A) \[0.8\times {{10}^{7}}\,ergs\]
B) 0.4 J
C) 8 J
D) 0.8 ergs
Correct Answer: A
Solution :
\[W=MB\,(cos\,\,{{\theta }_{1}}-cos\,\,{{\theta }_{2}})\] When the magnet is rotated from \[0{}^\circ \text{ }to\text{ }60{}^\circ \], then work done is 0.8 J \[0.8 =MB\left( cos\,0{}^\circ -cos\,60{}^\circ \right)\,\,=\,\,\frac{MB}{2}\] \[\Rightarrow \,\,\,MB\,\,=\,\,1.6\,\,N-m\] In order to rotate the magnet through an angle of\[\,30{}^\circ ,\text{ }i.e.,\text{ }from\text{ }60{}^\circ \text{ }to\text{ }90{}^\circ \], the work done is \[\operatorname{W}'=MB(cos\,60{}^\circ -cos\,90{}^\circ ) =MB\left( \frac{1}{2}-0 \right)\] \[\frac{MB}{2}\,\,=\,\,\frac{1.6}{2}\,\,=\,\,0.8\,J\,\,=\,\,0.8\,\,\times \,\,{{10}^{7}}\,ergs\]
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