AP EAMCET Engineering 2018 Apr 22 Shift 1 Paper

The force experienced by an electric charge is given as, $F = q E$ $m a = q E$ $a = \frac{q E}{m}$ Substitute the values in above expression. $a = \frac{2 \times 10^{-6} \times 4.2 \times 10^{4} \, \text{m}/\text{s}^{2}}{0.04}$ $a = 2.1 \, \text{m}/\text{s}^{2}$ The expression of time period in absence of an electric field is, $T_{0} = 2 \pi \sqrt{\frac{l}{g}} = 2 \pi \sqrt{\frac{l}{10}}$ The expression of time period in presence of an electric field is, $T_{1} = 2 \pi \sqrt{\frac{l}{a_{e}}} = 2 \pi \sqrt{\frac{l}{a_{e}}}$ $= 2 \pi \sqrt{\frac{l}{a+g}} = 2 \pi \sqrt{\frac{l}{12.1}}$ Here, the effective acceleration on bob is $a_{e}$ The ratio of time period is, $\frac{T_{0}}{T_{1}} = \sqrt{\frac{12.1}{10}} = \frac{11}{10}$ $T_{1} = \frac{10}{11} \times \frac{44}{20} = 2 \, \text{s}$ The time taken to complete 15 oscillation is, $T_{1} = 2 \, \text{s} \times 15 = 30 \, \text{s}$