A) \[4\lambda h=ab(1+\lambda )\]
B) \[\lambda h=ab{{(1+\lambda )}^{2}}\]
C) \[4\lambda {{h}^{2}}=ab{{(1+\lambda )}^{2}}\]
D) None of these
Correct Answer: C
Solution :
It is given that \[{{m}_{2}}=\lambda {{m}_{1}}\Rightarrow {{m}_{1}}+\lambda {{m}_{1}}=\frac{-2h}{b}\] \[\Rightarrow {{m}_{1}}=\frac{-2h}{b(1+\lambda )}\] .....(i) and \[{{m}_{1}}.\lambda {{m}_{1}}=\frac{a}{b}\Rightarrow {{m}_{1}}=\sqrt{\frac{a}{b\lambda }}\] ..?(ii) Hence, by (i) and (ii), \[\sqrt{\frac{a}{b\lambda }}=\frac{-2h}{b(1+\lambda )}\] On squaring both sides, we get \[4\lambda {{h}^{2}}=ab{{(1+\lambda )}^{2}}\].
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