A) \[\sqrt{2}\,{{l}_{AC}}={{l}_{EF}}\]
B) \[{{l}_{AD}}=4{{l}_{EF}}\]
C) \[{{l}_{AC}}={{l}_{EF}}\]
D) \[{{l}_{AC}}=\sqrt{2}\,{{l}_{EF}}\]
Correct Answer: B
Solution :
Let the each side of square lamina is d. So, \[{{l}_{EF}}={{l}_{GH}}\] (due to symmetry) and \[{{l}_{AC}}={{l}_{BD}}\] (due to symmetry) Now, according to theorem of perpendicular axes, \[{{l}_{AC}}+{{l}_{BD}}={{l}_{0}}\] \[x\] \[2{{l}_{AC}}\,={{l}_{0}}\] ...(i) and \[{{l}_{EF}}+{{l}_{GH}}={{l}_{0}}\] \[\pi /6\] \[2{{l}_{EF}}={{l}_{0}}\] ...(ii) From Eqs. (i) and (ii), we get \[\pi /4\] \[\therefore \,\,{{l}_{AD}}={{l}_{EF}}+\frac{m{{d}^{2}}}{4}\] (using theorem of parallel axis) \[{{\log }_{e}}x\] \[\left( as\,\,{{l}_{EF}}\,=\frac{m{{d}^{2}}}{12} \right)\] So, \[{{l}_{AD}}=\frac{m{{d}^{2}}}{3}=4{{l}_{EF}}\]You need to login to perform this action.
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