A) \[n=\frac{1}{2}\]
B) \[n=2\]
C) \[n=\frac{1}{4}\]
D) \[n=1\]
Correct Answer: B
Solution :
Work done \[W=MB\,(\cos {{\theta }_{1}}-\cos {{\theta }_{2}})\] In first case, \[{{\theta }_{1}}=0\] and \[{{\theta }_{1}}={{90}^{o}}\] \[\Rightarrow \] \[{{W}_{1}}=MB\,(\cos \,{{0}^{o}}-\cos {{90}^{o}})=MB\] In second case \[{{W}_{2}}=MB\,(\cos {{0}^{o}}-\cos \,{{60}^{o}})\] \[\Rightarrow \] \[{{W}_{2}}=MB\,(\cos \,{{0}^{o}}-\cos \,{{60}^{o}})\] \[{{W}_{2}}=MB\left( 1-\frac{1}{2} \right)=\frac{MB}{2}\] Given, \[{{W}_{1}}=n{{W}_{2}}\] \[\therefore \] \[MB=n\frac{MB}{2}\] \[\Rightarrow \] \[n=2\]
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