Fluid Mechanics

JEE Main level, numerical answer, medium

115 of 192

Question 115 of 192

6464 identical drops each charged upto potential of 10mV10mV are combined to form a bigger drop. The potential of the bigger drop will be _____ mVmV.

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Answer: 160

The formula to calculate the potential (V)\left(V\right) of each tiny drop is given by
V=kqr...(1)V=k\dfrac{q}{r}...\left(1\right)
When all the drops forms the bigger drop the radius of the new drop (r′)\left(r'\right) can be calculated as follows:
Volumeofbigdrop=64×volumeofeachtinydrop⇒43πr′3=64×43πr3⇒r′=4r\begin{matrix}Volumeofbigdrop & = & 64\times volumeofeachtinydrop \\ & \Rightarrow & \dfrac{4}{3}\pi r{'}^{3}=64\times \dfrac{4}{3}\pi {r}^{3} \\ & \Rightarrow & r'=4r\end{matrix}
So, the potential of the new big drop (V′)\left(V'\right) can be written as
V′=k64qr′=k64q4r=16kqr...(2)\begin{matrix}V' & = & k\dfrac{64q}{r'} \\ & = & k\dfrac{64q}{4r} \\ & = & 16k\dfrac{q}{r}...\left(2\right)\end{matrix}
Divide equation (2) by equation (1) and simplify to obtain the potential of the new drop.
V′V=16kqrkqr⇒V′=16V...(3)\begin{matrix}\dfrac{V'}{V} & = & \dfrac{16k\dfrac{q}{r}}{k\dfrac{q}{r}} \\ & \Rightarrow & V'=16V...\left(3\right)\end{matrix}
Substitute the value of the known parameter into equation (3) to calculate the required value of the potential for the new drop.
V′=16×10mV=160mV\begin{matrix}V' & = & 16\times 10mV \\ & = & 160mV\end{matrix}

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