Question

Under isothermal conditions, a single drop of radius \(R\) is formed from two mercury drops of radii \(R_{1}\) and \(R_{2}\). After combination, the relation between \(R\), \(R_{1}\) and \(R_{2}\) is

• A. \(R=\frac{R_{1}+R_{2}}{2}\)
• B. \(R^{3}=R_{1}^{3}+R_{2}^{3}\)
• C. \(R^{2}=R_{1}^{2}+R_{2}^{2}\)
• D. \(R=R_{1}+R_{2}\)

Hint:

Mass and hence volume is conserved when drops combine. Volume goes as the cube of the radius.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Under isothermal conditions the density does not change, so conservation of mass means conservation of volume.

Step 2: Write the Volumes
\(\dfrac{4}{3}\pi R^{3}=\dfrac{4}{3}\pi R_1^{3}+\dfrac{4}{3}\pi R_2^{3}\)

Step 3: Simplify
Cancel the common factor \(\dfrac{4}{3}\pi\): \(R^{3}=R_1^{3}+R_2^{3}\)

Step 4: Compare the options
This is the second option.

Final Answer:
The radii are related by R cubed equal to R1 cubed plus R2 cubed. \[ \boxed{\text{(B) }R^{3}=R_{1}^{3}+R_{2}^{3}} \]