The length of a given cylindrical wire is increased by 100%. Due to the consequent decrease in diame

Q: Why does the area decrease when the wire is stretched?

Because the volume of the material remains constant. If it gets longer, it must get thinner to compensate.

Q: Can I just use the formula $R = \rho l/A$ and assume $A$ is constant?

No, stretching a wire physically deforms it, changing both its length and its cross-sectional area simultaneously.

Q: What is the shortcut formula mentioned?

Since $V = A \cdot l$, we can write $A = V/l$. Substituting this into $R = \rho l/A$ gives $R = \rho l^2 / V$. Since $\rho$ and $V$ are constant during stretching, $R \propto l^2$.

Q: Why is the change $300\%$ and not $400\%$?

The new resistance is $400\%$ of the original ($4R$), but the *change* is the difference ($4R - R = 3R$), which corresponds to a $300\%$ increase.

Q: Does this apply if the wire is compressed instead of stretched?

Yes, the exact same principle of volume conservation applies. If compressed, the length decreases and the area increases proportionally.

Question

The length of a given cylindrical wire is increased by $100\%$. Due to the consequent decrease in diameter, the change in the resistance of the wire will be

Options:

(A) $200\%$

(B) $100\%$

(C) $50\%$

(D) $300\%$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a cylindrical wire. It has some initial length, let's call it $l$, and a cross-sectional area $A$. We know that the resistance of this wire depends on its material and dimensions. It is given by the fundamental formula:

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

Now, the question says the wire is stretched to increase its length by $100\%$. This means the new length $l'$ becomes twice the original length, that is, $2l$.

### The Catch: Conservation of Volume
Here is a catch. When you stretch a wire, it gets longer, but it also gets thinner. However, the total amount of material, or its volume, remains exactly the same. So, initial volume equals final volume.

$V_{	ext{initial}} = V_{	ext{final}}$

$A \cdot l = A' \cdot l'$

Let's substitute the new length into our volume equation. We get $A \cdot l = A' \cdot (2l)$. Solving this, we find that the new area $A'$ is exactly half of the original area $A$.

$A' = \frac{A}{2}$

### Calculating the New Resistance
Now, let's calculate the new resistance $R'$. Substituting the new length and new area into our resistance formula, we get:

$R' = ho \frac{l'}{A'} = ho \frac{2l}{A/2}$

This simplifies to $4$ times the original resistance $R$.

$R' = 4 \left( ho \frac{l}{A} ight) = 4R$

So, the resistance has become four times!

### Finding the Percentage Change
We need to find the percentage change. First, the absolute change is the final resistance minus the initial resistance, which is $4R - R$, giving us $3R$.

$\Delta R = R' - R = 4R - R = 3R$

The percentage change is this change divided by the original resistance, multiplied by one hundred.

$\% \Delta R = \frac{\Delta R}{R} 	imes 100\%$

Substituting the values, the $R$ cancels out, and we are left with $3 	imes 100\%$, which is **$300\%$**. That's our final answer.

### A Quick Pro-Tip
Here is a quick shortcut for your exams. Whenever a wire is stretched and volume is constant, resistance is directly proportional to the square of its length ($R \propto l^2$). Since length doubled, resistance becomes $2^2 = 4$ times. A four-fold increase means a $300\%$ change. Simple and fast!

Animated Solution for Physics - Current Electricity: The length of a given cylindrical wire is increased by $100%$ . Due to the consequent decrease in diameter, the change in the resistance of the wire will be

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Visualized Solution

$Initial State of the Wire$

$Resistance Formula$

$Stretching the Wire$

$Conservation of Volume$

$New Cross-sectional Area$

$New Resistance$

$Percentage Change in Resistance$

$Final Calculation$

$Shortcut Method$

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

Analyzing the Setup Imagine a cylindrical wire

The Catch

Calculating the New Resistance Now, let's calculate the new resistance $R^{'}$

Finding the Percentage Change We need to find the percentage change

A Quick Pro-Tip Here is a quick shortcut for your exams

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial State of the Wire

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resistance Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Stretching the Wire

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Volume

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } New Cross-sectional Area

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } New Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Percentage Change in Resistance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Shortcut Method