A) 325 K
B) 375 K
C) 300 K
D) 380 K
Correct Answer: C
Solution :
\[\Delta APQ,\] \[\tan {{30}^{o}}=\frac{PQ}{AP}\Rightarrow \frac{1}{\sqrt{3}}=\frac{100}{AP}\] \[=1-\frac{{{Q}_{L}}}{{{Q}_{H}}}\] Also we can show that \[AP=100\sqrt{3}\] \[\Delta BPQ,\] ...(i) \[\tan {{60}^{o}}=\frac{PQ}{PB}\Rightarrow \sqrt{3}=\frac{100}{PB}\] and \[\Rightarrow \] ...(ii) From Eq.(i) \[PB=\frac{100}{\sqrt{3}}\] \[AP+PB=100\left( \sqrt{3}+\frac{1}{\sqrt{3}} \right)\Rightarrow AB=\frac{400}{\sqrt{3}}m\] Substituting the value of \[{{\tan }^{-1}}x\] in Eq.(ii), we get \[{{\cot }^{-1}}x\] \[x\in R\] \[{{\cos }^{-1}}(2-x)\] \[-\text{1}\le \text{2}-x\le \text{1}\] \[\therefore \]
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