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MCQEasyJEE 2024Surface Tension & Capillarity

JEE Physics 2024 Question with Solution

A small liquid drop of radius RR is divided into 2727 identical liquid drops. If the surface tension is TT, then the work done in the process will be:

  • A

    8πR2T8\pi R^2 T

  • B

    3πR2T3\pi R^2 T

  • C

    18πR2T\frac{1}{8}\pi R^2 T

  • D

    4πR2T4\pi R^2 T

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A liquid drop of radius RR is divided into 2727 identical smaller drops, and surface tension is TT.

Find: The work done in dividing the drop.

The work done is equal to the increase in surface energy, so we first calculate the change in total surface area.

For the original drop, the surface area is

A1=4πR2A_1 = 4\pi R^2

Let the radius of each small drop be rr. Using volume conservation,

43πR3=27×43πr3\frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r^3

So,

R3=27r3R^3 = 27r^3 r=R3r = \frac{R}{3}

Surface area of each small drop is

Ar=4πr2=4π(R3)2=4πR29A_r = 4\pi r^2 = 4\pi \left(\frac{R}{3}\right)^2 = \frac{4\pi R^2}{9}

Therefore, the total surface area of 2727 smaller drops is

A2=27×4πR29=12πR2A_2 = 27 \times \frac{4\pi R^2}{9} = 12\pi R^2

Hence, the increase in surface area is

ΔA=A2−A1=12πR2−4πR2=8πR2\Delta A = A_2 - A_1 = 12\pi R^2 - 4\pi R^2 = 8\pi R^2

Work done equals change in surface energy:

W=TΔA=T×8πR2=8πR2TW = T\Delta A = T \times 8\pi R^2 = 8\pi R^2 T

Therefore, the work done in the process is 8πR2T8\pi R^2 T. The correct option is A.

Surface Energy Approach

Given: One spherical liquid drop of radius RR is broken into 2727 identical drops.

Find: The work done against surface tension.

Since the liquid volume remains constant, equate the initial and final volumes:

43πR3=27⋅43πr3\frac{4}{3}\pi R^3 = 27 \cdot \frac{4}{3}\pi r^3

This gives

r=R3r = \frac{R}{3}

Now compare the initial and final surface areas:

Ainitial=4πR2A_{\text{initial}} = 4\pi R^2 Afinal=27⋅4πr2=27⋅4π(R3)2=12πR2A_{\text{final}} = 27 \cdot 4\pi r^2 = 27 \cdot 4\pi \left(\frac{R}{3}\right)^2 = 12\pi R^2

So the extra area created is

ΔA=Afinal−Ainitial=12πR2−4πR2=8πR2\Delta A = A_{\text{final}} - A_{\text{initial}} = 12\pi R^2 - 4\pi R^2 = 8\pi R^2

Surface energy increase is

ΔU=TΔA=8πR2T\Delta U = T\Delta A = 8\pi R^2 T

Therefore, the required work done is 8πR2T8\pi R^2 T.

Common mistakes

  • Using radius of each small drop as R27\frac{R}{27} is incorrect because volume, not radius, is conserved. First write R3=27r3R^3 = 27r^3 and then find r=R3r = \frac{R}{3}.

  • Calculating work done from volume change is wrong here because surface tension is related to surface energy. Use the increase in surface area and apply W=TΔAW = T\Delta A.

  • Forgetting to multiply the surface area of one small drop by 2727 gives the wrong final area. After finding area of one drop, compute total area of all 2727 drops.

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