Logo
Jeeify
MCQMediumJEE 2025Surface Tension & Capillarity

JEE Physics 2025 Question with Solution

The amount of work done to break a big water drop of radius RR into 2727 small drops of equal radius is 10 J10 \, \text{J}. The work done required to break the same big drop into 6464 small drops of equal radius will be:

  • A

    15 J15 \, \text{J}

  • B

    10 J10 \, \text{J}

  • C

    20 J20 \, \text{J}

  • D

    5 J5 \, \text{J}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A big water drop of radius RR is broken into 2727 equal drops with work done 10 J10 \, \text{J}.

Find: The work required when the same drop is broken into 6464 equal drops.

Work done in breaking a drop is proportional to the increase in surface area:

W=T ΔAW = T\,\Delta A

where TT is the surface tension.

For a sphere, surface area is

A=4πr2A = 4\pi r^2

If the big drop is divided into nn small drops of radius rr, then conservation of volume gives

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

So,

R3=nr3R^3 = nr^3

which gives

r=Rn1/3r = \frac{R}{n^{1/3}}

For 2727 small drops,

r=R271/3=R3r = \frac{R}{27^{1/3}} = \frac{R}{3}

Hence the total new surface area is

A27=27⋅4π(R3)2=12πR2A_{27} = 27 \cdot 4\pi \left(\frac{R}{3}\right)^2 = 12\pi R^2

Therefore, the increase in surface area is

ΔA27=12πR2−4πR2=8πR2\Delta A_{27} = 12\pi R^2 - 4\pi R^2 = 8\pi R^2

Thus,

T⋅8πR2=10T \cdot 8\pi R^2 = 10

For 6464 small drops,

r=R641/3=R4r = \frac{R}{64^{1/3}} = \frac{R}{4}

The total new surface area is

A64=64⋅4π(R4)2=16πR2A_{64} = 64 \cdot 4\pi \left(\frac{R}{4}\right)^2 = 16\pi R^2

So the increase in surface area is

ΔA64=16πR2−4πR2=12πR2\Delta A_{64} = 16\pi R^2 - 4\pi R^2 = 12\pi R^2

Therefore,

W2=T⋅12πR2W_2 = T \cdot 12\pi R^2

Using

T⋅8πR2=10T \cdot 8\pi R^2 = 10

we get

W2=128×10=15 JW_2 = \frac{12}{8} \times 10 = 15 \, \text{J}

Therefore, the work required is 15 J15 \, \text{J} and the correct option is A.

Ratio Method

Given: Work done is proportional to increase in surface area.

Find: Work for 6464 drops using the given work for 2727 drops.

When a drop is broken into nn equal drops, the total surface area becomes proportional to n1/3n^{1/3} times the original surface area, so the increase in area is proportional to

n1/3−1n^{1/3} - 1

Hence,

W∝n1/3−1W \propto n^{1/3} - 1

So,

W2W1=641/3−1271/3−1=4−13−1=32\frac{W_2}{W_1} = \frac{64^{1/3} - 1}{27^{1/3} - 1} = \frac{4 - 1}{3 - 1} = \frac{3}{2}

Therefore,

W2=32×10=15 JW_2 = \frac{3}{2} \times 10 = 15 \, \text{J}

Therefore, the required work is 15 J15 \, \text{J}, so the correct option is A.

Common mistakes

  • Using the small-drop radius directly without conserving volume is incorrect, because the radius of each smaller drop depends on the number of drops. First apply

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

    and then find r=Rn1/3r = \frac{R}{n^{1/3}}.

  • Assuming work done is proportional to volume change is wrong, because the volume remains constant during breakup. The work is due to increase in surface energy, so use the increase in surface area instead.

  • Forgetting to subtract the initial surface area gives the final surface area instead of the increase in surface area. Use

    ΔA=Afinal−Ainitial\Delta A = A_{\text{final}} - A_{\text{initial}}

    not just the total area of the smaller drops.

Practice more Surface Tension & Capillarity questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Surface Tension & Capillarity

Related questions