Solution:

Ans. (i) If $R$ increases, current in circuit of potentiometer wire decreases making terminal potential difference between the poles to balance against large potentiometer wire length which shifts balance point towards $B$.

$2^{1 / 2}$

(ii) If $S$ decreases, current will increase due to Ohms law, decreasing the terminal potential difference among its poles. In such case, the balance point shifts towards point $A$. AI Q. 4. (i) State the principle of working of a potentiometer

(ii) In the following potentiometer circuit $A B$ is a uniform wire of length $1 \mathrm{~m}$ and resistance $10 \Omega$. Calculate the potential gradient along the wire and balance length $\mathrm{AO}(=l)$.

A [Delhi I, II, III 2016]

Ans. (i) Principle : Try yourself Similar to Q. 2 Short Answer Type Questions-II

(ii) Current flowing in the potentiometer wire

$$ \frac{E}{R_{\text {total }}}=\frac{2.0}{15+10}=\frac{2}{25} \mathrm{~A} \quad 1 / 2 $$

$\therefore$ Potential difference across the two ends of the wire

$$ V_{A B}=\frac{2}{25} \times 10 \mathrm{~V}=\frac{20}{25}=0.8 \text { volt } 1 / 2 $$

Hence potential gradient, $K=\frac{V_{A B}}{l_{A B}}$

$$ =\frac{0.8}{1.0}=0.8 \frac{\mathrm{V}}{\mathrm{m}} $$

Current flowing in the circuit containing experimental cell,

$$ =\frac{1.5}{1.2+0.3}=1 \mathrm{~A} $$

Hence, potential difference across length $A O$ of the wire

$$ \begin{align*} & =0.3 \times 1 \mathrm{~V}=0.3 \mathrm{~V} \ \Rightarrow \quad 0.3 & =K \times l_{\mathrm{AO}} \ & =0.8 \times l_{\mathrm{AO}} \ \Rightarrow \quad l_{A O} & =\frac{0.3}{0.8} \mathrm{~m}=0.375 \mathrm{~m} \ & =37.5 \mathrm{~cm} \end{align*} $$

[CBSE Marking Scheme 2016]