A) 2 m
B) 3 m
C) 1 m
D) 4 m
Correct Answer: A
Solution :
We know that, the surface area A of a cube of side x is given by \[=qEL\theta \] On differentiating w.r.t. x, we get \[\because \] Let the change in x be \[\theta \] \[\text{I=m}{{\left( \frac{\text{L}}{\text{2}} \right)}^{\text{2}}}\text{+m}{{\left( \frac{\text{L}}{\text{2}} \right)}^{\text{2}}}\text{=}\frac{\text{m}{{\text{L}}^{\text{2}}}}{\text{2}}\] Now, change in surface area, \[\tau =I\alpha \] \[\text{ }\!\!\alpha\!\!\text{ =}\frac{\text{ }\!\!\tau\!\!\text{ }}{\text{I}}\text{=}\frac{\text{qEL }\!\!\theta\!\!\text{ }}{\frac{\text{m}{{\text{L}}^{\text{2}}}}{\text{2}}}\] \[\Rightarrow \] The approximate change in surface area \[{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ =}\frac{\text{2qEL }\!\!\theta\!\!\text{ }}{\text{m}{{\text{L}}^{\text{2}}}}\] = 2m% of original surface area
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