Question 169: Current Electricity

Question

Wheatstone bridge principle is used to measure the specific resistance $$\left(S_1\right)$$ of given wire, having length $$L$$, radius $$r$$. If $$X$$ is the resistance of wire, then specific resistance is ; $$S_1=X\left(\frac{\pi r^2}{L}\right)$$. If the length of the wire gets doubled then the value of specific resistance will be :

Accepted Answer

The specific resistance (or resistivity) of a material is a fundamental property that describes how much the material resists the flow of electric current. The resistivity is typically denoted by the symbol $$\rho$$ (rho), and it can be calculated by using the resistance $$X$$ of a uniform specimen of the material, along with its physical dimensions. In the case of a wire, the resistivity formula in terms of its resistance $$X$$, length $$L$$, and cross-sectional area $$A=\pi r^2$$ is given by:
$$
\rho = \frac{X \cdot A}{L}
$$
This formula is a reinterpretation of Ohm's law, and it states that the specific resistance is proportional to the area of the cross-section of the wire and inversely proportional to its length.
Given that the specific resistance of the wire $$S_1$$ is determined using the formula:
$$
S_1 = X \left(\frac{\pi r^2}{L}\right)
$$
Now, let's see what happens to the specific resistance if the length of the wire is doubled. If we denote the new length as $$2L$$, the resistance of the wire with the new length will change because resistance is directly proportional to the length of the wire. However, resistivity (specific resistance) is an intrinsic property of the material and does not depend on its length or shape, only on its temperature. Therefore, even if we change the length of the wire, the specific resistance should remain the same.
If we calculate the new resistance $$X'$$ with the doubled length, it would be:
$$
X' = X \left(\frac{2L}{L}\right) = 2X
$$
Thus, we would use the new resistance $$X'$$ and the new length $$2L$$ to calculate the specific resistance again:
$$
S_1' = X' \left(\frac{\pi r^2}{2L}\right) = 2X \left(\frac{\pi r^2}{2L}\right) = X \left(\frac{\pi r^2}{L}\right)
$$
Since this formula is essentially the same as our original formula for $$S_1$$, we can conclude that:
$$
S_1' = S_1
$$
Therefore, the value of the specific resistance $$S_1$$ will remain the same even if the length of the wire gets doubled. The correct answer to the question is:
Option D: $$S_1$$

Answer

$$\frac{S_1}{4}$$

Answer

$$2 \mathrm{~S}_1$$

Answer

$$\frac{\mathrm{S}_1}{2}$$

Answer

$$S_1$$

Current Electricity

Choose the correct answer: