A) 4 : 1 exactly
B) 4 : 1 approx.
C) 8 : 1 exactly
D) 8 : 1 approx.
Correct Answer: D
Solution :
For a magnet \[{{W}_{1}}=MB(\cos {{0}^{o}}-\cos {{90}^{o}})=MB\left( 1-0 \right)=MB\] (Nearly) \[\Rightarrow \frac{{{B}_{1}}}{{{B}_{2}}}={{\left( \frac{{{x}_{1}}}{{{x}_{2}}} \right)}^{3}}={{\left( \frac{x}{2x} \right)}^{3}}=\frac{1}{8}\] (Approx.)You need to login to perform this action.
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