Question

In the circuit shown in fig., the potential difference between the points C and D is balanced against 40 cm length of potentiometer wire of total length 100 cm. In order to balance the potential difference between the points D and E, the jockey should be pressed on potentiometer wire at a distance of

• A. 16 cm
• B. 32 cm
• C. 56 cm
• D. 80 cm

Answer 1

The correct option is B 32 cm

Let, $V$ be the potential difference between the points $C$ and $D$ and $l_1$ is the length of potentiometer wire balanced against it. Hence the potential gradient is
$k=\frac{V}{l_1}=\frac{I{R}_1}{l_1}$
where,$I$ is the current through the wire and $R_1$ is the resistance between points $C$ and $D$.
$\therefore I{R}_1=k{l}_1$
Let, $l_2$ is the balancing length for potential difference between the points $D$ and $E$. Hence the potential gradient is
$k=\frac{V}{l_2}=\frac{I{R}_2}{l_2}$
where, $I$ is the current through the wire and $R_2$ is the resistance between points C and D.
$\therefore I{R}_2=k{l}_2$
comparing both equations
$\frac{I{R}_2}{I{R}_1}=\frac{k{l}_2}{k{l}_1}$
$l_2=\left(\frac{R_2}{R_1}\right)l_1$
Put $R_1=\frac{(10\times 10)}{10+10}=5\Omega ,R_2=4\Omega$ and $l_1=40cm$
$\therefore l_2=\left(\frac{4}{5}\right)40=32cm$