Question

Two small balls of same density \(\sigma\) and radii \(r\) and \(\frac{r}{3}\) are dropped into two different tubes of length \(L\), filled with liquids of densities \(\frac{\sigma }{2}\) and \(\frac{\sigma }{3}\) and coefficients of viscosity \(\eta _{1}\) and \(\eta _{2}\) respectively. If both balls reach the bottom of the tubes simultaneously, having moved at their respective terminal velocities, and the ratio \(\frac{\eta _{1}}{\eta _{2}}=\frac{x}{4}\), the value of \(x\) is

• A. \(27\)
• B. \(24\)
• C. \(26\)
• D. \(28\)

Hint:

Both balls cover the same length in the same time, so their terminal velocities are equal. Use the terminal velocity formula for each ball.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The terminal velocity is \(v=\dfrac{2r^{2}(\sigma_{ball}-\rho_{liquid})g}{9\eta}\). Equal length in equal time means the two terminal velocities are equal.

Step 2: Ball 1
\(v_{1}\propto\dfrac{r^{2}(\sigma-\sigma/2)}{\eta_{1}}=\dfrac{r^{2}\sigma}{2\eta_{1}}\)

Step 3: Ball 2
\(v_{2}\propto\dfrac{(r/3)^{2}(\sigma-\sigma/3)}{\eta_{2}}=\dfrac{r^{2}}{9}\cdot\dfrac{2\sigma}{3\eta_{2}}=\dfrac{2r^{2}\sigma}{27\eta_{2}}\)

Step 4: Equate the velocities
\(\dfrac{1}{2\eta_{1}}=\dfrac{2}{27\eta_{2}}\), so \(\dfrac{\eta_{1}}{\eta_{2}}=\dfrac{27}{4}\).

Step 5: Result
Comparing with \(\dfrac{x}{4}\) gives \(x=27\).

Final Answer:
The value of x is 27. \[ \boxed{\text{(A) }27} \]