Question

The bob of a pendulum of mass $8\, \mu g$ carries an electric charge of $39.2 \times 10^{-10}$ coulomb in an electric field of $20 \times 10^{3}$ volt/metre and it is at rest. The angle made by the pendulum with the vertical will be

• A. $27^{\circ}$
• B. $45^{\circ}$
• C. $87^{\circ}$
• D. $127^{\circ}$

Answer 1

Answer: (B)

$T \sin \theta=q E, T \cos \theta=m g$

$\therefore \tan \theta=\frac{q E}{m g}=\frac{39.2 \times 10^{-10} \times 20 \times 10^{3}}{8 \times 10^{-6} \times 9.8}=1$
$\Rightarrow \theta=45^{\circ}$