A uniform wire is cut into 10 segments of increasing length. Each segment is having a resistace of 8 `Omega ` more than that of previous segment. If the resistance of shortest segment is R and largest segment is 2 R. The original resistance of wrie in ohm.

Options

A. `72 `
B. `144`
C. `720 `
D. `1080`

Answer

To solve the problem, we need to determine the original resistance of the wire based on the given conditions. Here’s a step-by-step solution:

### Step 1: Define the resistances of the segments
Let the resistance of the shortest segment be \( R \). According to the problem, each subsequent segment has a resistance that is \( 8 \, \Omega \) more than the previous segment. Therefore, the resistances of the segments can be expressed as follows:

1. First segment: \( R \)
2. Second segment: \( R + 8 \)
3. Third segment: \( R + 16 \)
4. Fourth segment: \( R + 24 \)
5. Fifth segment: \( R + 32 \)
6. Sixth segment: \( R + 40 \)
7. Seventh segment: \( R + 48 \)
8. Eighth segment: \( R + 56 \)
9. Ninth segment: \( R + 64 \)
10. Tenth segment: \( R + 72 \)

### Step 2: Relate the largest segment to \( R \)
According to the problem, the resistance of the largest segment (the tenth segment) is \( 2R \). Thus, we can set up the equation:

\[ R + 72 = 2R \]

### Step 3: Solve for \( R \)
Rearranging the equation gives:

\[ 72 = 2R - R \]
\[ 72 = R \]

So, the resistance of the shortest segment \( R \) is \( 72 \, \Omega \).

### Step 4: Calculate the total resistance of the wire
Now, we need to find the total resistance of the wire, which is the sum of the resistances of all segments. The resistances of the segments can be expressed as an arithmetic progression (AP) where:

- First term \( a = R = 72 \)
- Common difference \( d = 8 \)
- Number of terms \( n = 10 \)

The last term (tenth term) can be calculated as:

\[ a_n = a + (n-1)d = 72 + (10-1) \cdot 8 = 72 + 72 = 144 \, \Omega \]

### Step 5: Use the formula for the sum of an arithmetic series
The sum \( S_n \) of the first \( n \) terms of an arithmetic series is given by:

\[ S_n = \frac{n}{2} \times (a + a_n) \]

Substituting the values we have:

\[ S_{10} = \frac{10}{2} \times (72 + 144) = 5 \times 216 = 1080 \, \Omega \]

### Final Answer
The original resistance of the wire is \( 1080 \, \Omega \).

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