Question

In an experiment of potentiometer for determining a small resistance, the balancing length for the potential difference across the large resistance is 320 cm and the balancing length for the potential difference across both resistances when connected in series is 360 cm. the ratio of large and small resistance is
(A) 1 : 8
(B) 8 : 1
(C) 8 : 9
(D) 9 : 8

Answer 1

Hint : For a constant current the voltage is directly proportional to the resistance of a conductor or resistor. The ratio of the voltage across the large resistance to the voltage across the two resistances at zero deflection is equal to the ratio of the lengths along the potentiometer for the large resistance to the length along the potentiometer for both the resistance.

Formula used: In this solution we will be using the following formula;
$\Rightarrow \dfrac{{{V_L}}}{{{V_B}}} = \dfrac{{{L_L}}}{{{L_B}}}$ , where ${V_L}$ is the voltage across the large resistance, ${V_B}$ is the voltage across both the resistances, ${L_L}$ is the length along the potentiometer at zero deflection for the large resistance, and ${L_B}$ is the length along the potentiometer at zero deflection for both resistances.
$V = IR$ , $V$ is the voltage across a resistance $R$ through the resistances. $I$ is the current.

Complete step by step answer
In general, at zero deflection, the potentiometer equation can be written as
$\Rightarrow \dfrac{{{V_L}}}{{{V_B}}} = \dfrac{{{L_L}}}{{{L_B}}}$ , where ${V_L}$ is the voltage across the large resistance, ${V_B}$ is the voltage across both the resistances, ${L_L}$ is the length along the potentiometer at zero deflection for the large resistance, and ${L_B}$ is the length along the potentiometer at zero deflection for both resistances.
Since at zero deflection current is same all through, then
Using $V = IR$ , we have
$\Rightarrow \dfrac{{{R_L}}}{{{R_B}}} = \dfrac{{{L_L}}}{{{L_B}}}$
The ratio of large resistance to small can be proven to be
$\Rightarrow \dfrac{{{R_L}}}{{{R_s}}} = \dfrac{{{L_L}}}{{{L_B} - {L_L}}}$
Hence,
$\Rightarrow \dfrac{{{R_L}}}{{{R_s}}} = \dfrac{{320}}{{360 - 320}} = \dfrac{{320}}{{40}} = \dfrac{8}{1}$
Hence, ${R_L}:{R_S} = 8:1$
Thus, the correct option is B.

Note
Alternatively, without extensive calculations, we could reason as follows:
Option A and C can be eliminated since a large resistance cannot be less than a small resistance.
Then by observing the equation $\dfrac{{{R_L}}}{{{R_B}}} = \dfrac{{{L_L}}}{{{L_B}}}$ we see that the ratio of the large resistance to both resistance is $\dfrac{{{R_L}}}{{{R_B}}} = \dfrac{{{L_L}}}{{{L_B}}} = \dfrac{{320}}{{360}} = \dfrac{8}{9}$
Hence, the ratio of both resistance to large resistance is
$\dfrac{{{R_B}}}{{{R_L}}} = \dfrac{9}{8}$
$\Rightarrow {R_B}:{R_L} = 9:8$
Hence, this can be eliminated from the possible answers, and we have only option B.