If 1000 droplets of water of surface tension 0.07 N / m, having same radius 1 mm each, combine to from a single drop. In the process the released surface energy is- (. Take .π=(22/7))

Question

If 1000 droplets of water of surface tension 0.07 N / m, having same radius 1 mm each, combine to from a single drop. In the process the released surface energy is- (. Take .π=(22/7)) (A) 8.8 × 10-5 J (B) 7.92 × 10-4 J (C) 7.92 × 10-6 J (D) 9.68 × 10-4 J

Tardigrade Admin · Accepted Answer

1000 × (4 π/3)(1)3=(4 π/3) R3 R=10 mm T × 1000 × 4 π(10-3)2- T × 4 π(10 × 10-3)2=Δ E Δ E =4 × π × 7 × 10-2[1000-100] × 10-6 Δ E =7.92 × 10-4 J

Q. If $1000$ droplets of water of surface tension $0.07 N / m$ , having same radius $1 mm$ each, combine to from a single drop. In the process the released surface energy is- $($ Take $π = \frac{22}{7})$

$8.8 \times 1 0^{- 5} J$

$7.92 \times 1 0^{- 4} J$

$7.92 \times 1 0^{- 6} J$

$9.68 \times 1 0^{- 4} J$

$1000 \times \frac{4 π}{3} (1)^{3} = \frac{4 π}{3} R^{3}$
$R = 10 mm$
$T \times 1000 \times 4 π (1 0^{- 3})^{2} - T \times 4 π (10 \times 1 0^{- 3})^{2} = Δ E$
$Δ E = 4 \times π \times 7 \times 1 0^{- 2} [1000 - 100] \times 1 0^{- 6}$
$Δ E = 7.92 \times 1 0^{- 4} J$

Q. If 1000 droplets of water of surface tension 0.07N/m, having same radius 1mm each, combine to from a single drop. In the process the released surface energy is- ( Take π=722​)

8.8×10−5J

7.92×10−4J

7.92×10−6J

9.68×10−4J