A) 4 F
B) 6 F
C) 9 F
D) F
Correct Answer: C
Solution :
[c] |
For wire 1, |
\[\Delta \text{l=}\left( \frac{\text{F}}{\text{AY}} \right)\text{3l}\] (i) |
For wire 2, |
\[\frac{\text{F }\!\!'\!\!\text{ }}{3A}=Y\frac{\Delta \text{I}}{\text{I}}\] |
\[\Rightarrow \] \[\Delta I=\left( \frac{F'}{3AY} \right)I\] (ii) |
From equation (i) & (ii), |
\[\Delta \text{I=}\left( \frac{\text{F}}{\text{AY}} \right)\text{3I=}\left( \frac{\text{F }\!\!'\!\!\text{ }}{\text{3AY}} \right)\text{I}\] |
\[\Rightarrow \] |
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