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Properties of Matter question

2022 · 26 Jul · Shift 1 · Q58
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Properties of Matter question

2022 · 26 Jul · Shift 1 · Q58

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A water drop of radius 1 cm1 \mathrm{~cm}1 cm is broken into 729 equal droplets. If surface tension of water is 75 dyne/cm\mathrm{cm}cm, then the gain in surface energy upto first decimal place will be : (Given π=3.14\pi=3.14π=3.14 )
  1. A
    8.5×10−4 J8.5 \times 10^{-4} \mathrm{~J}8.5×10−4 J
  2. B
    8.2×10−4 J8.2 \times 10^{-4} \mathrm{~J}8.2×10−4 J
  3. C
    7.5×10−4 J7.5 \times 10^{-4} \mathrm{~J}7.5×10−4 J
  4. D
    5.3×10−4 J5.3 \times 10^{-4} \mathrm{~J}5.3×10−4 J
View written solutionFree

Correct answer: C

  1. Given data
  • Radius of original drop: R=1 cmR = 1\,\text{cm}R=1cm
  • Number of small droplets: n=729=93n = 729 = 9^3n=729=93
  • Surface tension: T=75 dyne/cmT = 75\,\text{dyne/cm}T=75dyne/cm
  • π=3.14\pi = 3.14π=3.14

We need the gain in surface energy when one large drop breaks into many smaller drops.

  1. Find radius of each small droplet

Since volume is conserved,

43πR3=n⋅43πr3\frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^334​πR3=n⋅34​πr3

So,

r3=R3nr^3 = \frac{R^3}{n}r3=nR3​ r=Rn3=19 cmr = \frac{R}{\sqrt[3]{n}} = \frac{1}{9}\,\text{cm}r=3n​R​=91​cm
  1. Initial surface area
Ai=4πR2=4π(1)2=4πA_i = 4\pi R^2 = 4\pi (1)^2 = 4\piAi​=4πR2=4π(1)2=4π
  1. Final surface area
Af=n⋅4πr2A_f = n \cdot 4\pi r^2Af​=n⋅4πr2

Substitute n=729n=729n=729 and r=19r=\frac{1}{9}r=91​ cm:

Af=729⋅4π(19)2A_f = 729 \cdot 4\pi \left(\frac{1}{9}\right)^2Af​=729⋅4π(91​)2 Af=729⋅4π⋅181=9⋅4π=36πA_f = 729 \cdot 4\pi \cdot \frac{1}{81} = 9 \cdot 4\pi = 36\piAf​=729⋅4π⋅811​=9⋅4π=36π
  1. Increase in surface area
ΔA=Af−Ai=36π−4π=32π\Delta A = A_f - A_i = 36\pi - 4\pi = 32\piΔA=Af​−Ai​=36π−4π=32π

Using π=3.14\pi = 3.14π=3.14,

ΔA=32×3.14=100.48 cm2\Delta A = 32 \times 3.14 = 100.48\,\text{cm}^2ΔA=32×3.14=100.48cm2
  1. Gain in surface energy

Surface energy gained is

ΔE=T ΔA\Delta E = T\,\Delta AΔE=TΔA ΔE=75×100.48=7536 erg\Delta E = 75 \times 100.48 = 7536\,\text{erg}ΔE=75×100.48=7536erg

Now convert erg to joule:

1 erg=10−7 J1\,\text{erg} = 10^{-7}\,\text{J}1erg=10−7J

So,

ΔE=7536×10−7 J=7.536×10−4 J\Delta E = 7536 \times 10^{-7}\,\text{J} = 7.536 \times 10^{-4}\,\text{J}ΔE=7536×10−7J=7.536×10−4J

Up to first decimal place in the coefficient,

ΔE≈7.5×10−4 J\Delta E \approx 7.5 \times 10^{-4}\,\text{J}ΔE≈7.5×10−4J
  1. Option check
  • A: 8.5×10−4 J8.5 \times 10^{-4}\,\text{J}8.5×10−4J ❌
  • B: 8.2×10−4 J8.2 \times 10^{-4}\,\text{J}8.2×10−4J ❌
  • C: 7.5×10−4 J7.5 \times 10^{-4}\,\text{J}7.5×10−4J ✅
  • D: 5.3×10−4 J5.3 \times 10^{-4}\,\text{J}5.3×10−4J ❌

Therefore, the correct option is C.

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