TS EAMCET · Physics · Electrostatics
A spherical drop of liquid carrying charge, \(Q\) has potential \(V_0\) at its surface. If two drops of same charge and radius combine to form a single spherical drop, then the potential at the surface of new drop is (Assume, \(V=0\) at infinity.)
- A \(2^{\frac{1}{3}} V_0\)
- B \(4^{\frac{1}{3}} V_0\)
- C \(6^{\frac{1}{3}} V_0\)
- D \(2^{\frac{-1}{3}} V_0\)
Answer & Solution
Correct Answer
(B) \(4^{\frac{1}{3}} V_0\)
Step-by-step Solution
Detailed explanation
Given, charge of spherical drop of liquid \(=Q\) and potential at its surface \(=V_0\) If \(r\) be the radius of drop, \(V_0=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r}\) If \(R\) be the radius of big drop when two drops of radius \(r\) are combined, then volume of big drop \(=\)…
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