A) \[49{}^\circ \]to the horizontal
B) \[90{}^\circ \]with the vertical
C) \[49{}^\circ \]to the vertical
D) along the horizontal
E) in any direction
Correct Answer: C
Solution :
\[\sin \theta =\frac{1}{{{\mu }_{w}}}=\frac{1}{1.33}\Rightarrow \theta ={{\sin }^{-1}}\left( \frac{1}{1.33} \right)\] \[=49{}^\circ \]to the vertical
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