A wire of a certain material is stretched slowly by ten percent. Its new resistance and specific resistance become respectively

Question

In a meter bridge experiment, a resistance of \( 10 \, \Omega \) is balanced by a resistance \( X \) with a balance point at \( 40 \, \mathrm{cm} \). What is the value of \( X \)?

Answer 1

To solve this problem, we need to understand the relationship between the stretching of the wire and its resistance. Resistance (\(R\)) of a wire is related to its resistivity (\(\rho\)), length (\(L\)), and cross-sectional area (\(A\)) by the formula:

R = \frac{\rho L}{A}

When a wire is stretched, its length increases and its cross-sectional area decreases, while the volume remains constant.

Let's denote the original length of the wire as \(L_0\) and the original cross-sectional area as \(A_0\). After the wire is stretched by 10%, the new length (\(L_1\)) is:

L_1 = 1.1 L_0

Since the volume is conserved, the original volume equals the new volume:

V_0 = A_0 \cdot L_0 = A_1 \cdot L_1

Thus, the new cross-sectional area (\(A_1\)) can be given by:

A_1 = \frac{A_0 \cdot L_0}{L_1} = \frac{A_0 \cdot L_0}{1.1 L_0} = \frac{A_0}{1.1}

Now, the new resistance (\(R_1\)) is calculated as:

R_1 = \frac{\rho \cdot L_1}{A_1} = \frac{\rho \cdot 1.1L_0}{A_0/1.1} = \frac{1.1^2 \cdot \rho \cdot L_0}{A_0}

Since the original resistance is:

R_0 = \frac{\rho \cdot L_0}{A_0}

We find the new resistance \(R_1\) is:

R_1 = 1.21 R_0

Hence, the new resistance becomes 1.21 times the original resistance. Since resistivity (\(\rho\)) is an intrinsic property of the material and does not change with physical deformation, the specific resistance remains the same.

Therefore, the correct answer is: 1.21 times, same.

Answer 2