BITSAT 2024 Shift 1 Paper

An oil drop with a radius of 1µm is held stationary in a constant electric field of 3.65×104N∕C due to excess electrons on it. The oil drop's density is 1.26g∕cm3. To find the number of excess electrons on the oil drop, consider the following details:
Convert the given measurements:
Radius of the drop, r=1µm=10−6m.
Density of the oil, ρ‌oil ‌, is converted to:
ρ‌oil ‌=1.26g∕cm3=1.26×103‌kg∕m3
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 field
Mathematically, this can be expressed as:
‌

4

3

πr3ρ‌oil ‌⋅g=qE
where q is the charge due to excess electrons. If n is the number of excess electrons, then:
q=ne
Substitute this into the previous equation:
‌

4

3

πr3ρoil⋅g=neE
Solving for n :
n=‌

4πr3ρ‌oil ‌g

3eE

Substitute the known values into the equation to calculate n :

n‌=‌

4×3.14×(10−6)3×1.26×103×10

3×1.6×10−19×3.65×104

‌=9.03≈9
Thus, there are approximately 9 excess electrons on the oil drop.