question_answer

The bob of a pendulum carries an electric charge of \[39.2\,\,\times \,\,{{10}^{-}}^{10}\] coulomb in a horizontal electric field of \[20\text{ }\times \text{ }{{10}^{3}}\,V/m\] and it is at rest. The angle made by the pendulum with the vertical will be, if the mass of pendulum is \[8\,\,\times \,\,{{10}^{-6}}\text{ }kg\] and \[g\,\,=\,\,9.8\text{ }m/{{s}^{2}}\]-

A) \[30{}^\circ \]  

B)   \[45{}^\circ \]

C) \[60{}^\circ \]              

D)   \[120{}^\circ \]

Correct Answer: B

Solution :

  \[T\,\sin \theta \,\,=\,\,qE\] \[T\,\,cos\,\theta =mg\] \[\tan \,\theta \,\,\,=\,\,\frac{qE}{mg}\,\,=\,\,\,\frac{39.2\times {{10}^{-10}}\times 20\times {{10}^{3}}}{8\times {{10}^{-6}}\times 9.8}\] \[\tan \,\theta \,\,=\,\,1\] \[\theta \,\,=\,\,45{}^\circ \]