Electric current is passing through a solid conductor PQ from P to Q. The electric current densities at p and Q are in the ratio.

Options

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

Answer

To solve the problem of finding the ratio of electric current densities at points P and Q in a solid conductor, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Current Density**:
Current density (J) is defined as the amount of electric current (I) flowing per unit area (A) of the conductor. Mathematically, it is given by:
\[ J = \frac{I}{A} \]

2. **Identify the Cross-Sectional Area**:
For a cylindrical conductor, the cross-sectional area (A) can be calculated using the formula for the area of a circle:
\[ A = \pi r^2 \]
where \( r \) is the radius of the conductor.

3. **Define Current Densities at Points P and Q**:
- Let the current density at point P be \( J_P \) and at point Q be \( J_Q \).
- The current density at point P can be expressed as:
\[ J_P = \frac{I}{A_P} = \frac{I}{\pi r_P^2} \]
- The current density at point Q can be expressed as:
\[ J_Q = \frac{I}{A_Q} = \frac{I}{\pi r_Q^2} \]

4. **Set Up the Ratio of Current Densities**:
To find the ratio of current densities at points P and Q, we can write:
\[ \frac{J_P}{J_Q} = \frac{\frac{I}{\pi r_P^2}}{\frac{I}{\pi r_Q^2}} = \frac{r_Q^2}{r_P^2} \]
Here, the current (I) and the constant (\(\pi\)) cancel out.

5. **Assume a Relationship Between Radii**:
If we assume that the radius at point P is half of that at point Q, i.e., \( r_P = \frac{1}{2} r_Q \), we can substitute this into our ratio:
\[ \frac{J_P}{J_Q} = \frac{r_Q^2}{\left(\frac{1}{2} r_Q\right)^2} = \frac{r_Q^2}{\frac{1}{4} r_Q^2} = 4 \]

6. **Final Ratio of Current Densities**:
Thus, we can conclude that:
\[ \frac{J_P}{J_Q} = \frac{1}{4} \]
Therefore, the ratio of current densities at points P and Q is:
\[ J_P : J_Q = 1 : 4 \]

### Conclusion:
The ratio of electric current densities at points P and Q is \( 1 : 4 \).