Question

Terminal velocity of drop of radius 1 cm is 10 cm/sec. 64 such balls are combined to make a large drop. Find terminal velocity of this larger drop. :

• A. 160 cm/sec
• B. 140 cm/sec
• C. 180 cm/sec
• D. 150 cm/sec

Hint:

For problems involving coalescing drops (rain, mercury, etc.), always start with the conservation of volume to find the relationship between the radii of the small and large drops. Then, use the proportionality of the quantity in question (like terminal velocity, capacitance, potential) with the radius.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the terminal velocity of small, identical spherical drops. These drops coalesce to form a single large drop. We need to find the terminal velocity of this new, larger drop.
Step 2: Key Formula or Approach:
The terminal velocity (\(V_T\)) of a spherical object falling through a viscous fluid is given by Stokes' law:
\[ V_T = \frac{2}{9} \frac{r^2 g (\sigma - \rho)}{\eta} \] where \(r\) is the radius of the sphere, \(g\) is the acceleration due to gravity, \(\sigma\) is the density of the sphere, \(\rho\) is the density of the fluid, and \(\eta\) is the viscosity of the fluid.
From this formula, we can see that terminal velocity is directly proportional to the square of the radius, assuming all other factors are constant.
\[ V_T \propto r^2 \] When drops combine, the total volume is conserved.
Step 3: Detailed Explanation:
Let \(r\) be the radius of a small drop and \(R\) be the radius of the large drop.
Let \(V_T\) be the terminal velocity of a small drop and \(V'_T\) be the terminal velocity of the large drop.
Conservation of Volume:
The volume of the large drop is the sum of the volumes of the 64 small drops.
\[ \text{Volume of large drop} = 64 \times (\text{Volume of one small drop}) \] \[ \frac{4}{3}\pi R^3 = 64 \times \frac{4}{3}\pi r^3 \] \[ R^3 = 64 r^3 \] Taking the cube root of both sides:
\[ R = (64)^{1/3} r = 4r \] The radius of the large drop is 4 times the radius of a small drop.
Calculating the New Terminal Velocity:
Since \(V_T \propto r^2\), we can set up a ratio:
\[ \frac{V'_T}{V_T} = \frac{R^2}{r^2} \] Substitute \(R = 4r\):
\[ \frac{V'_T}{V_T} = \frac{(4r)^2}{r^2} = \frac{16r^2}{r^2} = 16 \] \[ V'_T = 16 \times V_T \] We are given \(V_T = 10\) cm/sec.
\[ V'_T = 16 \times 10 = 160 \text{ cm/sec} \] Step 4: Final Answer:
The terminal velocity of the larger drop is 160 cm/sec.