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ICSE Class 10 Physics 2013 Solved Paper

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Question : 58
Total: 70
A metal wire of resistance 6Ω 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
‌‌
l2
l1
=‌
A1
A2

‌R1=ρ‌
l1
A1

‌‌ New resistance, ‌R2=ρ‌
l2
A2

‌‌
R2
R1
=‌
l2
A2
×‌
A1
l1
=‌
l2
l1
×‌
l2
l1

‌R2=(‌
l2
l1
)2R1
‌[∵l2=2l1.‌ (Given)] ‌
‌=(‌
2l1
l1
)2R1
‌R2=4R1
‌=4×6
‌=24Ω
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