Question:
A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of 25.5 kVm\(^{-1}\). The density of liquid is \(1.26\times10^{3}\) kg m\(^{-3}\). The radius of the drop is (neglect buoyancy).
A liquid drop having 6 excess electrons is kept stationary under a uniform electric field of 25.5 kVm\(^{-1}\). The density of liquid is \(1.26\times10^{3}\) kg m\(^{-3}\). The radius of the drop is (neglect buoyancy).
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Balance electric force and weight: \(qE=mg\) with \(q=6e\) and \(m=\frac43\pi r^3\rho\).
Updated On: Oct 31, 2026
- \(4.3\times10^{-7}\) m
- \(7.8\times10^{-7}\) m
- \(0.078\times10^{-7}\) m
- \(3.4\times10^{-7}\) m
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The Correct Option is B
Solution and Explanation
Equilibrium: \(qE=mg\) with \(q=6e=6\times1.6\times10^{-19}=9.6\times10^{-19}\) C and \(E=25.5\times10^{3}\) V/m.
\(qE=9.6\times10^{-19}\times25.5\times10^{3}=2.448\times10^{-14}\) N.
Mass \(m=\dfrac{qE}{g}=\dfrac{2.448\times10^{-14}}{9.8}=2.5\times10^{-15}\) kg.
Volume \(=m/\rho=\dfrac{2.5\times10^{-15}}{1.26\times10^{3}}=1.98\times10^{-18}\) m\(^3\).
\(\frac43\pi r^3=1.98\times10^{-18}\Rightarrow r^3=4.73\times10^{-19}=473\times10^{-21}\Rightarrow r\approx7.8\times10^{-7}\) m.
Answer: (2) \(7.8\times10^{-7}\) m
\(qE=9.6\times10^{-19}\times25.5\times10^{3}=2.448\times10^{-14}\) N.
Mass \(m=\dfrac{qE}{g}=\dfrac{2.448\times10^{-14}}{9.8}=2.5\times10^{-15}\) kg.
Volume \(=m/\rho=\dfrac{2.5\times10^{-15}}{1.26\times10^{3}}=1.98\times10^{-18}\) m\(^3\).
\(\frac43\pi r^3=1.98\times10^{-18}\Rightarrow r^3=4.73\times10^{-19}=473\times10^{-21}\Rightarrow r\approx7.8\times10^{-7}\) m.
Answer: (2) \(7.8\times10^{-7}\) m
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