A) 1 : 2
B) \[2:\sqrt{3}\]
C) \[\sqrt{3}:1\]
D) \[1:\sqrt{2}\]
Correct Answer: D
Solution :
\[{{I}_{A}}=\frac{L}{{{(2r)}^{2}}}\] and \[{{I}_{B}}=\frac{L}{{{\left( r\sqrt{2} \right)}^{2}}}\cos \theta \] \[=\frac{L}{2{{r}^{2}}}.\frac{r}{r\sqrt{2}}\]\[=\frac{L}{2\sqrt{2}\ {{r}^{2}}}\] \[\therefore \frac{{{I}_{A}}}{{{I}_{B}}}=\frac{2\sqrt{2}}{4}=\frac{1}{\sqrt{2}}\]You need to login to perform this action.
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