A) 0.8 cm
B) 1.6 cm
C) 2.4 cm
D) 3.2 cm
Correct Answer: D
Solution :
\[l=\frac{FL}{AY}=\frac{F{{L}^{2}}}{(AL)Y}=\frac{F{{L}^{2}}}{VY}\] \ \[l\propto {{L}^{2}}\] If volume of the wire remains constant \[\frac{{{l}_{2}}}{{{l}_{1}}}={{\left( \frac{{{L}_{2}}}{{{L}_{1}}} \right)}^{2}}={{\left( \frac{8}{2} \right)}^{2}}=16\] \\[{{l}_{2}}=16\times {{l}_{1}}=16\times 2=32mm=3.2cm\]You need to login to perform this action.
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