JEE PYQ: Atomic Physics Question 10
Question 10 - 2021 (18 Mar Shift 1)
An oil drop of radius 2 mm with a density 3 g cm$^{-3}$ is held stationary under a constant electric field $3.55 \times 10^5$ 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/s$^2$)
(1) $48.8 \times 10^{11}$
(2) $1.73 \times 10^{10}$
(3) $17.3 \times 10^{10}$
(4) $1.73 \times 10^{12}$
Show Answer
Answer: (2)
Solution
$qE = Mg$. $neE = \rho\left(\frac{4}{3}\pi r^3\right) \times g$. $n \times 1.6 \times 10^{-19} \times 3.55 \times 10^5 = 3 \times 10^3 \times \frac{4}{3} \times \pi \times (2 \times 10^{-3})^3 \times 9.81$. $n = 1.73 \times 10^{10}$.
BETA
