Current Electricity 1 Question 9
9. A copper wire is stretched to make it $0.5 %$ longer. The percentage change in its electrical resistance, if its volume remains unchanged is
(2019 Main, 9 Jan I)
(a) $2.0 %$
(b) $1.0 %$
(c) $0.5 %$
(d) $2.5 %$
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Solution:
- Electrical resistance of wire of length ’ $l$ ‘, area of cross-section ’ $A$ ’ and resistivity ’ $\rho$ ’ is given as
$$ R=\rho \frac{l}{A} $$
Since we know, volume of the wire is
$$ V=A \times l $$
$\therefore$ From Eqs. (i) and (ii), we get
$$ R=o \frac{l^{2}}{V} $$
As, the length has been increased to $0.5 %$.
$\therefore$ New length of the wire,
$$ \begin{aligned} l^{\prime} & =l+0.5 % \text { of } l \\ & =l+0.005 l=1.005 l \end{aligned} $$
But $V$ and $\rho$ remains unchanged.
So, new resistance, $R^{\prime}=\frac{\rho[(1.005) l]^{2}}{V}$
Dividing Eq. (iv) and Eq. (iii), we get
$$ \frac{R^{\prime}}{R}=(1.005)^{2} $$
$\Rightarrow %$ change in the resistance
$$ \begin{aligned} & =\frac{R^{\prime}}{R}-1 \times 100 \\ & =\left[(1.005)^{2}-1\right] \times 100=1.0025 % \simeq 1 % \end{aligned} $$
