JEE MAIN - Physics (2019 - 8th April Morning Slot - No. 23)
The bob of a simple pendulum has mass 2g and
a charge of 5.0 μC. It is at rest in a uniform
horizontal electric field of intensity 2000 V/m.
At equilibrium, the angle that the pendulum
makes with the vertical is : (take g = 10 m/s2)
tan–1(5.0)
tan–1(2.0)
tan–1(0.5)
tan–1(0.2)
Explanation
Tcos$$\theta $$ = mg
Tsin$$\theta $$ = qE
tan$$\theta $$ = $${{qE} \over {mg}}$$
tan$$\theta $$ = $${{5 \times {{10}^{ - 16}} \times 2000} \over {2 \times {{10}^{ - 3}} \times 10}} = {1 \over 2}$$
$$ \Rightarrow $$ $${\tan ^{ - 1}}\left( {{1 \over 2}} \right) = {\tan ^{ - 1}}\left( {0.5} \right)$$
Tsin$$\theta $$ = qE
tan$$\theta $$ = $${{qE} \over {mg}}$$
tan$$\theta $$ = $${{5 \times {{10}^{ - 16}} \times 2000} \over {2 \times {{10}^{ - 3}} \times 10}} = {1 \over 2}$$
$$ \Rightarrow $$ $${\tan ^{ - 1}}\left( {{1 \over 2}} \right) = {\tan ^{ - 1}}\left( {0.5} \right)$$
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