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ICSE Class 10 Physics 2013 Paper

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Question : 58 of 70
Marks: +1, -0
A metal wire of resistance 6Ω6 \Omega is stretched so that its length is increased to twice its original length. Calculate its new resistance.
Solution:  
Volume of metal wire remains same.
∵A1l1=A2l2\because A_1 l_1 = A_2 l_2
l2l1=A1A2\frac{l_2}{l_1} = \frac{A_1}{A_2}
R1=ρl1A1R_1 = \rho \frac{l_1}{A_1}
R2=ρl2A2R_2 = \rho \frac{l_2}{A_2}
R2R1=l2A2×A1l1=l2l1×l2l1\frac{R_2}{R_1} = \frac{l_2}{A_2} \times \frac{A_1}{l_1} = \frac{l_2}{l_1} \times \frac{l_2}{l_1}
R2=(l2l1)2R1R_2 = \left(\frac{l_2}{l_1}\right)^2 R_1
[∵l2=2l1. (Given)][\because l_2 = 2l_1. \text{ (Given)}]
=(2l1l1)2R1= \left(\frac{2l_1}{l_1}\right)^2 R_1
R2=4R1R_2 = 4 R_1
=4×6= 4 \times 6
=24Ω= 24 \Omega
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