Two cells $P$ and $Q$ each of emf $2.16 \mathrm{~V}$ are connected in series with a resistor of $19.6 \Omega$. An ideal voltmeter reads $2 \mathrm{~V}$, when connected across the cell $P$ and $1.92 \mathrm{~V}$ when connected across the cell $Q$. The ratio of the internal resistances of the cell $P$ and $Q$ is

• A

  1 : 2
• Correct answer
  B

  2 : 3
• C

  3 : 4
• D

  1 : 3

Solution

Emf of each cell $=2.16 \mathrm{~V}$ Emf of circuit $=4.32 \mathrm{~V}$ Current in circuit, $$ I=\frac{4.32}{r_1+r_2+19.6} $$ Now, using $V=E-I \cdot r$ for both cells, we have For $P, 2=2.16-\frac{4.32 r_1}{r_1+r_2+19.6}$ and For $Q, 1.92=216-\frac{4.32 r_2}{r_1+r_2+19.6}$ So, we have,
From Eqs. (i) and (ii), we get $$ r_1: r_2=2: 3 $$

Hence, the correct answer is option: 'B'.