Question

In order to quadruple the resistan of a uniform wire, a part of it is uniformly stretched so that the final length of the wire becomes $1.5$ times the original length. The fractional length of the stretched part is

• A. $\dfrac{1}{6}$
• B. $\dfrac{1}{8}$
• C. $\dfrac{1}{4}$
• D. $\dfrac{1}{10}$

Hint:

Resistance after Stretching:

• $R = \rho \dfracLA$, and volume is conserved: $LA = L'A'$
• Stretching increases resistance by square of stretch factor.
• When partial stretch is involved, analyze each section separately.

Answer 1

The Correct Option is B

Let original length be $L_0$. Let fraction $x$ of it be stretched.
New length: $(1 - x)L_0 + (0.5 + x)L_0 = 1.5L_0$.
Volume constant $\Rightarrow A_0 xL_0 = A' (0.5 + x)L_0 \Rightarrow A' = A_0 \dfrac{x}{0.5 + x}$.
Resistance of unstretched part: $R_1 = \rho \dfrac{(1-x)L_0}{A_0}$
Resistance of stretched part: $R_2 = \rho \dfrac{(0.5+x)L_0}{A'} = \rho \dfrac{L_0}{A_0} \cdot \dfrac{(0.5 + x)^2}{x}$
Total resistance: \[ R = \rho \dfrac{L_0}{A_0} \left[ (1 - x) + \dfrac{(0.5 + x)^2}{x} \right] = 4 R_0 \] Solving: \[ (1 - x) + \frac{(0.5 + x)^2}{x} = 4 \Rightarrow 2x = 0.25 \Rightarrow x = \frac{1}{8} \]

Answer 2

Step 1: Understand the problem
The resistance of a wire depends on its length and cross-sectional area.
Given that part of the wire is stretched uniformly so that the final length is 1.5 times the original length.
We need to find the fractional length of the stretched part so that the overall resistance becomes quadruple.

Step 2: Recall the relation for resistance
Resistance \( R \) is proportional to \(\frac{L}{A}\), where \( L \) is length and \( A \) is cross-sectional area.
When the wire is stretched, volume remains constant:
\[ L \times A = \text{constant} \]
If length increases by factor \(k\), area decreases by factor \( \frac{1}{k} \).

Step 3: Define variables
Let total original length be \( L \).
Let \( x \) be the fractional length of the part stretched.
Length of stretched part: \( xL \), length of unstretched part: \( (1 - x)L \).

Step 4: After stretching
Stretched part length: \( 1.5 \times xL \).
Unstretched part length: remains \( (1 - x)L \).
Total final length:
\[ (1 - x)L + 1.5 xL = 1.5L \]
Check: the sum equals given total length after stretching, consistent with problem.

Step 5: Calculate resistance change
Resistance of unstretched part:
\[ R_1 = \rho \frac{(1 - x)L}{A} \]
Resistance of stretched part:
Length increased by 1.5 times → area decreases by \( \frac{1}{1.5} = \frac{2}{3} \).
New cross-sectional area for stretched part: \( \frac{A}{1.5} \).
So resistance of stretched part:
\[ R_2 = \rho \frac{1.5 x L}{A/1.5} = \rho \frac{1.5 x L \times 1.5}{A} = \rho \frac{2.25 x L}{A} \]

Step 6: Total resistance after stretching
\[ R_{final} = R_1 + R_2 = \rho \frac{(1 - x)L}{A} + \rho \frac{2.25 x L}{A} = \rho \frac{L}{A} \left(1 - x + 2.25 x \right) = R_{original} (1 + 1.25 x) \]

Step 7: Use the condition that resistance becomes quadruple
\[ R_{final} = 4 R_{original} \Rightarrow 1 + 1.25 x = 4 \]
\[ 1.25 x = 3 \Rightarrow x = \frac{3}{1.25} = 2.4 \]
This is more than 1, which is not possible since \(x\) is a fraction of the wire.

Step 8: Recheck step 5 (Correction)
The cross-sectional area decreases inversely with the increase in length.
If length increases by 1.5 times, cross-sectional area decreases by \( \frac{1}{1.5} = \frac{2}{3} \).
So new resistance:
\[ R_2 = \rho \frac{(1.5 x L)}{A \times (2/3)} = \rho \frac{1.5 x L}{\frac{2}{3} A} = \rho \frac{1.5 x L \times 3}{2 A} = \rho \frac{2.25 x L}{A} \]
Step 5 was correct.

Step 9: Consider that only part \(x\) is stretched and the final length is 1.5 times the original length
The total final length is:
\[ (1 - x)L + 1.5 x L = 1.5 L \]
Simplify:
\[ 1 - x + 1.5 x = 1.5 \Rightarrow 1 + 0.5 x = 1.5 \Rightarrow 0.5 x = 0.5 \Rightarrow x = 1 \]
This means entire wire is stretched, which contradicts the problem statement.

Step 10: Interpretation
The problem states that part of the wire is stretched so that final length is 1.5 times original length.
So, if part \(x\) is stretched by a factor \(k\), then:
\[ (1 - x)L + k x L = 1.5 L \]
Rearranged:
\[ 1 - x + k x = 1.5 \Rightarrow 1 + x(k - 1) = 1.5 \Rightarrow x(k - 1) = 0.5 \]

Step 11: Use resistance quadruple condition
Total resistance after stretching:
\[ R = R_{original} \left[(1 - x) + \frac{k^2 x}{1} \right] \]
Because resistance \( R \propto \frac{L}{A} \), and with volume constant:
\[ A_{new} = \frac{A}{k} \]
Resistance of stretched part:
\[ R_2 = \rho \frac{k x L}{A/k} = \rho \frac{k^2 x L}{A} \]
Hence:
\[ 4 = (1 - x) + k^2 x \]

Step 12: Solve system of equations
From length:
\[ x(k - 1) = 0.5 \]
From resistance:
\[ 4 = 1 - x + k^2 x = 1 + x(k^2 - 1) \]
Rearranged:
\[ x(k^2 - 1) = 3 \]

Step 13: Express \( x \) from length equation
\[ x = \frac{0.5}{k - 1} \]
Substitute into resistance equation:
\[ \frac{0.5}{k - 1} (k^2 - 1) = 3 \Rightarrow 0.5 (k + 1) = 3 \Rightarrow k + 1 = 6 \Rightarrow k = 5 \]

Step 14: Calculate \( x \)
\[ x = \frac{0.5}{5 - 1} = \frac{0.5}{4} = \frac{1}{8} \]

Final answer:
The fractional length of the stretched part is \( \boxed{\frac{1}{8}} \).