If a wire of 10 ohms is stretched to double its length keeping it's volume constant then what will be the new resistance?

80 Ω

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40 Ω

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100 Ω

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60 Ω

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Solution

The correct option is B $40 Ω$
When we stretch a wire to increase it's length without changing it's volume then:
$\frac{R_{1}}{R_{2}} = (\frac{l_{1}}{l_{2}})^{2} \Rightarrow \frac{10}{R_{2}} = (\frac{l}{2 l})^{2} = \frac{1}{4}$
$R_{2} = 40 Ω$

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If a wire of 10 ohms is stretched to double its length keeping it's volume constant then what will be the new resistance?

The correct option is B 40ΩWhen we stretch a wire to increase it's length without changing it's volume then: R1R2=(l1l2)2⇒10R2=(l2l)2=14 R2=40Ω

The correct option is B $40 Ω$
When we stretch a wire to increase it's length without changing it's volume then:
$\frac{R_{1}}{R_{2}} = (\frac{l_{1}}{l_{2}})^{2} \Rightarrow \frac{10}{R_{2}} = (\frac{l}{2 l})^{2} = \frac{1}{4}$
$R_{2} = 40 Ω$