Consider a rectangular block of wood moving with a velocity `v_(0)` in a gas at temperature T and mass density p. Assume the velocity is along x-axis and the are of cross-section of the block perpendicular to `v_(0)` is A. show that the drag force on the block is `4rAv_(0)sqrt((kT)/(m))` where,m is the mass of the gas molecule.

Options

A. `4pAv_(0)sqrt((kT)/(m))`
B. `2pAv_(0)sqrt((kT)/(3m))`
C. `(pA)/(2v_(0))sqrt((kT)/(m))`
D. `(v_(0))/(pA)sqrt((kT)/(2m))`

Explanation

To derive the expression for the drag force on a rectangular block of wood moving through a gas, we can follow these steps:

### Step 1: Understand the System
We have a rectangular block of wood moving with a velocity \( v_0 \) in a gas at temperature \( T \) and mass density \( \rho \). The area of cross-section of the block perpendicular to the velocity is \( A \).

### Step 2: Determine the Number Density of Gas Molecules
The number density \( n \) of gas molecules can be expressed as:
\[ n = \frac{\rho}{m} \]
where \( m \) is the mass of a gas molecule.

### Step 3: Calculate the Average Speed of Gas Molecules
The average speed \( v \) of gas molecules can be derived from the kinetic theory of gases:
\[ v = \sqrt{\frac{8kT}{\pi m}} \]
where \( k \) is the Boltzmann constant.

### Step 4: Calculate the Rate of Collisions
The number of collisions per unit time \( N \) that the block experiences can be calculated as:
\[ N = n \cdot A \cdot v \]
Substituting for \( n \):
\[ N = \left(\frac{\rho}{m}\right) A \cdot v \]

### Step 5: Determine the Change in Momentum
When the block moves through the gas, it collides with the gas molecules. The change in momentum during each collision can be considered. For a single molecule, the change in momentum when it collides with the block is:
\[ \Delta p = 2mv \]
Thus, the total change in momentum per unit time (which gives us the drag force \( F_d \)) is:
\[ F_d = N \cdot \Delta p = N \cdot 2mv \]
Substituting for \( N \):
\[ F_d = \left(\frac{\rho}{m}\right) A v \cdot 2mv = 2\rho A v^2 \]

### Step 6: Substitute the Average Speed
Now substituting the expression for \( v \):
\[ F_d = 2\rho A \left(\sqrt{\frac{8kT}{\pi m}}\right)^2 \]
This simplifies to:
\[ F_d = 2\rho A \cdot \frac{8kT}{\pi m} = \frac{16\rho A kT}{\pi m} \]

### Step 7: Relate Density to the Drag Force
Using the ideal gas law, we can relate the density \( \rho \) to the number density \( n \) and the temperature \( T \):
\[ \rho = n m \]
Substituting this back into the drag force equation gives:
\[ F_d = \frac{16 n m A kT}{\pi m} = \frac{16 n A kT}{\pi} \]

### Step 8: Final Expression for Drag Force
Now, we can express the drag force in terms of the velocity \( v_0 \):
\[ F_d = 4 \cdot A \cdot v_0 \cdot \sqrt{\frac{8kT}{\pi m}} = 4 A v_0 \sqrt{\frac{kT}{m}} \]

Thus, we arrive at the final expression for the drag force:
\[ F_d = 4 A v_0 \sqrt{\frac{kT}{m}} \]

### Final Result
The drag force on the block is given by:
\[ F_d = 4 A v_0 \sqrt{\frac{kT}{m}} \]