QJEEM23Phfed9: JEE Main 2023 Physics - Class 11 JEEM23Phfed9 Solution: The surface tension of soap solution is . The amount of work…

To calculate the work done to increase the radius of a soap bubble, we can use the formula:

$W = T \Delta A$

where W is the work done, T is the surface tension, and ΔA is the change in surface area.

For a soap bubble, we need to consider both the inner and outer surfaces, so the surface area is doubled. The surface area of a sphere is:

$A = 4\pi r^2$

The initial surface area of the soap bubble is:

$A_1 = 2 \cdot 4\pi (0.1\,\mathrm{m})^2 = 8\pi (0.1\,\mathrm{m})^2$

The final surface area of the soap bubble is:

$A_2 = 2 \cdot 4\pi (0.2\,\mathrm{m})^2 = 8\pi (0.2\,\mathrm{m})^2$

The change in surface area is:

$\Delta A = A_2 - A_1 = 8\pi(0.2\,\mathrm{m})^2 - 8\pi(0.1\,\mathrm{m})^2$

Now, we can calculate the work done:

$W = T \Delta A = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8\pi(0.2\,\mathrm{m})^2 - 8\pi(0.1\,\mathrm{m})^2]$

Using the given value of π:

$W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.2\,\mathrm{m})^2 - 8(22/7)(0.1\,\mathrm{m})^2]$

$W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.04\,\mathrm{m^2}) - 8(22/7)(0.01\,\mathrm{m^2})]$

$W = (3.5 \times 10^{-2}\,\mathrm{Nm}^{-1})[8(22/7)(0.03\,\mathrm{m^2})]$

$\begin{aligned}& W=2 \times 1.32 \times 10^{-2} \\\\&W =2 \times 132 \times 10^{-4} \mathrm{~J} \\\\& W=264 \times 10^{-4} \mathrm{~J}\end{aligned}$

The work done to increase the radius of the soap bubble is 264 × 10⁻⁴ J.