A) \[\text{y=}\frac{\text{-1}}{\text{2}}\text{ }\!\!\times\!\!\text{ -}\frac{\text{5}}{\text{6}}\]
B) \[y=x-\frac{7}{6}\]
C) \[y=2x+\frac{3}{7}\]
D) \[y=3x-\frac{3}{2}\]
Correct Answer: C
Solution :
\[{{E}_{R}}=\sqrt{E_{1}^{2}+E_{2}^{2}+2{{E}_{1}}{{E}_{2}}\cos {{60}^{\circ }}}\] \[=\sqrt{E_{1}^{2}+E_{1}^{2}+2E_{1}^{2}\times \frac{1}{2}=\sqrt{3}{{E}_{1}}}\] \[\therefore \] Now, \[{{E}_{R}}=\frac{\sqrt{3}q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[\tau =qEL\sin \theta \] \[=qEL\theta \] \[\because \] \[\theta \] \[\text{I=m}{{\left( \frac{\text{L}}{\text{2}} \right)}^{\text{2}}}\text{+m}{{\left( \frac{\text{L}}{\text{2}} \right)}^{\text{2}}}\text{=}\frac{\text{m}{{\text{L}}^{\text{2}}}}{\text{2}}\] So that, \[\tau =I\alpha \] Thus, c giving equation is \[\text{ }\!\!\alpha\!\!\text{ =}\frac{\text{ }\!\!\tau\!\!\text{ }}{\text{I}}\text{=}\frac{\text{qEL }\!\!\theta\!\!\text{ }}{\frac{\text{m}{{\text{L}}^{\text{2}}}}{\text{2}}}\] which in the present case becomes \[\Rightarrow \] \[{{\text{ }\!\!\omega\!\!\text{ }}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ =}\frac{\text{2qEL }\!\!\theta\!\!\text{ }}{\text{m}{{\text{L}}^{\text{2}}}}\] \[\because \] Thus, when \[\alpha ={{\omega }^{2}}\theta \] When \[\Rightarrow \] When \[{{\omega }^{2}}=\frac{2qE}{mL}\] Therefore, the asymptotes are \[T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{mL}{2qE}}\] \[\frac{T}{4}.\] and \[\text{t=}\frac{\text{T}}{\text{4}}\text{=}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\sqrt{\frac{\text{mL}}{\text{2qE}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
