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MCQMediumJEE 2023Surface Tension & Capillarity

JEE Physics 2023 Question with Solution

If 10001000 droplets of water of surface tension 0.07 N/m0.07 \, \text{N/m}, having same radius 1 mm1 \, \text{mm} each, combine to form a single drop. In the process, the released surface energy is: (Take π=227\pi = \frac{22}{7})

  • A

    7.92×10−6 J7.92 \times 10^{-6} \, \text{J}

  • B

    7.92×10−4 J7.92 \times 10^{-4} \, \text{J}

  • C

    9.68×10−4 J9.68 \times 10^{-4} \, \text{J}

  • D

    8.8×10−5 J8.8 \times 10^{-5} \, \text{J}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Number of droplets = 10001000, surface tension = T=0.07 N/mT = 0.07 \, \text{N/m}, radius of each droplet = r=1 mm=10−3 mr = 1 \, \text{mm} = 10^{-3} \, \text{m}.

Find: Released surface energy when all droplets combine into one larger drop.

When many droplets combine, total volume remains constant but total surface area decreases. The decrease in surface area releases surface energy.

Using volume conservation:

1000(4π3)r3=(4π3)R31000 \left( \frac{4\pi}{3} \right) r^3 = \left( \frac{4\pi}{3} \right) R^3

So,

R3=1000r3⇒R=10r=10−2 mR^3 = 1000r^3 \Rightarrow R = 10r = 10^{-2} \, \text{m}

Initial surface area of all small droplets:

Ai=1000×4πr2A_i = 1000 \times 4\pi r^2

Final surface area of the large drop:

Af=4πR2A_f = 4\pi R^2

Released surface energy:

ΔE=T(Ai−Af)\Delta E = T(A_i - A_f)

Hence,

ΔE=T[1000×4π(10−3)2−4π(10−2)2]\Delta E = T \left[1000 \times 4\pi (10^{-3})^2 - 4\pi (10^{-2})^2\right]

Substituting T=0.07 N/mT = 0.07 \, \text{N/m} and π=227\pi = \frac{22}{7}:

ΔE=0.07×4π[1000×10−6−10−4]\Delta E = 0.07 \times 4\pi \left[1000 \times 10^{-6} - 10^{-4}\right] ΔE=0.07×4π[10−3−10−4]\Delta E = 0.07 \times 4\pi \left[10^{-3} - 10^{-4}\right] ΔE=0.07×4π×9×10−4\Delta E = 0.07 \times 4\pi \times 9 \times 10^{-4} ΔE=7.92×10−4 J\Delta E = 7.92 \times 10^{-4} \, \text{J}

Therefore, the released surface energy is 7.92×10−4 J7.92 \times 10^{-4} \, \text{J}. The correct option is B.

Area Change Interpretation

Given: 10001000 identical droplets of radius rr merge to form one drop of radius RR.

Find: Surface energy released.

For identical droplets,

R=n1/3rR = n^{1/3}r

with n=1000n = 1000. Therefore,

R=10001/3r=10rR = 1000^{1/3}r = 10r

Now compare areas:

Ai=1000×4πr2A_i = 1000 \times 4\pi r^2 Af=4π(10r)2=400πr2A_f = 4\pi (10r)^2 = 400\pi r^2

So the decrease in area is

ΔA=Ai−Af=4000πr2−400πr2=3600πr2\Delta A = A_i - A_f = 4000\pi r^2 - 400\pi r^2 = 3600\pi r^2

Hence surface energy released is

ΔE=TΔA=0.07×3600π×(10−3)2\Delta E = T\Delta A = 0.07 \times 3600\pi \times (10^{-3})^2 ΔE=0.07×3600π×10−6\Delta E = 0.07 \times 3600\pi \times 10^{-6}

Using π=227\pi = \frac{22}{7},

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

Therefore, the correct option is B.

Common mistakes

  • Using volume directly for surface energy is incorrect because surface energy depends on surface area, not volume. First conserve volume to find the new radius, then compute the change in total surface area.

  • Forgetting to convert 1 mm1 \, \text{mm} into 10−3 m10^{-3} \, \text{m} gives a large numerical error. Surface tension is in N/m\text{N/m}, so SI units must be used throughout.

  • Assuming the final radius is 1000r1000r is wrong. Since volume is proportional to r3r^3, the correct relation is R=10001/3r=10rR = 1000^{1/3}r = 10r.

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