A resistance R is stretched to four times its length. Its new resistance will be

Options

A. 4R
B. 64R
C. `R//4`
D. `16R`

Answer

To find the new resistance when a resistance \( R \) is stretched to four times its length, we can follow these steps:

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

### Step 2: Set up the initial conditions
Let the initial length of the conductor be \( L_1 \) and the initial cross-sectional area be \( A_1 \). The initial resistance can be expressed as:
\[ R_1 = \frac{\rho L_1}{A_1} \]

### Step 3: Determine the new length and area after stretching
When the conductor is stretched to four times its length, the new length \( L_2 \) is:
\[ L_2 = 4L_1 \]
Since the volume of the conductor remains constant during stretching, we have:
\[ \text{Volume} = A_1 L_1 = A_2 L_2 \]
From this, we can express the new area \( A_2 \):
\[ A_2 = \frac{A_1 L_1}{L_2} = \frac{A_1 L_1}{4L_1} = \frac{A_1}{4} \]

### Step 4: Calculate the new resistance
Now we can find the new resistance \( R_2 \):
\[ R_2 = \frac{\rho L_2}{A_2} \]
Substituting \( L_2 \) and \( A_2 \):
\[ R_2 = \frac{\rho (4L_1)}{\frac{A_1}{4}} = \frac{4\rho L_1}{\frac{A_1}{4}} = 16 \frac{\rho L_1}{A_1} = 16 R_1 \]
Thus, the new resistance is:
\[ R_2 = 16R \]

### Final Answer
The new resistance when a resistance \( R \) is stretched to four times its length will be \( 16R \).

---