KCET · Physics · Capacitance
Eight drops of mercury of equal radii combine to form a big drop. The capacitance of a bigger drop as compared to each smaller drop is
- A 2 times
- B 8 times
- C 4 times
- D 16 times
Answer & Solution
Correct Answer
(A) 2 times
Step-by-step Solution
Detailed explanation
Let \(R\) and \(r\) be the radii of bigger and each smaller drop, respectively.
Volume of bigger drop \(=8 \times\) Volume of smaller drop
\(\Rightarrow \quad \frac{4}{3} \pi R^{3}=8 \times \frac{4}{3} \pi r^{3}\)
or \(R=2 r...(i)\)
As, capacitance of a smaller spherical drop is
\(\begin{aligned} C &=4 \pi \varepsilon_{0} r \\ \Rightarrow \quad C & \propto r \end{aligned}\)
So, \(\quad \frac{C_{\text {bigger }}}{C_{\text {smaller }}}=\frac{r_{\text {bigger }}}{r_{\text {smaller }}}=\frac{R}{r}=\frac{2 r}{r}=2\)[using Eq. (i)]
So, capacitance of bigger drop is two times as compared to each smaller drop.
Volume of bigger drop \(=8 \times\) Volume of smaller drop
\(\Rightarrow \quad \frac{4}{3} \pi R^{3}=8 \times \frac{4}{3} \pi r^{3}\)
or \(R=2 r...(i)\)
As, capacitance of a smaller spherical drop is
\(\begin{aligned} C &=4 \pi \varepsilon_{0} r \\ \Rightarrow \quad C & \propto r \end{aligned}\)
So, \(\quad \frac{C_{\text {bigger }}}{C_{\text {smaller }}}=\frac{r_{\text {bigger }}}{r_{\text {smaller }}}=\frac{R}{r}=\frac{2 r}{r}=2\)[using Eq. (i)]
So, capacitance of bigger drop is two times as compared to each smaller drop.
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