Surface Energy Questions in English

(A) Let $R$ be the radius of the original drop and $r$ be the radius of each of the $1000$ smaller drops.
Since the volume remains constant: $1000 \times (4/3 \pi r^3) = 4/3 \pi R^3$.
This simplifies to $1000r^3 = R^3$, which gives $R = 10r$ or $r = R/10$.
The initial surface energy is $u_i = T \times 4\pi R^2$, where $T$ is the surface tension.
The final total surface energy is $u_f = 1000 \times (T \times 4\pi r^2)$.
Substituting $r = R/10$: $u_f = 1000 \times T \times 4\pi (R/10)^2 = 1000 \times T \times 4\pi (R^2/100) = 10 \times (T \times 4\pi R^2) = 10 u_i$.
Thus, $u_f/u_i = 10$.
Given $u_f/u_i = 10/x$, we have $10 = 10/x$, which implies $x = 1$.

(B) Let the radius of each small drop be $r$ and the radius of the large drop be $R$.
Since the volume is conserved, the volume of $1000$ small drops equals the volume of $1$ large drop:
$1000 \cdot (\frac{4}{3}\pi r^3) = \frac{4}{3}\pi R^3$
$1000 r^3 = R^3 \implies R = 10r$.
The surface energy of a drop is given by $U = T \cdot A$, where $T$ is surface tension and $A$ is the surface area.
Surface energy of $1$ large drop: $U_{large} = T \cdot 4\pi R^2 = T \cdot 4\pi (10r)^2 = 100 T \cdot 4\pi r^2$.
Total surface energy of $1000$ small drops: $U_{small_total} = 1000 \cdot (T \cdot 4\pi r^2) = 1000 T \cdot 4\pi r^2$.
The ratio of the surface energy of $1$ large drop to $1000$ small drops is:
$\frac{U_{large}}{U_{small_total}} = \frac{100 T \cdot 4\pi r^2}{1000 T \cdot 4\pi r^2} = \frac{100}{1000} = \frac{1}{10}$.
Thus, the ratio is $1 : 10$.

(A) The volume of the large drop is equal to the sum of the volumes of the '$n$' small droplets.
$V_{large} = n \times V_{small}$
$\frac{4}{3} \pi R^3 = n \times \frac{4}{3} \pi r^3$
$R^3 = n r^3 \implies n = \frac{R^3}{r^3}$
The energy required is equal to the increase in surface area multiplied by the surface tension '$T$'.
$E = T \times (\text{Final Surface Area} - \text{Initial Surface Area})$
$E = T \times (n \times 4\pi r^2 - 4\pi R^2)$
Substitute $n = \frac{R^3}{r^3}$:
$E = 4\pi T (\frac{R^3}{r^3} \times r^2 - R^2)$
$E = 4\pi T (\frac{R^3}{r} - R^2)$
$E = 4\pi T R^2 (\frac{R}{r} - 1)$

(C) The work done in blowing a soap bubble of radius $r$ is given by $W_1 = T \times \Delta A \times 2$, where $T$ is the surface tension and $\Delta A$ is the change in surface area. Since a soap bubble has two surfaces, the area is $2 \times (4\pi r^2)$.
$W_1 = T \times 8\pi r^2 = 450 \text{ erg}$.
When the radius is increased to $3r$, the new work done $W_2$ is $T \times 8\pi (3r)^2 = T \times 8\pi (9r^2) = 9 \times (T \times 8\pi r^2)$.
Substituting the value of $W_1$, we get $W_2 = 9 \times 450 = 4050 \text{ erg}$.
The additional work required is $W_{\text{add}} = W_2 - W_1 = 4050 - 450 = 3600 \text{ erg}$.

(A) The work done $W$ 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 (inner and outer), the total surface area is $A = 2 \times (4\pi r^2) = 8\pi r^2$.
Thus, $W = T \times 8\pi r^2$.
For the first bubble of radius $R$, $W_1 = 8\pi R^2 T_1$.
For the second bubble of radius $2R$, $W_2 = 8\pi (2R)^2 T_2 = 32\pi R^2 T_2$.
When the solution is heated, the surface tension $T$ decreases, so $T_2 < T_1$.
The ratio is $\frac{W_2}{W_1} = \frac{32\pi R^2 T_2}{8\pi R^2 T_1} = 4 \times \frac{T_2}{T_1}$.
Since $T_2 < T_1$, the ratio $\frac{T_2}{T_1} < 1$, which implies $\frac{W_2}{W_1} < 4$.
Therefore, the work required is slightly less than $4$ times the work required for the first bubble.

(A) Let the radius of each small drop be $r$ and the radius of the big drop be $R$.
Since the volume remains constant, the volume of $1000$ small drops equals the volume of the big drop:
$1000 \times (\frac{4}{3} \pi r^3) = \frac{4}{3} \pi R^3$
$1000 r^3 = R^3$
$R = 10r$
Initial surface energy $(E_i)$ = $1000 \times (4 \pi r^2 T)$, where $T$ is the surface tension.
Final surface energy $(E_f)$ = $4 \pi R^2 T = 4 \pi (10r)^2 T = 400 \pi r^2 T$.
The ratio of final surface energy to initial surface energy is:
$\frac{E_f}{E_i} = \frac{400 \pi r^2 T}{1000 \times 4 \pi r^2 T} = \frac{400}{4000} = \frac{1}{10}$.

(C) The work done $(W)$ in expanding a soap bubble 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 surface area $A = 2 \times (4\pi r^2) = 8\pi r^2$.
Initial radius $r_1 = \frac{1}{\sqrt{\pi}} \text{ cm}$, so initial area $A_1 = 8\pi \left(\frac{1}{\sqrt{\pi}}\right)^2 = 8 \text{ cm}^2$.
Final radius $r_2 = \frac{3}{\sqrt{\pi}} \text{ cm}$, so final area $A_2 = 8\pi \left(\frac{3}{\sqrt{\pi}}\right)^2 = 8\pi \times \frac{9}{\pi} = 72 \text{ cm}^2$.
Change in area $\Delta A = A_2 - A_1 = 72 - 8 = 64 \text{ cm}^2$.
Work done $W = T \times \Delta A = 25 \times 64 = 1600 \text{ erg}$.