KCET · Physics · Electrostatics
The potential of a large liquid drop when eight liquid drops are combined is \(20 \mathrm{~V}\). Then the potential of each single drop was
- A \(10 \mathrm{~V}\)
- B \(7.5 \mathrm{~V}\)
- C \(5 \mathrm{~V}\)
- D \(2.5 \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(5 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Volume of 8 drops \(=\) volume of big drop
\(8 q=Q\)
Potential of one small drop \(\left(V^{\prime}\right)=\frac{q}{4 \pi \varepsilon_{0} r}\)
Similarly, potential of big drop \((V)=\frac{Q}{4 \pi \varepsilon_{0} R}\)
Now, \(\frac{V^{\prime}}{V}=\frac{q}{Q}+\frac{R}{r}\)
\(\Rightarrow \frac{V^{\prime}}{20} =\frac{q}{8 q} \times \frac{2 r}{r} \text { [from Eqs. (i) and (ii)] } \)
\(\therefore V^{\prime} =5 \text { volt }\)
\(8 q=Q\)
Potential of one small drop \(\left(V^{\prime}\right)=\frac{q}{4 \pi \varepsilon_{0} r}\)
Similarly, potential of big drop \((V)=\frac{Q}{4 \pi \varepsilon_{0} R}\)
Now, \(\frac{V^{\prime}}{V}=\frac{q}{Q}+\frac{R}{r}\)
\(\Rightarrow \frac{V^{\prime}}{20} =\frac{q}{8 q} \times \frac{2 r}{r} \text { [from Eqs. (i) and (ii)] } \)
\(\therefore V^{\prime} =5 \text { volt }\)
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