A soap bubble of surface tension 0.04 N/m is blown to a diameter of 7 cm. If (15000 - x) $$\mu J$$ of work is done in blowing it further to make its diameterl4 cm, then the value of x is_____.
$$\left(\pi=22/7\right)$$

We begin by noting that we need to find x where the work done in blowing a soap bubble from diameter 7 cm to 14 cm is (15000 - x) μJ.

We recall that a soap bubble has two surfaces, so the formula for the work done is:

$$W = T \times \Delta A \times 2 = 2T \times 4\pi(R_2^2 - R_1^2)$$

$$W = 8\pi T(R_2^2 - R_1^2)$$

Here, T = 0.04 N/m.

Also, $$R_1 = 3.5$$ cm = 0.035 m, $$R_2 = 7$$ cm = 0.07 m.

We take $$\pi = 22/7$$.

Substituting into the formula gives:

$$W = 8 \times \frac{22}{7} \times 0.04 \times (0.07^2 - 0.035^2)$$

$$= 8 \times \frac{22}{7} \times 0.04 \times (0.0049 - 0.001225)$$

$$= 8 \times \frac{22}{7} \times 0.04 \times 0.003675$$

$$= 8 \times \frac{22}{7} \times 0.000147$$

$$= 8 \times 22 \times \frac{0.000147}{7}$$

$$= 8 \times 22 \times 0.000021$$

$$= 8 \times 0.000462$$

$$= 0.003696 \text{ J} = 3696 \text{ μJ}$$

Thus, $$15000 - x = 3696$$

and $$x = 15000 - 3696 = 11304$$

Therefore, x = 11304.