Question

A large number of liquid drops each of radius a coalesce to form a single spherical drop of radius $b$. The energy released in the process is converted into the kinetic energy of the big drop formed, the speed of the big drop is

• A. $ \sqrt{\frac{4T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)} $
• B. $ \sqrt{\frac{2T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)} $
• C. $ \sqrt{\frac{T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)} $
• D. $ \sqrt{\frac{6T}{\rho }\left( \frac{1}{a}-\frac{1}{b} \right)} $

Answer 1

The Correct Option is D

According to question $\frac{4}{3} \pi a^{3}$ $ n=\frac{4}{3} \pi b^{3}$ $ n=\left(\frac{b}{a}\right)^{3}$ $W=T .4 \pi\left[n a^{2}-n^{2}\right] W$ $=\frac{1}{2} m v^{2}=\frac{1}{2} \cdot \frac{4}{3} \pi b^{3} p v^{2}$ $ \therefore \frac{1}{2} \times \frac{4}{3} \pi b^{3 p v^{2}=} T .4 \pi\left[n a^{2}-b^{2}\right]$ or $v=\sqrt{\frac{6 T}{\rho}\left(\frac{n a^{2}}{b^{3}}-\frac{b^{2}}{b^{3}}\right)} $ $\therefore v=\sqrt{\frac{6 T}{\rho}\left(\frac{1}{a}-\frac{1}{b}\right)}$