The length of a given cylindrical wire is increased by `100%`. Due to the consequent decrease in diameter the change in the resistance of the wire will be

Options

A. (a) 200%
B. (b) 100%
C. (c) 50%
D. (d) 300%

Explanation

To solve the problem, we need to analyze how the resistance of a cylindrical wire changes when its length is increased and its diameter decreases. Let's break it down step by step.

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

The cross-sectional area \( A \) of a cylindrical wire can be expressed as:
\[ A = \pi \left( \frac{d}{2} \right)^2 = \frac{\pi d^2}{4} \]
where \( d \) is the diameter of the wire.

### Step 2: Analyze the change in length
The problem states that the length of the wire is increased by 100%. If the original length is \( L \), the new length \( L' \) will be:
\[ L' = L + 100\% \text{ of } L = 2L \]

### Step 3: Analyze the change in diameter
Since the problem states that the diameter decreases due to the increase in length, we need to find the new diameter. The volume of the wire remains constant (assuming no material is added or removed). The volume \( V \) of a cylinder is given by:
\[ V = A \times L = \frac{\pi d^2}{4} \times L \]
Let the new diameter be \( d' \). The volume after the change in length will be:
\[ V' = A' \times L' = \frac{\pi (d')^2}{4} \times (2L) \]
Setting the volumes equal gives:
\[ \frac{\pi d^2}{4} \times L = \frac{\pi (d')^2}{4} \times (2L) \]
Canceling common terms:
\[ d^2 = 2(d')^2 \]
Thus, we find:
\[ d' = \frac{d}{\sqrt{2}} \]

### Step 4: Calculate the new area
Now, we can find the new cross-sectional area \( A' \):
\[ A' = \frac{\pi (d')^2}{4} = \frac{\pi \left( \frac{d}{\sqrt{2}} \right)^2}{4} = \frac{\pi \frac{d^2}{2}}{4} = \frac{\pi d^2}{8} \]

### Step 5: Calculate the new resistance
Now we can calculate the new resistance \( R' \):
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (2L)}{\frac{\pi d^2}{8}} = \frac{16 \rho L}{\pi d^2} = 16 \left( \frac{\rho L}{\pi d^2} \right) = 16 R \]

### Step 6: Find the percentage change in resistance
Now we can find the percentage change in resistance:
\[ \text{Percentage Change} = \frac{R' - R}{R} \times 100\% \]
Substituting the values:
\[ = \frac{16R - R}{R} \times 100\% = \frac{15R}{R} \times 100\% = 1500\% \]

### Final Answer
The change in the resistance of the wire will be **1500%**.