BITSAT 2024 Shift 1 Paper

When eight small liquid drops combine to form one large drop, the potential of the large drop is given as 20 V . We need to find the potential of each individual small drop.
Solution
Volume Relationship:
The total volume of the eight small drops is equal to the volume of the one large drop. This relationship is expressed as:
(‌

4

3

πr3)×8=‌

4

3

πR3
Simplifying gives:
2r=R‌‌...‌ (i) ‌
This equation indicates the radius of the large drop R is twice the radius of one small drop r.
Charge Conservation:
The total charge of the eight small drops is equal to the charge of the large drop:
8q=Q‌‌...‌ (ii) ‌
Potential Derivation:
The potential for one small drop (V′) and the large drop (V) are given by:
‌V′=‌

q

4πε0r

‌V=‌

Q

4πε0R

Finding the Ratio:
Using the above expressions, we derive:
‌

V′

V

=‌

q

Q

×‌

R

r

Substituting from equations (i) and (ii), we get:

‌

V′

20

=‌

q

8q

×‌

2r

r

This simplifies to:
‌

V′

20

=‌

1

4

Calculating V′ :
Solving for V′, we find:
V′=5v
Thus, the potential of each single small liquid drop is 5 V .