times 3 \times 10^{-4}$.
$W = 263.76 \times 10^{-6} \ J \approx 264 \times 10^{-6} \ J$.

(D) The work done in blowing a soap bubble is given by $W = T \cdot \Delta A$,where $T$ is the surface tension and $\Delta A$ is the change in surface area. For a soap bubble,the surface area is $A = 2 \times (4 \pi r^2) = 8 \pi r^2$. Thus,$W \propto r^2$.
Since the volume of the bubble is $V = \frac{4}{3} \pi r^3$,we have $r \propto V^{1/3}$.
Substituting this into the work relation: $W \propto (V^{1/3})^2 = V^{2/3}$.
Let $W_1 = W$ for volume $V_1 = V$,and $W_2$ be the work for volume $V_2 = 2V$.
Then,$\frac{W_2}{W_1} = \left( \frac{V_2}{V_1} \right)^{2/3} = \left( \frac{2V}{V} \right)^{2/3} = (2)^{2/3}$.
Therefore,$W_2 = (2)^{2/3} W = (2^2)^{1/3} W = (4)^{1/3} W$.

(D) Let $R$ be the radius of the large drop and $r$ be the radius of each small droplet. The surface area of each small droplet is $A_s = 4\pi r^2 = 10^{-7} \,m^2$.
Since the volume is conserved, $\frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3$, which gives $R^3 = 64r^3$, so $R = 4r$.
The surface area of the large drop is $A_L = 4\pi R^2 = 4\pi (4r)^2 = 16(4\pi r^2) = 16 \times 10^{-7} \,m^2$.
The total surface area of $64$ droplets is $A_{total} = 64 \times 10^{-7} \,m^2$.
The increase in surface area is $\Delta A = A_{total} - A_L = (64 - 16) \times 10^{-7} = 48 \times 10^{-7} \,m^2$.
The increase in surface energy is $\Delta U = T \times \Delta A$, where $T = 0.07 \,N/m$.
$\Delta U = 0.07 \times 48 \times 10^{-7} = 3.36 \times 10^{-7} \,J = 336 \times 10^{-9} \,J$.

(D) soap bubble has two surfaces (inner and outer),so the surface area is $2 \times 4 \pi r^2 = 8 \pi r^2$.
Initial radius $r_i = R$.
Initial surface energy $U_i = T \times (8 \pi R^2) = 8 \pi R^2 T$.
Final diameter is doubled,so final radius $r_f = 2R$.
Final surface energy $U_f = T \times (8 \pi (2R)^2) = T \times (8 \pi \times 4R^2) = 32 \pi R^2 T$.
The surface energy needed is $\Delta U = U_f - U_i$.
$\Delta U = 32 \pi R^2 T - 8 \pi R^2 T = 24 \pi R^2 T$.

$(A)$ Radius of the large drop, $R = 1 \,cm = 10^{-2} \,m$.
Number of smaller droplets, $n = 10^6$.
Surface tension of mercury, $T = 435 \times 10^{-3} \,N/m$.
The work done in splitting a large drop of radius $R$ into $n$ smaller droplets is given by the change in surface energy: $W = T \times \Delta A$, where $\Delta A$ is the increase in surface area.
The surface area of the large drop is $A_1 = 4 \pi R^2$.
If $r$ is the radius of each small droplet, then by conservation of volume: $\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$, which gives $r = R / n^{1/3}$.
The total surface area of $n$ small droplets is $A_2 = n \times 4 \pi r^2 = n \times 4 \pi (R / n^{1/3})^2 = 4 \pi R^2 n^{1/3}$.
The increase in area is $\Delta A = A_2 - A_1 = 4 \pi R^2 (n^{1/3} - 1)$.
Substituting the values:
$W = 4 \times \pi \times (10^{-2})^2 \times 435 \times 10^{-3} \times ((10^6)^{1/3} - 1)$
$W = 4 \times 3.14159 \times 10^{-4} \times 435 \times 10^{-3} \times (100 - 1)$
$W = 4 \times 3.14159 \times 435 \times 10^{-7} \times 99$
$W \approx 54.1 \times 10^{-3} \,J$.

(B) The volume of the big drop is equal to the total volume of $n$ small droplets: $\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$.
From this,we get $r^3 = \frac{R^3}{n}$,which implies $r = \frac{R}{n^{1/3}}$.
The surface area of the big drop is $A = 4 \pi R^2$.
The total surface area of $n$ small droplets is $A' = n \times 4 \pi r^2$.
Substituting $r = R n^{-1/3}$,we get $A' = n \times 4 \pi (R n^{-1/3})^2 = n \times 4 \pi R^2 n^{-2/3} = 4 \pi R^2 n^{1/3}$.
The change in surface area is $\Delta A = A' - A = 4 \pi R^2 n^{1/3} - 4 \pi R^2 = 4 \pi R^2 (n^{1/3} - 1)$.
The change in surface energy is $\Delta U = T \times \Delta A = 4 \pi R^2 (n^{1/3} - 1) T$.

(D) soap bubble has two surfaces (inner and outer).
Initially, the surface area of the soap bubble is $A_1 = 4 \pi r^2$.
Since it has two surfaces, the effective initial area is $S_1 = 2 \times (4 \pi r^2) = 8 \pi r^2$.
Under isothermal conditions, the radius becomes $2r$.
The new surface area is $A_2 = 4 \pi (2r)^2 = 16 \pi r^2$.
The effective final area is $S_2 = 2 \times (16 \pi r^2) = 32 \pi r^2$.
The increase in surface area is $\Delta S = S_2 - S_1 = 32 \pi r^2 - 8 \pi r^2 = 24 \pi r^2$.
The energy spent $(W)$ is given by $W = T \times \Delta S$.
Therefore, $W = T \times 24 \pi r^2 = 24 \pi T r^2$.

(B) The surface tension of the soap solution is given as $T = 0.03 \,N/m$.
The surface area of the soap bubble is $A = 40 \,cm^2 = 40 \times 10^{-4} \,m^2$.
$A$ soap bubble has two surfaces (inner and outer), so the change in surface area is $\Delta A = 2 \times A$.
The work done $W$ in blowing the soap bubble is given by the formula:
$W = T \times \Delta A_{total} = T \times 2 \times A$.
Substituting the values:
$W = 0.03 \times 2 \times 40 \times 10^{-4} \,J$.
$W = 0.06 \times 40 \times 10^{-4} \,J$.
$W = 2.4 \times 10^{-4} \,J$.

(D) Given:
Diameter of the soap bubble $d = 5 \text{ mm} = 5 \times 10^{-3} \text{ m}$.
Radius $R = \frac{d}{2} = 2.5 \times 10^{-3} \text{ m}$.
Surface tension $T = \frac{1}{10 \pi} \text{ N m}^{-1}$.
$A$ soap bubble has two surfaces (inner and outer),so the total surface area $A_{total} = 2 \times (4 \pi R^2) = 8 \pi R^2$.
The free energy $E$ is given by $E = T \times A_{total}$.
Substituting the values:
$E = \left( \frac{1}{10 \pi} \right) \times 8 \pi \times (2.5 \times 10^{-3})^2$
$E = \frac{8}{10} \times 6.25 \times 10^{-6}$
$E = 0.8 \times 6.25 \times 10^{-6} = 5 \times 10^{-6} \text{ J}$.

(C) The work done in blowing a soap bubble is equal to the surface energy stored in the bubble.
Since a soap bubble has two surfaces (inner and outer), the work done $W$ is given by:
$W = T \times \Delta A = T \times 2 \times (4 \pi R^2) = 8 \pi R^2 T$
From this expression, we can see that $W \propto R^2$.
Let $W_1$ be the work done for radius $R$, so $W_1 = W$.
Let $W_2$ be the work done for radius $2R$.
Then, $\frac{W_2}{W_1} = \frac{(2R)^2}{R^2} = \frac{4R^2}{R^2} = 4$.
Therefore, $W_2 = 4W$.

(A) The work done in blowing a soap bubble is given by the change in surface energy: $W = \Delta U = S \times \Delta A$.
Since a soap bubble has two surfaces,the change in area is $\Delta A = 2 \times (4 \pi r_2^2 - 4 \pi r_1^2) = 8 \pi (r_2^2 - r_1^2)$.
Given: $S = 0.04 \ N/m$,$r_1 = 3.5 \ cm = 0.035 \ m$,$r_2 = 7 \ cm = 0.07 \ m$.
$W = 0.04 \times 8 \times \frac{22}{7} \times [(0.07)^2 - (0.035)^2]$.
$W = 0.32 \times \frac{22}{7} \times [0.0049 - 0.001225] = 0.32 \times \frac{22}{7} \times 0.003675$.
$W = 0.32 \times 22 \times 0.000525 = 0.003696 \ J = 3696 \ \mu J$.
Equating to the given expression: $15000 - x = 3696$.
$x = 15000 - 3696 = 11304$.

(B) Let $R$ be the radius of the original drop and $r$ be the radius of each of the $512$ small droplets.
By conservation of volume,$\frac{4}{3}\pi R^3 = 512 \times \frac{4}{3}\pi r^3$,which simplifies to $R^3 = 512r^3$,so $r = \frac{R}{8}$.
The change in surface energy $\Delta U = T(A_{\text{final}} - A_{\text{initial}}) = T(512 \times 4\pi r^2 - 4\pi R^2)$.
Substituting $r = \frac{R}{8}$,we get $\Delta U = 4\pi T (512 \times (\frac{R}{8})^2 - R^2) = 4\pi T (8R^2 - R^2) = 28\pi T R^2$.
Given $D = 2 \text{ mm}$,so $R = 1 \text{ mm} = 10^{-3} \text{ m}$ and $T = 0.08 \text{ N/m}$.
$\Delta U = 28 \times 3.14159 \times 0.08 \times (10^{-3})^2 \approx 7.036 \times 10^{-6} \text{ J}$.
Comparing with $\alpha \times 10^{-6} \text{ J}$,the value of $\alpha$ is approximately $7$.

(D) The work done $W$ in increasing the surface area of a soap bubble is given by $W = T \times \Delta A$.
Since a soap bubble has two surfaces (inner and outer),the change in surface area is $\Delta A = 2 \times 4\pi (r_2^2 - r_1^2) = 8\pi (r_2^2 - r_1^2)$.
Given $T = 3.5 \times 10^{-2} \text{ N/m}$,$r_1 = 1 \text{ cm} = 0.01 \text{ m}$,and $r_2 = 2 \text{ cm} = 0.02 \text{ m}$.
Substituting the values: $W = 8 \times (22/7) \times 3.5 \times 10^{-2} \times ((0.02)^2 - (0.01)^2)$.
$W = 8 \times (22/7) \times 3.5 \times 10^{-2} \times (4 \times 10^{-4} - 1 \times 10^{-4})$.
$W = 8 \times 22 \times 0.5 \times 10^{-2} \times 3 \times 10^{-4}$.
$W = 88 \times 3 \times 10^{-6} = 264 \times 10^{-6} \text{ J}$.
Thus,$\alpha = 264$.