question_answer

Two points masses, m each carrying charges -q and +g are attached to the ends of a massless rigid non-conducting wire of length ?L?. When this arrangement is placed in a |      uniform electric field, then it deflects       through an angle i. The minimum time      needed by rod to align itself along the field is

A)  \[2\pi \sqrt{\frac{mL}{qE}}\]                     

B)  \[\frac{\pi }{2}\sqrt{\frac{mL}{2qE}}\]

C) \[\pi \sqrt{\frac{2mL}{qE}}\]                      

D) \[\pi \sqrt{\frac{3mL}{qE}}\]

Correct Answer: B

Solution :

When the wire is brought in a uniform field E. Then the torque is given by                    \[\begin{align}   & Zn{{\left( OH \right)}_{2}}+2\overset{-}{\mathop{OH}}\,\to ZnO_{2}^{2-}+2{{H}_{2}}O \\  & \text{AcidBaseSaltWater} \\ \end{align}\]                     \[\begin{align}   & Zn{{\left( OH \right)}_{2}}+2{{H}^{+}}\to Z{{n}^{2+}}+2{{H}_{2}}O \\  & \text{BaseAcidSaltWater} \\ \end{align}\]              [ \[{{\left[ \text{Cu}{{\left( \text{N}{{\text{H}}_{\text{3}}} \right)}_{\text{4}}} \right]}^{\text{2+}}}\text{}{{\left[ \text{Cu}{{\left( \text{en} \right)}_{\text{2}}} \right]}^{\text{2+}}}\text{}{{\left[ \text{Cu}\left( \text{trien} \right) \right]}^{\text{2+}}}\text{.}\] \[{{\left[ \text{Fe}{{\left( {{\text{H}}_{\text{2}}}\text{O} \right)}_{\text{6}}} \right]}^{\text{3+}}}\text{}{{\left[ \text{Fe}{{\left( \text{N}{{\text{O}}_{\text{2}}} \right)}_{\text{6}}} \right]}^{\text{3-}}}\text{}{{\left[ \text{Fe}{{\left( \text{N}{{\text{H}}_{\text{3}}} \right)}_{\text{6}}} \right]}^{\text{3-}}}\] is very small] The moment of inertia of rod AB about O is                   \[\because \] As,                      \[{{\text{E}}_{\text{1}}}\text{=}{{\text{E}}_{\text{2}}}\text{=}\frac{\text{1}}{\text{4 }\!\!\pi\!\!\text{ }{{\text{ }\!\!\varepsilon\!\!\text{ }}_{\text{0}}}}\text{.}\frac{\text{q}}{{{\text{r}}^{\text{2}}}}\]                            So           \[{{E}_{R}}=\sqrt{E_{1}^{2}+E_{2}^{2}+2{{E}_{1}}{{E}_{2}}\cos {{60}^{\circ }}}\] \[=\sqrt{E_{1}^{2}+E_{1}^{2}+2E_{1}^{2}\times \frac{1}{2}=\sqrt{3}{{E}_{1}}}\]           \[\therefore \]               [\[{{E}_{R}}=\frac{\sqrt{3}q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}}\] \[\tau =qEL\sin \theta \]] \[=qEL\theta \]             \[\because \] The time period of the wire is                   \[\theta \] The rod will becomes parallel to the field in time \[\text{I=m}{{\left( \frac{\text{L}}{\text{2}} \right)}^{\text{2}}}\text{+m}{{\left( \frac{\text{L}}{\text{2}} \right)}^{\text{2}}}\text{=}\frac{\text{m}{{\text{L}}^{\text{2}}}}{\text{2}}\] So,        \[\tau =I\alpha \]