A) \[\text{0}\text{.8 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{7}}}\,\text{erg}\]
B) 4.0 J
C) 8 J
D) 0.8 erg
Correct Answer: A
Solution :
\[W=MB(\cos {{\theta }_{1}}-\cos {{\theta }_{2}})\] When the magnet is rotated from \[{{0}^{o}}\] to\[{{60}^{o}}\], then work done is\[0.8\,\,J\]. \[0.8=MB(\cos {{0}^{o}}-\cos {{60}^{o}})=\frac{MB}{2}\] \[\Rightarrow \] \[MB=1.6\,\,N-m\] In order to rotate the magnet through an angle of\[{{30}^{o}}\],\[ie,\] from\[{{60}^{o}}\]to\[{{90}^{o}}\], the work done is \[W=MB(\cos {{60}^{o}}-\cos {{90}^{o}})=MB\left( \frac{1}{2}-0 \right)\] \[=\frac{MB}{2}=\frac{1.6}{2}=0.8\,\,J=0.8\times {{10}^{7}}erg\]
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