A) 1.2 times, 1.1 times
B) 1.21 times, same
C) both remain the same
D) 1.1 times, 1.1 times
Correct Answer: B
Solution :
Key Idea: In stretching, specific resistance remains unchanged. After stretching, specific resistance (p) will remain same. Original resistance of the wire, \[R=\frac{\rho l}{A}\] or\[R\propto \frac{l}{A}\]or\[R\propto \frac{{{l}^{2}}}{V}\] \[(as\,V=Al)\] and\[R'\propto \frac{{{(l+10%l)}^{2}}}{V}\] Therefore,\[\frac{R'}{R}=\frac{{{\left( l+\frac{10}{100}l \right)}^{2}}}{{{l}^{2}}}\] or\[\frac{R'}{R}=\frac{{{\left( \frac{11l}{10} \right)}^{2}}}{{{l}^{2}}}=\frac{121}{100}\]or\[R'=1.21R\]
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