A copper wire is stretched to make it 0.5% longer. The percentage change in its electrical resistanc

Q: Why can't I just say the resistance increases by $0.5\%$ since $R$ is proportional to $l$?

Because the cross-sectional area $A$ doesn't stay constant! When you stretch a wire, it gets longer but also thinner to conserve volume. You must account for the decrease in area as well.

Q: How did we get the formula $R = \rho \frac{l^2}{V}$?

We know $R = \rho \frac{l}{A}$ and Volume $V = A \cdot l$. From the volume equation, $A = \frac{V}{l}$. Substituting this into the resistance formula gives $R = \rho \frac{l}{\frac{V}{l}} = \rho \frac{l^2}{V}$.

Q: Why is the percentage change in resistance exactly double the percentage change in length?

This is a result of the binomial approximation or calculus for small changes. Since $R \propto l^2$, taking the derivative gives $\frac{dR}{R} = 2 \frac{dl}{l}$. This means the fractional change in $R$ is twice the fractional change in $l$.

Q: Does this shortcut work for a $20\%$ increase in length?

No! The error approximation shortcut is only accurate for very small changes, typically less than $5\%$. For a $20\%$ increase, you must use the exact calculation: $(1.2)^2 - 1 = 0.44$, which is a $44\%$ increase, not $40\%$.

Q: What if the wire was compressed to be $0.5\%$ shorter?

The same logic applies! The length decreases, so the resistance decreases. Using the shortcut, the percentage change in resistance would be $2 \times (-0.5\%) = -1.0\%$. The resistance would decrease by $1\%$.

Question

A copper wire is stretched to make it $0.5\%$ longer. The percentage change in its electrical resistance, if its volume remains unchanged is

Options:

(A) $2.0\%$

(B) $1.0\%$

(C) $0.5\%$

(D) $2.5\%$

Sigma Percentile · Accepted Answer

The Art of Stretching Wires: A Tale of Resistance and Volume

Have you ever wondered what happens to a wire when you stretch it? It gets longer, sure, but it also gets thinner! This simple physical reality is the heart of many classic physics problems. Let's dive into the fascinating relationship between a wire's dimensions and its electrical resistance.

### Analyzing the Setup

Imagine a cylindrical copper wire. It has an initial length $l$ and a cross-sectional area $A$. The resistance $R$ of this wire is given by the fundamental formula:

$$R = ho \frac{l}{A}$$

where $ho$ is the resistivity of copper.

Now, the problem states that the wire is stretched to make it $0.5\%$ longer. The crucial piece of information here is that its **volume remains unchanged**. When you stretch the wire, you aren't adding or removing any copper; you are simply reshaping it.

### The Master Equation

Since the volume $V$ of a cylinder is the product of its cross-sectional area and length ($V = A \cdot l$), we can express the area as $A = \frac{V}{l}$.

Let's substitute this into our resistance formula to eliminate the area variable, which we don't explicitly know:

$$R = ho \frac{l}{\frac{V}{l}} = ho \frac{l^2}{V}$$

This is our master equation! It tells us that for a constant volume, the resistance is directly proportional to the square of the length ($R \propto l^2$).

### Final Calculation

The wire is stretched by $0.5\%$, so the new length $l'$ is:

$$l' = l + 0.005l = 1.005l$$

Using our master equation, the new resistance $R'$ is:

$$R' = ho \frac{(1.005l)^2}{V}$$

To find the percentage change, we first find the ratio of the new resistance to the old resistance:

$$\frac{R'}{R} = \frac{ho \frac{(1.005l)^2}{V}}{ho \frac{l^2}{V}} = (1.005)^2$$

The percentage change in resistance is:

$$\% 	ext{ Change} = \left( \frac{R'}{R} - 1 ight) 	imes 100$$

$$\% 	ext{ Change} = [(1.005)^2 - 1] 	imes 100 = [1.010025 - 1] 	imes 100 = 1.0025\%$$

Rounding to the nearest given option, we get **$1.0\%$**.

### The Way Forward: The Error Approximation Shortcut

For competitive exams like JEE and NEET, time is of the essence. When dealing with very small percentage changes (typically less than $5\%$), we can use a powerful shortcut derived from calculus (error approximation).

Since $R \propto l^2$, taking the natural logarithm and differentiating gives us:

$$\frac{\Delta R}{R} = 2 \frac{\Delta l}{l}$$

Multiplying by 100 to get percentages:

$$\% 	ext{ Change in } R = 2 	imes (\% 	ext{ Change in } l)$$

Applying this to our problem:

$$\% 	ext{ Change in } R = 2 	imes 0.5\% = 1.0\%$$

This shortcut gets us to the exact answer in seconds! Always keep an eye out for small percentage changes to deploy this trick.

Animated Solution for Physics - Current Electricity: A copper wire is stretched to make it $0.5%$ longer. The percentage change in its electrical resistance, if its volume remains unchanged is

Select Answer:

Visualized Solution

Initial Resistance

Resistance in terms of Volume

New Length

New Resistance

Ratio of Resistances

Percentage Change

Shortcut for Small Changes

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

Analyzing the Setup

The Master Equation

Final Calculation

The Way Forward

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