1.

Q = A1V1 = π × \(\frac{0.1^2}{4}\) × 5
= 0.0393 m3 /s
V2 = \(\frac{A_1V_1}{A_2}\) (continuity equation)
= 2.22 m/s
2. Here, s = 0.06 N/m
r1 = 2 cm = 0.02 m ;
r2 = 5 cm = 0.05 m
Since bubble has 2 surfaces, initial surface area of the bubble.
= 2 × \(4πr_1^2\)= 2 × 4π(0.02)2
= 32π × 10-4m2
Final surface area of the bubble
= 2 × \(4πr_2^2\) = 2 × 4π(0.05)2
= 200π × 10-4m2
Increase in surface area
= 200π × 10-4 – 32π × 10-4
= 168π × 10-4m2
∴ work done = S × increase in surface area
= 0.06 × 168π × 10-4 = 0.003168 J