Solution:

Ans. } \quad \begin{aligned} \frac{X}{Y} & =\frac{40}{60}=\frac{2}{3} \ X & =4 \Omega . \end{aligned} $$

$4 \Omega$ and $6 \Omega$ are in series, $=10 \Omega$

$40 \Omega$ and $60 \Omega$ are in series, $=100 \Omega$

$10 \Omega$ and $100 \Omega$ are in parallel, $=\frac{1000}{110} \Omega=9.09 \Omega 1$

There will be no change in the balancing length. $1 / 2$ Formulae for series and parallel :

For parallel

$$ R^{\prime}=\frac{R_{1} R_{2}}{R_{1}+R_{2}} $$

For series

$$ R^{\prime}=R_{1}+R_{2} $$

[CBSE Marking Scheme 2016]

[AI Q. 8.In the figure a long uniform potentiometer wire $A B$ is having a constant potential gradient along its length. The null points for the two primary cells of emf $E_{1}$ and $E_{2}$ connected in the manner shown are obtained at a distance of $120 \mathrm{~cm}$ and $300 \mathrm{~cm}$ from the end $A$. Find (i) $E_{1} / E_{2}$ and (ii) position of null point for the cell $E_{1}$.

How is the sensitivity of a potentiometer increased?

A [Delhi I, II, III 2012]

Ans. (i) $E_{1}+E_{2}=300 \mathrm{~K}$

( $K$ is potential gradient in volt $/ \mathrm{cm}$ )

$$ \Rightarrow \quad \frac{E_{1}}{E_{2}}=\frac{7}{3} $$

(ii)

$$ \begin{aligned} E_{1}+E_{2} & =300 \mathrm{~K} \ E_{1}+\frac{3}{7} E_{1} & =300 \mathrm{~K} \end{aligned} $$

$$ \Rightarrow \quad E_{1}=210 \mathrm{~K} $$

Therefore, balancing length for cell $E_{1}$ is $210 \mathrm{~cm} . \frac{1 / 2}{2}$ (Award this $1 / 2$ mark even if the student writes $\left.E_{1}=240 \mathrm{~K}\right)$

(Award full marks for any other correct method)

By decreasing potential gradient, the sensitivity of a potentiometer is increased.

[Or through increasing length, reducing potential drop across wire, increasing resistance input in series with the main cell etc.]

[CBSE Marking Scheme 2012]