If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to form a single drop, In the process the released surface energy is (Take ) [2023]

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Q.
If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to form a single drop, In the process the released surface energy is (Take $π = \frac{22}{7}$ ) [2023]

1 $7.92 \times 10^{- 6} J$

2 $7.92 \times 10^{- 4} J$

3 $9.68 \times 10^{- 4} J$

4 $8.8 \times 10^{- 5} J$

Ans.
(2)

$1000 \times \frac{4 π}{3} {(1)}^{3} = \frac{4 π}{3} R^{3}$

$R = 10 mm$

$T \times 1000 \times 4 π {(10^{- 3})}^{2} - T \times 4 π {(10 \times 10^{- 3})}^{2} = Δ E$

$Δ E = 4 \times π \times 7 \times 10^{- 2} [1000 - 100] \times 10^{- 6}$

$Δ E = 7.92 \times 10^{- 4} J$

Solution

Q.
If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to form a single drop, In the process the released surface energy is (Take $π = \frac{22}{7}$ ) [2023]

1 $7.92 \times 10^{- 6} J$

2 $7.92 \times 10^{- 4} J$

3 $9.68 \times 10^{- 4} J$

4 $8.8 \times 10^{- 5} J$

Ans.
(2)

$1000 \times \frac{4 π}{3} {(1)}^{3} = \frac{4 π}{3} R^{3}$

$R = 10 mm$

$T \times 1000 \times 4 π {(10^{- 3})}^{2} - T \times 4 π {(10 \times 10^{- 3})}^{2} = Δ E$

$Δ E = 4 \times π \times 7 \times 10^{- 2} [1000 - 100] \times 10^{- 6}$

$Δ E = 7.92 \times 10^{- 4} J$

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Solution

Q. If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to form a single drop, In the process the released surface energy is (Take π=227) [2023]

1 7.92×10-6 J

2 7.92×10-4 J

3 9.68×10-4 J

4 8.8×10-5 J

Ans. (2) 1000×4π3(1)3=4π3R3 R=10 mm T×1000×4π(10-3)2-T×4π(10×10-3)2=ΔE ΔE=4×π×7×10-2[1000-100]×10-6 ΔE=7.92×10-4 J

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Q.
If 1000 droplets of water of surface tension 0.07 N/m, having same radius 1 mm each, combine to form a single drop, In the process the released surface energy is (Take $π = \frac{22}{7}$ ) [2023]

1 $7.92 \times 10^{- 6} J$

2 $7.92 \times 10^{- 4} J$

3 $9.68 \times 10^{- 4} J$

4 $8.8 \times 10^{- 5} J$

Ans.
(2)

$1000 \times \frac{4 π}{3} {(1)}^{3} = \frac{4 π}{3} R^{3}$

$R = 10 mm$

$T \times 1000 \times 4 π {(10^{- 3})}^{2} - T \times 4 π {(10 \times 10^{- 3})}^{2} = Δ E$

$Δ E = 4 \times π \times 7 \times 10^{- 2} [1000 - 100] \times 10^{- 6}$

$Δ E = 7.92 \times 10^{- 4} J$