JEE Physics 2026 Question with Solution

Given: Surface tension of the soap bubble is $T = 0.04\,\text{N/m}$. Initial diameter is $7\,\text{cm}$ and final diameter is $14\,\text{cm}$. The work done is given as $\left(15000 - x\right)\,\mu \text{J}$.

Find: The value of $x$.

A soap bubble has two surfaces, so surface energy is

$$E = 2 \times T \times 4\pi r^2 = 8\pi T r^2$$

Initial radius:

$$r_1 = 3.5\,\text{cm} = 0.035\,\text{m}$$

Final radius:

$$r_2 = 7\,\text{cm} = 0.07\,\text{m}$$

Work done in expansion equals increase in surface energy:

$$W = 8\pi T \left(r_2^2 - r_1^2\right)$$

Substituting values:

$$W = 8 \times \frac{22}{7} \times 0.04 \left(0.07^2 - 0.035^2\right)$$ $$W = 0.006304\,\text{J} = 6304\,\mu\text{J}$$

Now compare with the given expression:

$$15000 - x = 6304$$

So,

$$x = 11304$$

Therefore, the value of $x$ is $11304$.

Given: A soap bubble has surface tension $0.04\,\text{N/m}$ and its diameter increases from $7\,\text{cm}$ to $14\,\text{cm}$.

Find: The numerical value of $x$.

The hint says that a soap bubble has two surfaces. Therefore, the total surface area contributing to energy is twice the outer spherical area. Hence,

$$E = 8\pi T r^2$$

For the initial state,

$$r_1 = \frac{7}{2}\,\text{cm} = 3.5\,\text{cm} = 0.035\,\text{m}$$

For the final state,

$$r_2 = \frac{14}{2}\,\text{cm} = 7\,\text{cm} = 0.07\,\text{m}$$

Increase in surface energy is

$$\Delta E = 8\pi T \left(r_2^2 - r_1^2\right)$$

Now,

$$r_2^2 = 0.07^2 = 0.0049$$ $$r_1^2 = 0.035^2 = 0.001225$$

So,

$$r_2^2 - r_1^2 = 0.0049 - 0.001225 = 0.003675$$

Thus,

$$W = 8 \times \frac{22}{7} \times 0.04 \times 0.003675$$ $$W = 0.006304\,\text{J}$$

Convert into microjoule:

$$0.006304\,\text{J} = 6304\,\mu\text{J}$$

Given,

$$15000 - x = 6304$$

Hence,

$$x = 15000 - 6304 = 11304$$

Therefore, the required answer is $11304$.