Current Electricity Ques 27
- 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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Answer:
Correct Answer: 27.(b)
Solution:
Formula:
- Electrical resistance of wire of length ’ $l$ ‘, area of cross-section ’ $A$ ’ and resistivity ’ $\rho$ ’ is given as
$ R=\rho \frac{l}{A} $ $\quad$ …….(i)
Since we know, volume of the wire is
$ V=A \times l $ $\quad$ …….(ii)
$\therefore$ From Eqs. (i) and (ii), we get
$ R=o \frac{l^{2}}{V} $ $\quad$ …….(iii)
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, $R^{\prime}=\frac{\rho[(1.005) l]^{2}}{V}$ $\quad$ …….(iv)
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} $
BETA
