Two spheres of equal masses but radii R and 2 R are allowed to fall in a liquid, the ratio of their terminal velocities is

Options

A. `1 : 4`
B. `1 : 2`
C. `1 : 32`
D. `2 : 1`

Explanation

To solve the problem of finding the ratio of the terminal velocities of two spheres with equal masses but different radii (R and 2R) falling in a liquid, we can follow these steps:

### Step 1: Understand the concept of terminal velocity
Terminal velocity (V_t) is the constant speed that a freely falling object eventually reaches when the resistance of the medium (in this case, a liquid) prevents further acceleration. For a sphere falling through a fluid, the terminal velocity is given by the equation:
\[ V_t \propto r^2 \cdot \rho \]
where \( r \) is the radius of the sphere and \( \rho \) is the density of the sphere.

### Step 2: Identify the variables
Let:
- \( r_1 = R \) (radius of the first sphere)
- \( r_2 = 2R \) (radius of the second sphere)
- \( \rho_1 \) and \( \rho_2 \) are the densities of the spheres.

Since both spheres have equal masses, we can express their masses in terms of their densities and volumes.

### Step 3: Relate mass to density and volume
The mass of a sphere can be expressed as:
\[ m = \text{density} \times \text{volume} \]
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, for the first sphere:
\[ m_1 = \rho_1 \cdot \frac{4}{3} \pi R^3 \]
And for the second sphere:
\[ m_2 = \rho_2 \cdot \frac{4}{3} \pi (2R)^3 = \rho_2 \cdot \frac{4}{3} \pi (8R^3) \]

### Step 4: Set the masses equal
Since \( m_1 = m_2 \):
\[ \rho_1 \cdot \frac{4}{3} \pi R^3 = \rho_2 \cdot \frac{4}{3} \pi (8R^3) \]
Cancelling out common terms:
\[ \rho_1 \cdot R^3 = 8 \rho_2 \cdot R^3 \]
This simplifies to:
\[ \rho_1 = 8 \rho_2 \]

### Step 5: Calculate the ratio of terminal velocities
Using the relationship for terminal velocity:
\[ \frac{V_{t1}}{V_{t2}} = \frac{r_1^2 \cdot \rho_1}{r_2^2 \cdot \rho_2} \]
Substituting the values:
- \( r_1 = R \)
- \( r_2 = 2R \)
- \( \rho_1 = 8 \rho_2 \)

We get:
\[ \frac{V_{t1}}{V_{t2}} = \frac{R^2 \cdot (8 \rho_2)}{(2R)^2 \cdot \rho_2} \]
This simplifies to:
\[ \frac{V_{t1}}{V_{t2}} = \frac{R^2 \cdot 8 \rho_2}{4R^2 \cdot \rho_2} = \frac{8}{4} = 2 \]

### Step 6: Final ratio
Thus, the ratio of the terminal velocities is:
\[ V_{t1} : V_{t2} = 2 : 1 \]

### Conclusion
The ratio of the terminal velocities of the two spheres is \( 2:1 \).

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