ICSE Class 10 Physics 2013 Solved Paper

Question : 58 of 70

Marks: +1, -0

A metal wire of resistance

6 Ω

is stretched so that its length is increased to twice its original length. Calculate its new resistance.

Solution:

Volume of metal wire remains same.

\; ∵ \;\; {\A}_1 l_1= {\A}_2 l_2

\;\;{l_2}/{l_1}=\;{ {\A}_1}/{ {\A}_2}

\; {\R}_1=ρ \;{l_1}/{ {\A}_1}

\;\;\text " New resistance, "\; {\R}_2=ρ \;{l_2}/{ {\A}_2}

\;\;{ {\R}_2}/{ {\R}_1}=\;{l_2}/{ {\A}_2} × \;{ {\A}_1}/{l_1}=\;{l_2}/{l_1} × \;{l_2}/{l_1}

\; {\R}_2= (\;{l_2}/{l_1}) ^2 {\R}_1

\;{ [ ∵ l_2=2 l_1. \;\text " (Given)] "\;}

\;= (\;{2 l_1}/{l_1}) ^2 {\R}_1

\; {\R}_2=4 {\R}_1

\;=4 × 6

\;=24 Ω

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