question_answer

For the given uniform square lamina ABCD, whose centre is O     AIEEE  Solved  Paper-2007

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, \[{{l}_{AD}}=\frac{m{{d}^{2}}}{3}=4{{l}_{EF}}\]