A wire of 1$\Omega$ has a length of 1 m. It is stretched till its length increases by 25%. The percentage change in resistance to the nearest integer is :
Solution
R<sub>0</sub> = 1$\Omega$<br><br>R<sub>1</sub> = ?<br><br>l<sub>0</sub> = 1m<br><br>l<sub>1</sub> = 1.25 m<br><br>A<sub>0</sub> = A<br><br>As volume of wire remains constant so<br><br>A<sub>0</sub>l<sub>0</sub> = A<sub>1</sub>l<sub>1</sub> $\Rightarrow$ A<sub>1</sub> = ${{{l_0}{A_0}} \over {{l_1}}}$<br><br>Now<br><br>Resistance (R) = ${{pl} \over A}$<br><br>$${{{R_0}} \over {{R_1}}} = {{{l_0}} \over {{A_0}}}\left( {{{{l_0}{A_0}} \over {{l_1} \times {l_1}}}} \right)$$
<br><br>$\Rightarrow$ ${R_1} = {{l_1^2} \over {l_0^2}} = 1.5625 \,\Omega$<br><br>So % change in resistance <br><br>$= {{{R_1} - {R_0}} \over {{R_0}}} \times 100\%$<br><br>$= {{1.5625 - 1} \over 1} \times 100\%$<br><br>$= 56.25\%$
About this question
Subject: Physics · Chapter: Current Electricity · Topic: Ohm's Law and Resistance
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