An electric of `5A` is passing through a circuit contaning three arrengement in parallel if the length and radius of the wires are in the ratio `2:3:4 and 3:4:5` then the ratio of current passing through wires should be

Options

A. `3 : 6 : 10`
B. `4 : 9 : 16`
C. `9: 16 : 25`
D. `54 : 64 : 75`

Answer

To solve the problem step by step, we need to find the ratio of the currents passing through three wires arranged in parallel, given the ratios of their lengths and areas.

### Step 1: Understand the relationship between resistance, length, and area
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{L}{A} \]
where \( L \) is the length of the wire and \( A \) is the cross-sectional area.

### Step 2: Define the lengths and areas based on the given ratios
Let the lengths of the three wires be:
- \( L_1 = 2k \)
- \( L_2 = 3k \)
- \( L_3 = 4k \)

Let the areas of the three wires be:
- \( A_1 = 3m \)
- \( A_2 = 4m \)
- \( A_3 = 5m \)

### Step 3: Calculate the resistances of the wires
Using the formula for resistance:
- For wire 1:
\[ R_1 = \frac{L_1}{A_1} = \frac{2k}{3m} \]
- For wire 2:
\[ R_2 = \frac{L_2}{A_2} = \frac{3k}{4m} \]
- For wire 3:
\[ R_3 = \frac{L_3}{A_3} = \frac{4k}{5m} \]

### Step 4: Find the equivalent resistance for the parallel arrangement
The formula for the equivalent resistance \( R_{eq} \) of resistors in parallel is:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \]

Substituting the values:
\[ \frac{1}{R_{eq}} = \frac{3m}{2k} + \frac{4m}{3k} + \frac{5m}{4k} \]

### Step 5: Calculate the common denominator and simplify
The common denominator for the fractions is \( 12k \):
\[ \frac{1}{R_{eq}} = \frac{18m}{12k} + \frac{16m}{12k} + \frac{15m}{12k} = \frac{(18 + 16 + 15)m}{12k} = \frac{49m}{12k} \]

Thus,
\[ R_{eq} = \frac{12k}{49m} \]

### Step 6: Use the current division rule
In a parallel circuit, the current through each resistor can be found using the current division rule:
\[ I_n = I_{total} \cdot \frac{R_{eq}}{R_n} \]
where \( I_{total} = 5A \).

Calculating the currents:
- For wire 1:
\[ I_1 = 5A \cdot \frac{R_{eq}}{R_1} = 5A \cdot \frac{\frac{12k}{49m}}{\frac{2k}{3m}} = 5A \cdot \frac{12 \cdot 3}{49 \cdot 2} = 5A \cdot \frac{36}{98} = \frac{180}{98} = \frac{90}{49} A \]

- For wire 2:
\[ I_2 = 5A \cdot \frac{R_{eq}}{R_2} = 5A \cdot \frac{\frac{12k}{49m}}{\frac{3k}{4m}} = 5A \cdot \frac{12 \cdot 4}{49 \cdot 3} = 5A \cdot \frac{48}{147} = \frac{240}{147} = \frac{80}{49} A \]

- For wire 3:
\[ I_3 = 5A \cdot \frac{R_{eq}}{R_3} = 5A \cdot \frac{\frac{12k}{49m}}{\frac{4k}{5m}} = 5A \cdot \frac{12 \cdot 5}{49 \cdot 4} = 5A \cdot \frac{60}{196} = \frac{300}{196} = \frac{75}{49} A \]

### Step 7: Find the ratio of currents
Now, we can find the ratio of the currents \( I_1 : I_2 : I_3 \):
\[ I_1 : I_2 : I_3 = \frac{90}{49} : \frac{80}{49} : \frac{75}{49} = 90 : 80 : 75 \]

### Step 8: Simplify the ratio
To simplify:
\[ 90 : 80 : 75 = 18 : 16 : 15 \]

### Final Answer
Thus, the ratio of the currents passing through the wires is:
\[ \boxed{18 : 16 : 15} \]