A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is done in blowing it further. If surface tension of soap is 40 dyne/ cm then the new radius is ________ cm. Take (π = 22/7). Numerical answer type. Work it out, then open the answer.

Answer: 7. In CGS: W = 8π T (R² - r²) with r = 3.5 cm, T = 40 dyne/cm. 36960 = 8 × 22/7 × 40 × (R² - 12.25) R² - 12.25 = 36.75 ⇒ R² = 49 New radius R = 7 cm.

A soap bubble is blown to a diameter of 7 cm. 36960 erg of work is… | JEE Main 2024 (4 Apr, Shift 1)

Q14JEE Main 2024 · 4 APR · SHIFT 1Medium

Surface Energy & Work Done

A soap bubble is blown to a diameter of $7\text{ cm}.\ 36960\text{ erg}$ of work is done in blowing it further. If surface tension of soap is 40 dyne/ $\text{cm}$ then the new radius is ________ cm. Take $\left(\pi = \frac{22}{7}\right)$ .

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 7

In CGS: $W = 8\pi T (R^2 - r^2)$ with $r = 3.5\text{ cm}$ , $T = 40$ dyne/cm.
$36960 = 8 \times \frac{22}{7} \times 40 \times (R^2 - 12.25)$
$R^2 - 12.25 = 36.75 \Rightarrow R^2 = 49$
New radius $R = 7\text{ cm}$ .

Report a mistake

Surface Energy & Work Done question from JEE Main 2024 (4 Apr, Shift 1)

More questions on Surface Energy & Work Done