A) \[w\sin \alpha \]
B) \[1\]
C) \[1/Q\]
D) \[1/{{Q}^{2}}\]
Correct Answer: D
Solution :
From figure (i)\[P\cos \alpha =w\sin \alpha \] \[\Rightarrow \] \[P=\frac{w\sin \alpha }{\cos \alpha }\] And from figure (ii) Q = w sin a \[\therefore \] \[\frac{1}{{{p}^{2}}}+\frac{1}{{{w}^{2}}}=\frac{{{\cos }^{2}}\alpha }{{{w}^{2}}{{\sin }^{2}}\alpha }+\frac{1}{{{w}^{2}}}\] \[=\frac{{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha }{{{w}^{2}}{{\sin }^{2}}\alpha }={{\left( \frac{1}{w\sin \alpha } \right)}^{2}}\] \[=\frac{1}{{{Q}^{2}}}\]You need to login to perform this action.
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