JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Fluid Mechanics

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A spherical soap bubble of radius 3 cm is formed inside another spherical soap bubble of radius 6 cm. If the internal pressure of the smaller bubble of radius 3 cm in the above system is equal to the internal pressure of the another single soap bubble of radius r cm. The value of r is. ______

Answer: 2

Solution

For a soap bubble, the excess pressure is

Δ P = \frac{4 T}{R}

(two surfaces).

For the outer bubble (radius 6 cm):

P_{2} - P_{0} = \frac{4 T}{6}

For the inner bubble (radius 3 cm), the pressure difference from

P_{2}

P_{1}

is:

P_{1} - P_{2} = \frac{4 T}{3}

Adding these:

P_{1} - P_{0} = \frac{4 T}{6} + \frac{4 T}{3} = \frac{4 T}{6} + \frac{8 T}{6} = \frac{12 T}{6} = 2 T

But

P_{1} - P_{0} = \frac{4 T}{2} = 2 T

, so:

\frac{4 T}{r} = 2 T ⟹ r = 2 cm

Answer: 2

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer

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