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Properties of Matter question

2023 · 1 Feb · Shift 1 · Q62
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Properties of Matter question

2023 · 1 Feb · Shift 1 · Q62

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A mercury drop of radius 10−3 m10^{-3}~\mathrm{m}10−3 m is broken into 125 equal size droplets. Surface tension of mercury is 0.45 Nm−10.45~\mathrm{Nm}^{-1}0.45 Nm−1. The gain in surface energy is :
  1. A
    28×10−5 J28\times10^{-5}~\mathrm{J}28×10−5 J
  2. B
    17.5×10−5 J17.5\times10^{-5}~\mathrm{J}17.5×10−5 J
  3. C
    5×10−5 J5\times10^{-5}~\mathrm{J}5×10−5 J
  4. D
    2.26×10−5 J2.26\times10^{-5}~\mathrm{J}2.26×10−5 J
View written solutionFree

Correct answer: D

  1. Given data
  • Radius of original mercury drop: R=10−3 mR = 10^{-3}\,\text{m}R=10−3m
  • Number of छोटे droplets formed: n=125n = 125n=125
  • Surface tension of mercury: T=0.45 N m−1T = 0.45\,\text{N m}^{-1}T=0.45N m−1

We need the gain in surface energy.

  1. Relation between radii using volume conservation

When one big drop breaks into nnn equal droplets, total volume remains constant:

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

So,

R3=nr3R^3 = n r^3R3=nr3

r3=R3125r^3 = \frac{R^3}{125}r3=125R3​

Since 125=53125 = 5^3125=53,

r=R5=10−35=2×10−4 mr = \frac{R}{5} = \frac{10^{-3}}{5} = 2\times 10^{-4}\,\text{m}r=5R​=510−3​=2×10−4m

  1. Initial surface area

Ai=4πR2A_i = 4\pi R^2Ai​=4πR2

  1. Final surface area

Each small droplet has area 4πr24\pi r^24πr2, so total final area is

Af=125⋅4πr2A_f = 125 \cdot 4\pi r^2Af​=125⋅4πr2

Using r=R/5r = R/5r=R/5,

Af=125⋅4π(R5)2A_f = 125 \cdot 4\pi \left(\frac{R}{5}\right)^2Af​=125⋅4π(5R​)2

Af=125⋅4πR225=5⋅4πR2=20πR2A_f = 125 \cdot 4\pi \frac{R^2}{25} = 5\cdot 4\pi R^2 = 20\pi R^2Af​=125⋅4π25R2​=5⋅4πR2=20πR2

  1. Increase in surface area

ΔA=Af−Ai=20πR2−4πR2=16πR2\Delta A = A_f - A_i = 20\pi R^2 - 4\pi R^2 = 16\pi R^2ΔA=Af​−Ai​=20πR2−4πR2=16πR2

Substitute R=10−3 mR = 10^{-3}\,\text{m}R=10−3m:

ΔA=16π(10−3)2=16π×10−6 m2\Delta A = 16\pi (10^{-3})^2 = 16\pi \times 10^{-6}\,\text{m}^2ΔA=16π(10−3)2=16π×10−6m2

  1. Gain in surface energy

Surface energy gained:

ΔE=T ΔA\Delta E = T\,\Delta AΔE=TΔA

ΔE=0.45×16π×10−6\Delta E = 0.45 \times 16\pi \times 10^{-6}ΔE=0.45×16π×10−6

ΔE=7.2π×10−6\Delta E = 7.2\pi \times 10^{-6}ΔE=7.2π×10−6

Using π≈3.14\pi \approx 3.14π≈3.14,

ΔE≈7.2×3.14×10−6\Delta E \approx 7.2 \times 3.14 \times 10^{-6}ΔE≈7.2×3.14×10−6

ΔE≈22.6×10−6 J\Delta E \approx 22.6 \times 10^{-6} \text{ J}ΔE≈22.6×10−6 J

ΔE=2.26×10−5 J\Delta E = 2.26 \times 10^{-5}\,\text{J}ΔE=2.26×10−5J

  1. Option check
  • A: 28×10−5 J28\times10^{-5}\,\text{J}28×10−5J ✗
  • B: 17.5×10−5 J17.5\times10^{-5}\,\text{J}17.5×10−5J ✗
  • C: 5×10−5 J5\times10^{-5}\,\text{J}5×10−5J ✗
  • D: 2.26×10−5 J2.26\times10^{-5}\,\text{J}2.26×10−5J ✓

Therefore, the correct option is D.

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