Question

When a mercury drop of radius 'R' breaks into n droplets of equal size, the radius (r) of each droplet is 'xR'. The value of x is

• A. \((n)^{\frac{-1}{3}}\)
• B. \((n)^{\frac{-1}{2}}\)
• C. \((n)^{\frac{1}{2}}\)
• D. \((n)^{\frac{1}{3}}\)

Hint:

The total volume of mercury stays the same when the drop splits into n equal droplets.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Splitting a drop does not change the volume of liquid, so the big drop volume equals n times one droplet volume.

Step 2: Volume balance
\(\dfrac{4}{3}\pi R^3=n\cdot\dfrac{4}{3}\pi r^3\)

Step 3: Solve for r
\(r^3=\dfrac{R^3}{n}\Rightarrow r=n^{-1/3}R\)

Step 4: Read x
Since \(r=xR\), \(x=n^{-1/3}\).

Final Answer:
The ratio is \(x=(n)^{-1/3}\). \[ \boxed{\text{(A) }(n)^{\frac{-1}{3}}} \]