Current Electricity Ques 28
28. If a wire of resistance $R$ is melted and recasted to half of its length, then the new resistance of the wire will be
(a) $R / 4$
(b) $R / 2$
(c) $R$
(d) $2 R$
[1995]
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Answer:
Correct Answer: 28.(a)
Solution:
- (a) Initial resistance $(R_1)=R$; Initial length is $\ell_1$ and final length $(\ell_2)=0.5 \ell$. Volume of a wire $=\ell . A$. Since the volume of the wire remains the same after recasting, therefore $\ell_1 \cdot A_1=\ell_2$ $A_2$
or $\frac{\ell_1}{\ell_2}=\frac{A_2}{A_1}$ or $\frac{\ell}{0.5 \ell}=\frac{A_2}{A_1}$ or $\frac{A_2}{A_1}=2$.
We also know that resistance of a wire $(R)$
$R=\rho \times \frac{\ell}{A} . ; R \propto \frac{\ell}{A}$
$\therefore \frac{R_1}{R_2}=\frac{\ell_1}{\ell_2} \times \frac{A_2}{A_1}=\frac{\ell}{0.5 \ell} \times 2=4$
or , $R_2=\frac{R_1}{4}=\frac{R}{4}$.
When wires are drawn from same volume but with different area of cross-section, then
$R \propto \frac{1}{(\text{ Area of cross-section })^{2}}$
BETA
