The resistance of a wire is `20 ohm`. It is so stretched that the length becomes three times, then the new resistance of the wire will be

Options

A. 6.67 ohm
B. 60 ohm
C. 120 ohm
D. 180 ohm

Explanation

To solve the problem, we need to understand how the resistance of a wire changes when its length is altered. 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 1: Understand the initial conditions
The initial resistance of the wire is given as \( R_1 = 20 \, \Omega \). Let's denote the initial length of the wire as \( L \) and the initial cross-sectional area as \( A \).

### Step 2: Determine the new length
When the wire is stretched to three times its original length, the new length \( L' \) becomes:
\[ L' = 3L \]

### Step 3: Determine the new cross-sectional area
When a wire is stretched, its volume remains constant (assuming no material is added or removed). The volume \( V \) of the wire can be expressed as:
\[ V = L \times A \]
After stretching, the volume is still the same:
\[ V' = L' \times A' = 3L \times A' \]
Setting the initial and final volumes equal gives us:
\[ L \times A = 3L \times A' \]
From this, we can solve for the new cross-sectional area \( A' \):
\[ A' = \frac{A}{3} \]

### Step 4: Calculate the new resistance
Now we can find the new resistance \( R' \) using the resistance formula:
\[ R' = \rho \frac{L'}{A'} = \rho \frac{3L}{A/3} = \rho \frac{3L \times 3}{A} = 9 \rho \frac{L}{A} \]
Since \( R = \rho \frac{L}{A} \), we can substitute:
\[ R' = 9R \]
Given that \( R = 20 \, \Omega \):
\[ R' = 9 \times 20 \, \Omega = 180 \, \Omega \]

### Final Answer
The new resistance of the wire after it has been stretched is \( 180 \, \Omega \).

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