Question 16: Current Electricity

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

Accepted Answer

When a conducting wire is stretched due to a deforming force, its diameter decreases to 40% of its original value. We need to determine the percentage change in its resistance.
Conceptual Framework:
Volume Conservation:
When the wire is stretched, its volume ($V$) remains constant. Therefore, the relationship between volume, cross-sectional area ($A$), and length ($l$) is given by:
$ V = A 	imes l $
Solving for $l$, we have:
$ l = \frac{V}{A} $
Resistance Formula:
The resistance ($R$) of a wire is determined by its resistivity ($ho$), length, and cross-sectional area:
$ R = ho \cdot \frac{l}{A} = \frac{ho V}{A^2} $
Given the relation between area and diameter ($D$), we express $A$ in terms of $D$:
$ A = \frac{\pi D^2}{4} $
Thus, resistance can be reformulated as:
$ R = \frac{ho V}{A^2} = \frac{16 ho V}{\pi^2 D^4} $
Change in Resistance:
The percentage change in resistance can be calculated using the change in diameter:
$ \frac{\Delta R}{R} = -4 \frac{\Delta D}{D} $
Given the diameter decreases to 40% of its original value, the change in diameter is $\Delta D = -0.6D$.
Substituting this into our percentage change formula, we get:
$ \frac{\Delta R}{R} = -4(-0.6) = 4 	imes 0.6 = 1.6\% $
Therefore, the percentage change in resistance as a result of the stretching is 1.6%.

Answer

$ 0.9 \% $

Answer

$ 0.12 \% $

Answer

$ 1.6 \% $

Answer

$ 0.5 \% $

Current Electricity

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

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