question_answer

A block weighing \[w\], is supported on an inclined surface with the help of a horizontal force \[P\]. The same block can be supported with the help of another force \[Q\] acting parallel to the inclined surface, then \[\frac{1}{{{p}^{2}}}+\frac{1}{{{w}^{2}}}\] is equal to:

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 \[\therefore \]  \[\frac{1}{{{p}^{2}}}+\frac{1}{{{w}^{2}}}=\frac{{{\cos }^{2}}\alpha }{{{w}^{2}}{{\sin }^{2}}\alpha }+\frac{1}{{{w}^{2}}}\]                 \[=\frac{{{\cos }^{2}}\alpha +{{\sin }^{2}}\alpha }{{{w}^{2}}{{\sin }^{2}}\alpha }={{\left( \frac{1}{w\sin \alpha } \right)}^{2}}\]                 \[=\frac{1}{{{Q}^{2}}}\]