The resistance of wire is `20 Omega`. The wire is stretched to three times its length. Then the resistance will now be

Options

A. `6.67 Omega`
B. `60 Omega`
C. `120 Omega`
D. `180 Omega`

Answer

To solve the problem of finding the new resistance of a wire when it is stretched to three times its length, we can follow these steps:

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

### Step 2: Determine the initial conditions
We know the initial resistance \( R_1 \) is \( 20 \, \Omega \) and the initial length \( L_1 \) is \( L \). The initial area \( A_1 \) can be expressed as:
\[ R_1 = \rho \frac{L_1}{A_1} \implies 20 = \rho \frac{L}{A_1} \]

### Step 3: Analyze the change in length
When the wire is stretched to three times its original length, the new length \( L_2 \) becomes:
\[ L_2 = 3L \]

### Step 4: Understand the volume conservation
The volume of the wire remains constant during stretching. The volume \( V \) is given by:
\[ V = A_1 L_1 = A_2 L_2 \]
where \( A_2 \) is the new cross-sectional area after stretching. Thus, we can express \( A_2 \) as:
\[ A_2 = \frac{A_1 L_1}{L_2} = \frac{A_1 L}{3L} = \frac{A_1}{3} \]

### Step 5: Calculate the new resistance
Now we can find the new resistance \( R_2 \) using the new length and area:
\[ R_2 = \rho \frac{L_2}{A_2} = \rho \frac{3L}{A_2} \]
Substituting \( A_2 \):
\[ R_2 = \rho \frac{3L}{\frac{A_1}{3}} = \rho \frac{3L \cdot 3}{A_1} = \rho \frac{9L}{A_1} \]

### Step 6: Relate the new resistance to the initial resistance
From the initial resistance equation, we know:
\[ R_1 = \rho \frac{L}{A_1} \implies \rho = R_1 \frac{A_1}{L} \]
Substituting this into the equation for \( R_2 \):
\[ R_2 = 9 \cdot R_1 \]
Given \( R_1 = 20 \, \Omega \):
\[ R_2 = 9 \cdot 20 = 180 \, \Omega \]

### Final Answer
The new resistance of the wire after being stretched to three times its length is:
\[ \boxed{180 \, \Omega} \]