A) \[64\times {{10}^{-4}}\operatorname{N}-m and 64\times 1{{0}^{-4}}\operatorname{J}\]
B) \[64\times {{10}^{-4}}\operatorname{N}-m and 32\times 1{{0}^{-4}}\operatorname{J}\]
C) \[32\times {{10}^{-4}}\operatorname{N}-m and 32\times 1{{0}^{-4}}\operatorname{J}\]
D) \[32\times {{10}^{-4}}\operatorname{N}-m and 64\times 1{{0}^{-4}}\operatorname{J}\]
Correct Answer: D
Solution :
Torque will be maximum when 6 = 90 \[{{\tau }_{max}}=PES\,\,in\,9{{0}^{o}}\] \[=q\ell \operatorname{E} \,\,\,\therefore \left( P=q\times \ell \right)\] \[=4\times 1{{0}^{-8}}\times 2\times 1{{0}^{-4}}\left( m \right)\times 4\times {{10}^{8}}\] \[=32\times {{10}^{-4}}N-m\] \[\operatorname{W}=PE\left[ cos{{\theta }_{1}}-cos{{\theta }_{2}} \right]\] Note: When nothing is specified assume \[{{\theta }_{1}} =0{}^\circ ,\,\,{{\theta }_{2}} =180{}^\circ \] \[=PE\left[ cos0{}^\circ -cos180{}^\circ \right]\] \[=PE\left[ 1-cos180{}^\circ \right]\Rightarrow 2PE\] \[\operatorname{W}=64\times {{10}^{-4}}N-m\]
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