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.SolutionVolume Relationship:The total volume of the eight small drops is equal to the volume of the one large drop. This relationship is expressed as:$\left( \;\frac{4}{3} \pi r^3 \right) \times 8 = \;\frac{4}{3} \pi R^3$Simplifying gives:$2r = R \;\; \dots \;\text{(i)} \;$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:$8q = Q \;\; \dots \;\text{(ii)} \;$Potential Derivation:The potential for one small drop $(V')$ and the large drop $(V)$ are given by:$\; V' = \; \frac{q}{4 \pi \varepsilon_0 r}$$\; V = \; \frac{Q}{4 \pi \varepsilon_0 R}$Finding the Ratio:Using the above expressions, we derive:$\; \frac{V'}{V} = \; \frac{q}{Q} \times \; \frac{R}{r}$Substituting from equations (i) and (ii), we get: $\; \frac{V'}{20} = \; \frac{q}{8q} \times \; \frac{2r}{r}$This simplifies to:$\; \frac{V'}{20} = \; \frac{1}{4}$Calculating $V'$ :Solving for $V'$, we find:$V' = 5 v$Thus, the potential of each single small liquid drop is 5 V .