Q147 / 160 · AP EAMCET 2022 (07 Jul Shift 2) · Physics

Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time and the balancing lengths are respectively $360 \mathrm{~cm}$ and $420 \mathrm{~cm}$. If emf of $A$ is $2.4 \mathrm{~V}$, the emf of the second cell $B$ is

  • Correct answer
    A
    $2.8 \mathrm{~V}$
  • B
    $3.2 \mathrm{~V}$
  • C
    $3.0 \mathrm{~V}$
  • D
    $2.6 \mathrm{~V}$

Solution

If $E_1$ and $E_2$ are emf's of cells with $l_1$ and $l_2$ balance lengths in a potentiometer experiment, then, $\frac{E_1}{E_2}=\frac{l_1}{l_2}$ or $\quad E_2=E_1 \cdot \frac{l_2}{l_1}$ Here, $E_1=24 \mathrm{~V}, l_1=360 \mathrm{~cm}$ and $l_2=420 \mathrm{~cm}$ Substituting values in Eq. (i), we get $$ E_2=24 \times \frac{420}{360}=28 \mathrm{~V} $$ So, emf of cell $B$ is $2.8 \mathrm{~V}$.

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

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