Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Two large circular discs separated by a distance of 0.01 m are connected to a battery via a switch as shown in the figure. Charged oil drops of density 900 kg m$$^{−3}$$ are released through a tiny hole at the center of the top disc. Once some oil drops achieve terminal velocity, the switch is closed to apply a voltage of 200 V across the discs. As a result, an oil drop of radius $$8 \times 10^{−7}$$ m stops moving vertically and floats between the discs. The number of electrons present in this oil drop is ________. (neglect the buoyancy force, take acceleration due to gravity = 10 ms$$^{−2}$$ and charge on an electron (e) = $$1.6 \times 10^{-19}$$ C)
Correct Answer: e
The electric field between two parallel plates separated by distance $$d$$ and connected to a potential difference $$V$$ is
$$E = \frac{V}{d}$$
Here $$V = 200 \text{ V}$$ and $$d = 0.01 \text{ m}$$, so
$$E=\frac{200}{0.01}=2.0\times10^{4}\,\text{V m}^{-1}$$ $$-(1)$$
The oil drop is spherical with radius $$r = 8\times10^{-7}\,\text{m}$$.
Volume of a sphere: $$V_{\text{drop}}=\frac{4}{3}\pi r^{3}$$.
First evaluate $$r^{3}$$:
$$(8\times10^{-7})^{3}=512\times10^{-21}=5.12\times10^{-19}\,\text{m}^3$$
Thus
$$V_{\text{drop}}=\frac{4}{3}\pi(5.12\times10^{-19})$$
$$=4.1888\times5.12\times10^{-19}$$
$$\approx2.14\times10^{-18}\,\text{m}^3$$ $$-(2)$$
Density of oil, $$\rho = 900\,\text{kg m}^{-3}$$, so mass
$$m=\rho V_{\text{drop}}=900\times2.14\times10^{-18}\approx1.93\times10^{-15}\,\text{kg}$$ $$-(3)$$
Weight of the drop (buoyancy neglected):
$$W = mg = (1.93\times10^{-15})\times10 \approx 1.93\times10^{-14}\,\text{N}$$ $$-(4)$$
When the drop floats stationary, the upward electric force equals its weight:
$$qE = W$$
$$\Rightarrow q = \frac{W}{E} = \frac{1.93\times10^{-14}}{2.0\times10^{4}}$$
$$\approx 9.65\times10^{-19}\,\text{C}$$ $$-(5)$$
Charge on one electron, $$e = 1.6\times10^{-19}\,\text{C}$$. Therefore the number of excess electrons on the drop is
$$n = \frac{q}{e} = \frac{9.65\times10^{-19}}{1.6\times10^{-19}}\approx 6.0$$
Since the number of electrons must be an integer,
the oil drop carries 6 electrons.
Create a FREE account and get:
Educational materials for JEE preparation