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Q.

A $10 Ω$ thick wire is stretched so that its length becomes three times then the new resistance of the wire is

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a

10 Ω

b

30 Ω

c

\frac{10}{9} Ω

d

90 Ω

answer is C.

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Detailed Solution

V_{1} = V_{2}

$A_{1} l_{1} = A_{2} l_{2}$

$∴ A_{2} = \frac{A_{1}}{3}$

$\frac{R_{2}}{R_{1}} = \frac{l_{2} A_{1}}{A_{2} l_{1}} = 9$

$R_{2} = 9 R_{1} = 90 Ω$

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