A) \[\frac{\pi }{2}>{{\theta }_{1}}>{{\theta }_{2}}>{{\theta }_{3}}\ge 0\]
B) \[0\le {{\theta }_{1}}<{{\theta }_{2}}<{{\theta }_{3}}<\frac{\pi }{2}\]
C) \[\frac{\pi }{2}\le {{\theta }_{1}}<{{\theta }_{2}}<{{\theta }_{3}}<\pi \]
D) \[\pi >{{\theta }_{1}}>{{\theta }_{2}}>{{\theta }_{3}}>\frac{\pi }{2}\]
Correct Answer: B
Solution :
\[h=\frac{2T\cos \theta }{rpg}\] \[\Rightarrow r\,\alpha \,\cos \theta \] (as T, h and r are constants) \[\rho \uparrow \Rightarrow \theta \downarrow \] \[{{\theta }_{1}}<{{\theta }_{2}}<{{\theta }_{3}}\] Its rise so \[0\le {{\theta }_{1}}<{{\theta }_{2}}<{{\theta }_{3}}<\frac{\pi }{2}\]You need to login to perform this action.
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