JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 26)
Two small spheres each of mass 10 mg are suspended from a point by threads 0.5 m long. They are equally charged and repel each other to a distance of 0.20 m. The charge on each of the sphere is $${a \over {21}} \times {10^{ - 8}}$$C. The value of 'a' will be ___________. [Given g = 10 ms$$-$$2]
Answer
20
Explanation
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T sin$$\theta$$ = $${{k{q^2}} \over {{r^2}}}$$
T cos$$\theta$$ = mg
tan$$\theta$$ = $${{k{q^2}} \over {mg{r^2}}}$$
q2 = $${{\tan \theta mg{r^2}} \over k}$$
$$ \because $$ tan$$\theta$$ = $${{0.1} \over {0.5}} = {1 \over 5}$$
$${q^2} = {1 \over 5} \times {{10 \times {{10}^{ - 6}} \times 10 \times 0.2 \times 0.2} \over {9 \times {{10}^9}}}$$
$$q = {{2\sqrt 2 } \over 3} \times {10^{ - 8}}$$
after comparison from the given equation a = 20
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