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

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