BITSAT 2024 Shift 1 Paper

An oil drop with a radius of $1 \mu \mathrm{m}$ is held stationary in a constant electric field of $3.65 \times 10^{4}\ \mathrm{N/C}$ due to excess electrons on it. The oil drop's density is $1.26\ \mathrm{g/cm}^{3}$. To find the number of excess electrons on the oil drop, consider the following details:Convert the given measurements:Radius of the drop, $r=1 \mu \mathrm{m}=10^{-6}\ \mathrm{m}$.Density of the oil, $\rho_{\;\text{oil}\;}$, is converted to:$\rho_{\;\text{oil}\;} = 1.26\ \mathrm{g/cm}^{3} = 1.26 \times 10^{3}\ \mathrm{kg/m}^{3}$Since the droplet is stationary, the gravitational force acting downward is balanced by the electric force acting upward:Weight of the droplet $=$ Force due to the electric fieldMathematically, this can be expressed as:$\;\frac{4}{3} \pi r^{3} \rho_{\;\text{oil}\;} \cdot g = q E$where $q$ is the charge due to excess electrons. If $n$ is the number of excess electrons, then:$q = n e$Substitute this into the previous equation:$\;\frac{4}{3} \pi r^{3} \rho_{\mathrm{oil}} \cdot g = n e E$Solving for $n$ :$n = \frac{4 \pi r^{3} \rho_{\;\text{oil}\;} g}{3 e E}$Substitute the known values into the equation to calculate $n$ : $n\;=\;\frac{4 \times 3.14 \times (10^{-6})^{3} \times 1.26 \times 10^{3} \times 10}{3 \times 1.6 \times 10^{-19} \times 3.65 \times 10^{4}}$$\; = 9.03 \approx 9$Thus, there are approximately 9 excess electrons on the oil drop.