JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Fluid Mechanics

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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 solution is 40 dyne/cm then the new radius is ______ cm. Take:

(π = \frac{22}{7})

Answer: 7

Solution

Work done in blowing a soap bubble (two surfaces):

W = T Δ A = T \cdot 2 \cdot 4 π (r_{2}^{2} - r_{1}^{2}) .

Initial radius

r_{1} = \frac{7}{2} = 3.5

cm. With

T = 40

dyne/cm,

π = \frac{22}{7}

36960 = 40 \cdot 8 \cdot \frac{22}{7} (r_{2}^{2} - 12.25) .

r_{2}^{2} - 12.25 = \frac{36960 \times 7}{40 \times 8 \times 22} = 36.75 \Rightarrow r_{2}^{2} = 49 \Rightarrow r_{2} = 7 cm .

Answer: 7.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Integer answer

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