Question 31: Properties of Matter

Question

The amount of work done to break a big water drop of radius ' $R$ ' into 27 small drops of equal radius is 10 J . The work done required to break the same big drop into 64 small drops of equal radius will be

Accepted Answer

To solve the problem, we need to compare the increase in surface area when the big drop is split into small drops.
Step 1: Determine the radius of the small drops
For 27 small drops:
Total volume is conserved:
$     27 	imes \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies r^3 = \frac{R^3}{27} \implies r = \frac{R}{3}.     $
For 64 small drops:
Similarly,
$     64 	imes \frac{4}{3}\pi r'^3 = \frac{4}{3}\pi R^3 \implies r'^3 = \frac{R^3}{64} \implies r' = \frac{R}{4}.     $
Step 2: Calculate the surface area before and after the break-up
Initial surface area of the big drop:
$   A_{	ext{initial}} = 4\pi R^2.   $
For 27 drops:
Surface area of one small drop:
$     4\pi\left(\frac{R}{3}ight)^2 = \frac{4\pi R^2}{9}.     $
Total surface area:
$     A_{27} = 27 	imes \frac{4\pi R^2}{9} = 3 	imes 4\pi R^2 = 12\pi R^2.     $
Increase in surface area:
$     \Delta A_{27} = 12\pi R^2 - 4\pi R^2 = 8\pi R^2.     $
For 64 drops:
Surface area of one small drop:
$     4\pi\left(\frac{R}{4}ight)^2 = 4\pi \frac{R^2}{16} = \frac{\pi R^2}{4}.     $
Total surface area:
$     A_{64} = 64 	imes \frac{\pi R^2}{4} = 16\pi R^2.     $
Increase in surface area:
$     \Delta A_{64} = 16\pi R^2 - 4\pi R^2 = 12\pi R^2.     $
Step 3: Relate work done to the change in surface area
Since the work done is proportional to the increase in surface area:
$ 	ext{Work} \propto \Delta A. $
We know that breaking into 27 drops requires 10 J corresponding to an increase of $8\pi R^2$.
Thus, if $W$ is the work done,
$ W_{27} = k \cdot 8\pi R^2 = 10\, 	ext{J}, $
where $k$ is the proportionality constant.
For 64 drops:
$ W_{64} = k \cdot 12\pi R^2. $
Dividing the two equations:
$ \frac{W_{64}}{10} = \frac{12\pi R^2}{8\pi R^2} = \frac{12}{8} = \frac{3}{2}. $
Thus,
$ W_{64} = 10 	imes \frac{3}{2} = 15\, 	ext{J}. $
Final Answer: 15 J (Option D)

Answer

20 J

Answer

10 J

Answer

5 J

Answer

15 J

Properties of Matter

The amount of work done to break a big water drop of radius ' R ' into 27 small drops of equal radius is 10 J . The work done required to break the same big drop into 64 small drops of equal radius will be

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