JEE MAIN - Physics (2021 - 18th March Morning Shift - No. 1)
An oil drop of radius 2 mm with a density 3g cm$$-$$3 is held stationary under a constant electric field 3.55 $$\times$$ 105 V m$$-$$1 in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (consider g = 9.81 m/s2)
48.8 $$\times$$ 1011
1.73 $$\times$$ 1010
17.3 $$\times$$ 1010
1.73 $$\times$$ 1012
Explanation
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Fe = qE = (ne)E
Fe = mg
(ne)E = mg
$$n = {{mg} \over {eE}} = {{\rho {4 \over 3}\pi {R^3} \times g} \over {eE}}$$
$$n = {{3000 \times {4 \over 3} \times 3.14 \times 8 \times {{10}^{ - 9}} \times 9.8} \over {1.6 \times {{10}^{ - 19}} \times 3.55 \times {{10}^5}}}$$
$$n = {{984704 \times {{10}^5}} \over {5.68}} = 1.73 \times {10^{10}}$$
$$n = 1.73 \times {10^{10}}$$
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