A) \[{{\alpha }_{2}}=3{{\alpha }_{1}}\]
B) \[{{\alpha }_{2}}=4{{\alpha }_{1}}\]
C) \[{{\alpha }_{1}}=3{{\alpha }_{2}}\]
D) \[{{\alpha }_{1}}=4{{\alpha }_{2}}\]
Correct Answer: D
Solution :
[d] \[{{(OR)}^{2}}={{(PR)}^{2}}-{{(PO)}^{2}}={{l}^{2}}-{{\left( \frac{l}{2} \right)}^{2}}\] \[{{[l(1+{{\alpha }_{2}}t)]}^{2}}-{{\left[ \frac{1}{2}(1+{{\alpha }_{1}}t) \right]}^{2}}\] \[{{l}^{2}}-\frac{{{l}^{2}}}{4}={{l}^{2}}(1+{{a}_{2}}^{2}{{t}^{2}}+2{{\alpha }_{2}}t)-\frac{{{l}^{2}}}{4}(1+{{a}_{1}}^{2}{{t}^{2}}+2{{\alpha }_{1}}t)\] Neglecting \[{{\alpha }_{2}}^{2}{{t}^{2}}\] and \[{{\alpha }_{1}}^{2}{{t}^{2}}\] \[0={{l}^{2}}(2{{\alpha }_{2}}t)-\frac{{{l}^{2}}}{4}(2{{\alpha }_{1}}t)\Rightarrow 2{{\alpha }_{2}}=\frac{2{{\alpha }_{1}}}{4}\Rightarrow ;{{\alpha }_{1}}=4{{\alpha }_{2}}\]You need to login to perform this action.
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