Current Electricity Ques 84
84. A wire of resistance $4 \Omega$ is stretched to twice its original length. The resistance of stretched wire would be
[2013]
(a) $4 \Omega$
(b) $8 \Omega$
(c) $16 \Omega$
(d) $2 \Omega$
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Answer:
Correct Answer: 84.(c)
Solution:
- (c) Resistance $R=\frac{\rho \ell}{A}=4 \Omega$
$\because \ell^{\prime}=2 \ell$
On stretching, volume of wire remains constant $\therefore \ell A^{\prime}=\ell^{\prime} A^{\prime}$
$\therefore A^{\prime}=\frac{A}{2}$
$\therefore R^{\prime} \rho \frac{2 \ell}{\frac{A}{2}}=4 R=4 \times 4 \Omega=16 \Omega \Rightarrow \ell A=2 \ell A^{\prime}$
Therefore the resistance of new wire becomes $16 \Omega$
BETA
