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  • 64 small drops of mercury, each of radius  and charge <img alt="q" src="https://entrancecorner.oncodecogs.com/gif.latex?q

64 small drops of mercury, each of radius r and charge q, come together to form a big drop. Therefore, the ratio of surface density of charge of each small drop with that of the big drop is,

Option: 1

64: 1


Option: 2

1: 4


Option: 3

1.64


Option: 4

4: 1


Answers (1)

best_answer

One of the 64 small drops has a surface area of A=4 \pi r^2
Where r is its radius. Now these small drops coalesce into a large drop, the total volume hence must be the same as in

V=64 \times \frac{4}{3} \pi r^3=\frac{4}{3} \pi R^3

where \mathrm{R} must be the radius of the larger sphere. 4 / 3 \pi r^3 is the volume of one small drop. Hence, 

\begin{aligned} &64 r^3=R^3\\ &\Rightarrow R=\sqrt[3]{64 r^3}=4 r \end{aligned}

Also, the total charge on the larger drop is the sum of the charge on the individual drop, hence 

\Rightarrow Q=64 q

where Q is the charge on the larger drop and q is the charge on the smaller drop. Hence, the surface densities of the two surfaces are:

\Rightarrow \sigma_s=\frac{q}{4 \pi r^2} \text { ( for small surface) and } \sigma_l=\frac{64 q}{4 \pi R^2} \text { ( for large surface) }

Hence, the ratio of small drop to large drop is 

\Rightarrow \frac{\sigma_s}{\sigma_l}=\frac{q}{4 \pi r^2} \div \frac{64 q}{4 \pi R^2}=\frac{q}{4 \pi r^2} \times \frac{4 \pi(4 r)^2}{64 q}

By elimination and reduction, we have

\begin{aligned} & \Rightarrow \frac{\sigma_s}{\sigma_l}=\frac{1}{4} \text { hence, } \\ & \Rightarrow \sigma_s: \sigma_l=1: 4 \end{aligned}

Therefore, option (B) is the correct answer.

Posted by

sudhir.kumar

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