A) \[1:\sqrt{2}\]
B) l : 3
C) 2: l
D) \[\sqrt{5}:\sqrt{6}\]
Correct Answer: D
Solution :
[d] \[{{I}_{{{y}_{1}}}}=\frac{M{{R}^{2}}}{4}\] \[\therefore \,\,I{{'}_{{{y}_{1}}}}=\frac{M{{R}^{2}}}{4}+M{{R}^{2}}=\frac{5}{4}M{{R}^{2}}\] . \[{{I}_{{{y}_{2}}}}=\frac{M{{R}^{2}}}{2}\] \[\therefore \,\,I{{'}_{{{y}_{2}}}}=\frac{M{{R}^{2}}}{2}+M{{R}^{2}}=\frac{3}{2}M{{R}^{2}}\] \[I{{'}_{{{y}_{1}}}}=MK_{1}^{2},\,\,I{{'}_{{{y}_{2}}}}=MK_{2}^{2}\] \[\therefore \,\,\,\frac{K_{1}^{2}}{K_{2}^{2}}=\frac{I{{'}_{{{y}_{1}}}}}{I{{'}_{{{y}_{2}}}}}\Rightarrow {{K}_{1}}:{{K}_{2}}=\sqrt{5}:\sqrt{6}\]You need to login to perform this action.
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