A) 1000 N
B) 200 N
C) 2000 N
D) 400 N
Correct Answer: D
Solution :
\[Y\,=\frac{F}{A}.\frac{L}{l}\] \[\therefore \]\[F\propto \frac{l}{K}\] ( Y and A are same for both wires) \[\therefore \]\[\frac{{{F}_{2}}}{{{F}_{1}}}=\frac{{{l}_{2}}}{{{l}_{1}}}\times \frac{{{L}_{1}}}{{{L}_{2}}}=\frac{0.02}{1\times {{10}^{-3}}}\times \frac{1}{10}=2\] \[{{F}_{2}}=2{{F}_{1}}=2\times 200=400N\]You need to login to perform this action.
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