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Properties of Matter question

2023 · 31 Jan · Shift 1 · Q60
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Properties of Matter question

2023 · 31 Jan · Shift 1 · Q60

JEE MainPhysicsProperties of MatterMCQ+4 / −1
If 1000 droplets of water of surface tension 0.07 N/m0.07 \mathrm{~N} / \mathrm{m}0.07 N/m, having same radius 1 mm1 \mathrm{~mm}1 mm each, combine to from a single drop. In the process the released surface energy is :-(Take π=227)\left( {\mathrm{Take}\,\pi = {{22} \over 7}} \right)(Takeπ=722​)
  1. A
    7.92×10−4 J7 .92 \times 10^{-4} \mathrm{~J}7.92×10−4 J
  2. B
    7.92×10−6 J7 .92 \times 10^{-6} \mathrm{~J}7.92×10−6 J
  3. C
    8.8×10−5 J8 .8 \times 10^{-5} \mathrm{~J}8.8×10−5 J
  4. D
    9.68×10−4 J9 .68 \times 10^{-4} \mathrm{~J}9.68×10−4 J
View written solutionFree

Correct answer: A

  1. Surface energy of a liquid drop

For a liquid drop, there is only one surface, so surface energy is

E=T×AE = T \times AE=T×A

where TTT is surface tension and AAA is surface area.

  1. Given data
  • Number of small droplets: n=1000n = 1000n=1000
  • Radius of each small droplet: r=1 mm=10−3 mr = 1\text{ mm} = 10^{-3}\text{ m}r=1 mm=10−3 m
  • Surface tension: T=0.07 N/mT = 0.07\,\text{N/m}T=0.07N/m
  1. Radius of the final big drop

When droplets combine, volume is conserved.

If RRR is the radius of the big drop, then

1000⋅43πr3=43πR31000 \cdot \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^31000⋅34​πr3=34​πR3

R3=1000r3R^3 = 1000r^3R3=1000r3

R=10r=10×10−3=10−2 mR = 10r = 10 \times 10^{-3} = 10^{-2}\text{ m}R=10r=10×10−3=10−2 m

  1. Initial total surface area

Each small drop has surface area

4πr24\pi r^24πr2

So total initial area is

Ai=1000⋅4πr2A_i = 1000 \cdot 4\pi r^2Ai​=1000⋅4πr2

Substitute r=10−3r = 10^{-3}r=10−3 m:

Ai=1000⋅4π(10−3)2A_i = 1000 \cdot 4\pi (10^{-3})^2Ai​=1000⋅4π(10−3)2

Ai=1000⋅4π×10−6A_i = 1000 \cdot 4\pi \times 10^{-6}Ai​=1000⋅4π×10−6

Ai=4π×10−3A_i = 4\pi \times 10^{-3}Ai​=4π×10−3

Using π=227\pi = \frac{22}{7}π=722​,

Ai=4⋅227⋅10−3=887×10−3 m2A_i = 4 \cdot \frac{22}{7} \cdot 10^{-3} = \frac{88}{7} \times 10^{-3}\,\text{m}^2Ai​=4⋅722​⋅10−3=788​×10−3m2

  1. Final surface area

Af=4πR2=4π(10−2)2=4π×10−4A_f = 4\pi R^2 = 4\pi (10^{-2})^2 = 4\pi \times 10^{-4}Af​=4πR2=4π(10−2)2=4π×10−4

Using π=227\pi = \frac{22}{7}π=722​,

Af=4⋅227⋅10−4=887×10−4 m2A_f = 4 \cdot \frac{22}{7} \cdot 10^{-4} = \frac{88}{7} \times 10^{-4}\,\text{m}^2Af​=4⋅722​⋅10−4=788​×10−4m2

  1. Decrease in surface area

ΔA=Ai−Af\Delta A = A_i - A_fΔA=Ai​−Af​

ΔA=4π×10−3−4π×10−4\Delta A = 4\pi \times 10^{-3} - 4\pi \times 10^{-4}ΔA=4π×10−3−4π×10−4

ΔA=4π(9×10−4)\Delta A = 4\pi (9 \times 10^{-4})ΔA=4π(9×10−4)

ΔA=36π×10−4\Delta A = 36\pi \times 10^{-4}ΔA=36π×10−4

Using π=227\pi = \frac{22}{7}π=722​,

ΔA=36⋅227×10−4\Delta A = 36 \cdot \frac{22}{7} \times 10^{-4}ΔA=36⋅722​×10−4

ΔA=7927×10−4 m2\Delta A = \frac{792}{7} \times 10^{-4}\,\text{m}^2ΔA=7792​×10−4m2

  1. Released surface energy

Released energy equals decrease in surface energy:

ΔE=TΔA\Delta E = T\Delta AΔE=TΔA

ΔE=0.07×7927×10−4\Delta E = 0.07 \times \frac{792}{7} \times 10^{-4}ΔE=0.07×7792​×10−4

Since 0.07=71000.07 = \frac{7}{100}0.07=1007​,

ΔE=7100×7927×10−4\Delta E = \frac{7}{100} \times \frac{792}{7} \times 10^{-4}ΔE=1007​×7792​×10−4

ΔE=792100×10−4\Delta E = \frac{792}{100} \times 10^{-4}ΔE=100792​×10−4

ΔE=7.92×10−4 J\Delta E = 7.92 \times 10^{-4}\,\text{J}ΔE=7.92×10−4J

  1. Option check
  • A: 7.92×10−4 J7.92 \times 10^{-4}\,\text{J}7.92×10−4J ✅
  • B: 7.92×10−6 J7.92 \times 10^{-6}\,\text{J}7.92×10−6J ❌
  • C: 8.8×10−5 J8.8 \times 10^{-5}\,\text{J}8.8×10−5J ❌
  • D: 9.68×10−4 J9.68 \times 10^{-4}\,\text{J}9.68×10−4J ❌

So the correct option is A.

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