Question

A conducting wire is stretched by applying a deforming force, so that its diameter decreases to 40% of original value. The percentage change in its resistance will be

• A. 0.9%
• B. 0.12%
• C. 1.6%
• D. 0.5%

Answer 1

Answer: (C)

1.6%

Answer 2

Let $R_0$, $L_0$, $A_0$, and $d_0$ be the original resistance, length, cross-sectional area, and diameter of the wire, respectively. Let $R_f$, $L_f$, $A_f$, and $d_f$ be the final resistance, length, cross-sectional area, and diameter after stretching. The resistance of a wire is given by $R = \rho \frac{L}{A}$, where $\rho$ is the resistivity of the material. The cross-sectional area of a cylindrical wire is $A = \frac{\pi d^2}{4}$. So, $R = \rho \frac{L}{\frac{\pi d^2}{4}} = \frac{4 \rho L}{\pi d^2}$.

When the wire is stretched, the volume of the material remains constant. Original volume $V_0 = A_0 L_0 = \frac{\pi d_0^2}{4} L_0$. Final volume $V_f = A_f L_f = \frac{\pi d_f^2}{4} L_f$. Since $V_0 = V_f$, we have $\frac{\pi d_0^2}{4} L_0 = \frac{\pi d_f^2}{4} L_f$, which simplifies to $d_0^2 L_0 = d_f^2 L_f$. This gives the relationship between length and diameter: $L \propto \frac{1}{d^2}$.

The problem states that the diameter decreases to 40% of the original value. So, $d_f = 0.40 d_0$.

Now, let's express the final resistance $R_f$ in terms of the original resistance $R_0$. $R_0 = \frac{4 \rho L_0}{\pi d_0^2}$. $R_f = \frac{4 \rho L_f}{\pi d_f^2}$. The ratio of the resistances is $\frac{R_f}{R_0} = \frac{\frac{4 \rho L_f}{\pi d_f^2}}{\frac{4 \rho L_0}{\pi d_0^2}} = \frac{L_f}{L_0} \left(\frac{d_0}{d_f}\right)^2$.

From the volume conservation, $d_0^2 L_0 = d_f^2 L_f$, we get $\frac{L_f}{L_0} = \left(\frac{d_0}{d_f}\right)^2$. Substituting this into the resistance ratio equation: $\frac{R_f}{R_0} = \left(\frac{d_0}{d_f}\right)^2 \left(\frac{d_0}{d_f}\right)^2 = \left(\frac{d_0}{d_f}\right)^4$.

We are given $d_f = 0.40 d_0$. So, $\frac{d_0}{d_f} = \frac{d_0}{0.40 d_0} = \frac{1}{0.40} = \frac{10}{4} = 2.5$.

Now, calculate the ratio of resistances: $\frac{R_f}{R_0} = (2.5)^4$. $2.5^2 = 6.25$. $2.5^4 = (6.25)^2 = 39.0625$.

So, $R_f = 39.0625 R_0$.

The percentage change in resistance is given by $\frac{R_f - R_0}{R_0} \times 100\%$. Percentage change $= \frac{39.0625 R_0 - R_0}{R_0} \times 100\% = \frac{(39.0625 - 1) R_0}{R_0} \times 100\%$. Percentage change $= (38.0625) \times 100\% = 3806.25\%$.

This result (3806.25%) is not among the given options (0.9%, 0.12%, 1.6%, 0.5%).

There seems to be a significant discrepancy between the calculated value and the options provided. It is highly probable that there is a typo in the question (e.g., diameter decreases by 40%, or area decreases to 40%, or length increases to/by a certain percentage) or in the options.

Assuming the problem statement is correct as written ($d_f = 0.4 d_0$), the calculated percentage change is 3806.25%. Since this does not match any option, we cannot select any of the given options based on the direct calculation.

Given the standard nature of such problems in physics, the calculation $R \propto 1/d^4$ for constant volume is correct. The result 3806.25% is also correct based on $d_f = 0.4 d_0$. The discrepancy with the options strongly suggests an error in the question or options.

However, if forced to provide one of the given options as the correct answer, and knowing that C is the indicated solution, it implies the intended question, despite being written as "diameter decreases to 40% of original value", should somehow lead to a 1.6% change in resistance. This is impossible with the standard physics principles applied correctly to the stated condition.