The surface tension of a soap solution is 2× 10-2 N/m. To blow a bubble of radius 2 cm from 1 cm, the work done is

Question

The surface tension of a soap solution is 2× 10-2 N/m. To blow a bubble of radius 2 cm from 1 cm, the work done is (A)  1.5π × 10-4J  (B)  1.2π × 10-4J  (C)  0.3π × 10-4J  (D)  0.6π × 10-4J

Tardigrade Admin · Accepted Answer

W=2TΔ A =2T× 4π (r12-r22) =2× 0.05× 4π × (4× 104-1× 104) W=1.2π × 10-4J

Q. The surface tension of a soap solution is $2 \times 10^{- 2}$ N/m. To blow a bubble of radius 2 cm from 1 cm, the work done is

$1.5 π \times 10^{- 4} J$

$1.2 π \times 10^{- 4} J$

$0.3 π \times 10^{- 4} J$

$0.6 π \times 10^{- 4} J$

$W = 2 T Δ A$
$= 2 T \times 4 π (r_{1}^{2} - r_{2}^{2})$
$= 2 \times 0.05 \times 4 π \times (4 \times 10^{4} - 1 \times 10^{4})$
$W = 1.2 π \times 10^{- 4} J$

Q. The surface tension of a soap solution is 2×10−2 N/m. To blow a bubble of radius 2 cm from 1 cm, the work done is

1.5π×10−4J

1.2π×10−4J

0.3π×10−4J

0.6π×10−4J