A wire of resistanc e`9Omega` is broken in two parts. The length ratio being `1:2`. The two pieces are connected in parallel. The net resistance will be

Options

A. `2Omega`
B. `3Omega`
C. `4Omega`
D. `6Omega`

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understanding the Problem
We have a wire with a total resistance of \( R = 9 \, \Omega \) that is broken into two parts in the ratio of \( 1:2 \). We need to find the equivalent resistance when these two parts are connected in parallel.

**Hint:** Remember that the resistance of a wire is directly proportional to its length when the material and cross-sectional area remain constant.

### Step 2: Determine the Lengths of the Two Parts
Let the lengths of the two parts be \( L_1 \) and \( L_2 \). Given the ratio \( L_1 : L_2 = 1 : 2 \), we can express the lengths as:
- \( L_1 = x \)
- \( L_2 = 2x \)

### Step 3: Calculate the Resistance of Each Part
Using the relationship between resistance and length, we know that:
- \( R_1 \) (for length \( L_1 \)) is proportional to \( L_1 \)
- \( R_2 \) (for length \( L_2 \)) is proportional to \( L_2 \)

Since the total resistance \( R_1 + R_2 = 9 \, \Omega \) and the lengths are in the ratio \( 1:2 \), we can express the resistances as:
- \( R_1 = k \cdot L_1 = k \cdot x \)
- \( R_2 = k \cdot L_2 = k \cdot 2x \)

Thus, we have:
\[ R_1 + R_2 = kx + 2kx = 3kx = 9 \, \Omega \]
From this, we can find \( kx \):
\[ kx = \frac{9}{3} = 3 \, \Omega \]

### Step 4: Find Individual Resistances
Now we can find \( R_1 \) and \( R_2 \):
- \( R_1 = kx = 3 \, \Omega \)
- \( R_2 = k \cdot 2x = 2 \cdot 3 = 6 \, \Omega \)

### Step 5: Calculate the Equivalent Resistance in Parallel
The equivalent resistance \( R_{eq} \) of two resistances \( R_1 \) and \( R_2 \) connected in parallel is given by the formula:
\[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} \]
Substituting the values we found:
\[ \frac{1}{R_{eq}} = \frac{1}{3} + \frac{1}{6} \]

### Step 6: Simplify the Expression
To add the fractions:
\[ \frac{1}{R_{eq}} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]
Thus, taking the reciprocal gives:
\[ R_{eq} = 2 \, \Omega \]

### Conclusion
The net resistance when the two pieces are connected in parallel is \( R_{eq} = 2 \, \Omega \).

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