A) 1/2
B) 2
C) 1/4
D) 1
Correct Answer: B
Solution :
\[{{W}_{1}}=MB(\cos {{0}^{o}}-\cos {{90}^{o}})=MB(1-0)=MB\] \[{{W}_{2}}=MB(\cos {{0}^{o}}-\cos {{60}^{o}})=MB\left( 1-\frac{1}{2} \right)=\frac{MB}{2}\] \[\therefore \ {{W}_{1}}=2{{W}_{2}}\Rightarrow n=2\]
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