A conducting wire is stretched by applying a deforming force, so that its diameter decreases to 40% of the original value. The percentage change in its resistance will be:
A conducting wire is stretched by applying a deforming force, so that its diameter decreases to 40% of the original value. The percentage change in its resistance will be:
Show Hint
- \(0.9\%\)
- \(0.12\%\)
- \(1.6\%\)
- \(0.5\%\)
The Correct Option is C
Approach Solution - 1
Step 1: {Understanding the effect of stretching}
Since the volume of the wire remains constant, we use the relation: \[ V = A l \] where \(A\) is the cross-sectional area and \(l\) is the length.
Step 2: {Deriving the new resistance}
We use the resistance formula: \[ R = \rho \frac{l}{A} \] Since \(A\) decreases as \(d^2\) and \(l\) increases proportionally: \[ \frac{\Delta R}{R} = -4 \frac{\Delta D}{D} \] Substituting \(\Delta D = -0.4\), \[ \frac{\Delta R}{R} = -4(-0.4) = 1.6\% \] Thus, the percentage change in resistance is \(1.6\%\).
Approach Solution -2
Resistance of a wire is given by:
\( R = \rho \frac{L}{A} \)
where:
- \( \rho \) is the resistivity
- \( L \) is the length
- \( A \) is the cross-sectional area
Step 2: Volume of the wire remains constant when stretched
Since volume \( V = A \times L \), and volume remains constant,
\( A_1 L_1 = A_2 L_2 \)
Step 3: New diameter is 40% of the original
Let the original diameter be \( d \), so the new diameter is \( 0.4d \)
Then the new area becomes:
\( A_2 = \frac{\pi (0.4d)^2}{4} = 0.16 A_1 \)
Step 4: Use volume conservation to find new length
From \( A_1 L_1 = A_2 L_2 \),
\( L_2 = \frac{A_1}{A_2} L_1 = \frac{1}{0.16} L_1 = 6.25L_1 \)
Step 5: New resistance
\[ R_2 = \rho \frac{L_2}{A_2} = \rho \frac{6.25L}{0.16A} = \frac{6.25}{0.16} R = 39.0625R \]
Step 6: Percentage change in resistance
\[ \% \text{ change} = \left( \frac{R_2 - R_1}{R_1} \right) \times 100 = \left( \frac{39.0625R - R}{R} \right) \times 100 = 3806.25\% \]
Clarification
The given answer is \( 1.6\% \), which would correspond to a very small reduction in diameter, not a reduction to 40%.
If the diameter truly reduces to 40% of the original, the percentage increase in resistance is about 3806.25%, not 1.6%.
Final Answer (as per given condition of diameter becoming 0.4d):
Resistance increases by approximately 3806.25%
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