A wire of 3 ohm resistance and 10 cm length is stretched to 30 cm length. Let us assume that it has a uniform cross-section, what can be its new resistance?

A) 27 Ω
B) 9 Ω
C) 3 Ω
D) 30 Ω

To determine the new resistance of the wire when it is stretched, let's analyze the situation step by step.

### Step 1: Relationship Between Resistance and Length
The resistance \( R \) of a wire is given by:
\[
R = \rho \frac{L}{A}
\]
where:
- \( \rho \) 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.

When the wire is stretched, its length changes, and its cross-sectional area decreases. However, the volume of the wire remains constant.

### Step 2: Volume Conservation
The volume of the wire is given by:
\[
V = A \cdot L
\]
Since the volume remains constant, the relationship is:
\[
A_1 L_1 = A_2 L_2
\]
where:
- \( A_1 \) and \( L_1 \) are the initial area and length,
- \( A_2 \) and \( L_2 \) are the final area and length.

Rearranging for \( A_2 \):
\[
A_2 = A_1 \frac{L_1}{L_2}
\]

### Step 3: New Resistance
The new resistance \( R_2 \) is expressed as:
\[
R_2 = \rho \frac{L_2}{A_2}
\]
Substitute \( A_2 = A_1 \frac{L_1}{L_2} \) into the equation:
\[
R_2 = \rho \frac{L_2}{A_1 \frac{L_1}{L_2}} = \rho \frac{L_2^2}{A_1 L_1}
\]
Using the initial resistance \( R_1 = \rho \frac{L_1}{A_1} \), rewrite \( R_2 \) as:
\[
R_2 = R_1 \frac{L_2^2}{L_1^2}
\]

### Step 4: Substitute the Values
Given:
- Initial resistance \( R_1 = 3 \, \Omega \),
- Initial length \( L_1 = 10 \, \text{cm} \),
- Final length \( L_2 = 30 \, \text{cm} \),

Substitute into the formula:
\[
R_2 = 3 \cdot \frac{30^2}{10^2} = 3 \cdot \frac{900}{100} = 3 \cdot 9 = 27 \, \Omega
\]

### Final Answer:
The new resistance of the wire is **27 ohms**.

Answer: A. 27Ω