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Properties of Matter question

2022 · 28 Jun · Shift 1 · Q48
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Properties of Matter question

2022 · 28 Jun · Shift 1 · Q48

JEE MainPhysicsProperties of MatterMCQ+4 / −1
A water drop of diameter 2 cm is broken into 64 equal droplets. The surface tension of water is 0.075 N/m. In this process the gain in surface energy will be :
  1. A
    2.8 ×\times× 10 −-− 4 J
  2. B
    1.5 ×\times× 10 −-− 3 J
  3. C
    1.9 ×\times× 10 −-− 4 J
  4. D
    9.4 ×\times× 10 −-− 5 J
View written solutionFree

Correct answer: A

  1. Given data
  • Diameter of original drop =2 cm= 2\text{ cm}=2 cm
  • So, radius of original drop: R=1 cm=10−2 mR = 1\text{ cm} = 10^{-2}\text{ m}R=1 cm=10−2 m
  • Number of छोटे droplets: n=64n = 64n=64
  • Surface tension of water: T=0.075 N/mT = 0.075\,\text{N/m}T=0.075N/m
  1. Find radius of each small droplet

Since volume is conserved, 43πR3=64(43πr3)\frac{4}{3}\pi R^3 = 64\left(\frac{4}{3}\pi r^3\right)34​πR3=64(34​πr3)

So, R3=64r3R^3 = 64r^3R3=64r3 r=R4r = \frac{R}{4}r=4R​

Thus, r=10−24=2.5×10−3 mr = \frac{10^{-2}}{4} = 2.5\times 10^{-3}\text{ m}r=410−2​=2.5×10−3 m

  1. Initial surface area

Original drop is one sphere, so Ai=4πR2A_i = 4\pi R^2Ai​=4πR2

  1. Final surface area

There are 646464 droplets, each of radius rrr: Af=64⋅4πr2A_f = 64\cdot 4\pi r^2Af​=64⋅4πr2

Using r=R/4r = R/4r=R/4, Af=64⋅4π(R4)2A_f = 64\cdot 4\pi \left(\frac{R}{4}\right)^2Af​=64⋅4π(4R​)2 Af=64⋅4πR216A_f = 64\cdot 4\pi \frac{R^2}{16}Af​=64⋅4π16R2​ Af=16πR2A_f = 16\pi R^2Af​=16πR2

  1. Increase in surface area

ΔA=Af−Ai\Delta A = A_f - A_iΔA=Af​−Ai​ ΔA=16πR2−4πR2\Delta A = 16\pi R^2 - 4\pi R^2ΔA=16πR2−4πR2 ΔA=12πR2\Delta A = 12\pi R^2ΔA=12πR2

Now substitute R=10−2 mR = 10^{-2}\,\text{m}R=10−2m: ΔA=12π(10−2)2\Delta A = 12\pi (10^{-2})^2ΔA=12π(10−2)2 ΔA=12π×10−4 m2\Delta A = 12\pi \times 10^{-4}\,\text{m}^2ΔA=12π×10−4m2

  1. Gain in surface energy

Surface energy gained: ΔE=TΔA\Delta E = T\Delta AΔE=TΔA ΔE=0.075×12π×10−4\Delta E = 0.075 \times 12\pi \times 10^{-4}ΔE=0.075×12π×10−4

Now, 0.075×12=0.90.075 \times 12 = 0.90.075×12=0.9

So, ΔE=0.9π×10−4\Delta E = 0.9\pi \times 10^{-4}ΔE=0.9π×10−4

Using π≈3.14\pi \approx 3.14π≈3.14, ΔE≈2.826×10−4 J\Delta E \approx 2.826\times 10^{-4}\,\text{J}ΔE≈2.826×10−4J

Therefore, ΔE≈2.8×10−4 J\boxed{\Delta E \approx 2.8\times 10^{-4}\,\text{J}}ΔE≈2.8×10−4J​

  1. Option check
  • A: 2.8×10−4 J2.8\times 10^{-4}\,\text{J}2.8×10−4J ✅
  • B: 1.5×10−3 J1.5\times 10^{-3}\,\text{J}1.5×10−3J ❌
  • C: 1.9×10−4 J1.9\times 10^{-4}\,\text{J}1.9×10−4J ❌
  • D: 9.4×10−5 J9.4\times 10^{-5}\,\text{J}9.4×10−5J ❌

Hence, the correct option is A.

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