Q84 / 100 · BITSAT 2024 (Memory Based Paper 3) · Physics
An oil drop of radius $1 \mu \mathrm{~m}$ is held stationary under a constant electric field of $3.65 \times 10^4 \mathrm{~N} / \mathrm{C}$ due to some excess electrons presence on it. If the density of oil drop is $1.26 \mathrm{~g} / \mathrm{cm}^3$, then number of excess electrons on the oil drop approximately are
[Take, $g=10 \mathrm{~ms}^{-2}$ ]
- A7
- B12
- Correct answerC9
- D8
Solution
$E=3.65 \times 10^4 \mathrm{~N} / \mathrm{C}, r=1 \mu \mathrm{~m}=10^{-6} \mathrm{~m}$
$\begin{aligned}
\rho_{\text {oil }} & =1.26 \mathrm{~g} / \mathrm{cm}^3 \\
& =1.26 \times 10^3 \mathrm{~kg} / \mathrm{m}^3
\end{aligned}$
Since, droplet is stationary hence weight of droplet $=$ force due to electric field
$\Rightarrow \quad \frac{4}{3} \pi r^3 \cdot \rho_{\text {oil }} \cdot g=q E...(i)$
If $n$ be the number of excess electrons in the oil drop, then
$q=n e$
Hence, from Eq. (i), $\frac{4}{3} \pi r^3 \rho_{\text {oil }} \cdot g=n e E$
$\begin{aligned}
\Rightarrow n & =\frac{4 \pi r^3 \rho_{\text {oill }} g}{3 e E} \\
& =\frac{4 \times 3.14 \times\left(10^{-6}\right)^3 \times 1.26 \times 10^3 \times 10}{3 \times 1.6 \times 10^{-19} \times 3.65 \times 10^4} \\
& =9.03 \simeq 9
\end{aligned}$
Hence, the correct answer is option: 'C'.