Two soap bubbles of radius 5cm and 12cm coalesce to form a single bubble under isothermal conditions. Then the radius of bigger bubble(in cm) is(Assume, surrounding pressure is equal to zero)
Question
Two soap bubbles of radius 5cm and 12cm coalesce to form a single bubble under isothermal conditions. Then the radius of bigger bubble(in cm) is(Assume, surrounding pressure is equal to zero)
Solution
The problem is based on the concept of conservation of surface area. The surface area of a sphere (or a bubble in this case) is given by the formula 4πr², where r is the radius of the sphere.
Step 1: Calculate the total surface area of the two soap bubbles before they coalesce.
Surface area of the first bubble = 4π(5)² = 100π cm² Surface area of the second bubble = 4π(12)² = 576π cm²
Total surface area = 100π + 576π = 676π cm²
Step 2: The total surface area of the two bubbles before they coalesce will be equal to the surface area of the single bubble formed after they coalesce.
Let the radius of the bigger bubble formed after coalescence be R.
So, 4πR² = 676π
Step 3: Solve for R
R² = 676/4 = 169 R = √169 = 13 cm
So, the radius of the bigger bubble formed after the coalescence of the two soap bubbles is 13 cm.
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