Mechanical Properties Of Fluids Question 3
Question 3 - 29 January - Shift 1
Surface tension of a soap bubble is $2.0 \times 10^{-2} Nm^{-1}$. Work done to increase the radius of soap bubble from $3.5 cm$ to $7 cm$ will be : [Take $\pi=\frac{22}{7}$ ]
(1) $0.72 \times 10^{-4} J$
(2) $5.76 \times 10^{-4} J$
(3) $18.48 \times 10^{-4} J$
(4) $9.24 \times 10^{-4} J$
Show Answer
Answer: (3)
Solution:
Formula: Surface Tension
Surface area of soap bubble $=2 \times 4 \pi R^{2}$
Work done $=$ change in surface energy $\times T_S$
$=T_S \times 8 \pi \times(R_2^{2}-R_1^{2})$
$=2 \times 10^{-2} \times 8 \times \frac{22}{7} \times 49 \times \frac{3}{4} \times 10^{-4}$
$=18.48 \times 10^{-4} J$
