Solution:

Ans. (i) Decreases

(The potential gradient would increase.) $1/2+1/2$

(ii) Increases

(The terminal p.d. across the cell would increase.)

$1/2+1/2$

[CBSE Marking Scheme 2013]

Detailed Answer :

(i) If resistance $R_1$ decreases, potential drop across the wire $AB$ will increase which increases the potential gradient and hence decreases balancing length. $\mathbf{1}$

(ii) If resistance $R_2$ increases, potential drop across the wire $AB$ will decrease which decreases the potential gradient and hence increases balancing length. $\mathbf{1}$

(AI Q. 2. In the metre bridge experiment, balance point was observed at $J$ with $AJ=l$.

(i) The values of $R$ and $X$ were doubled and then interchanged. What would be the new position of the balance point?

(ii) If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected ?

U [O.D. I, 2011; OD Comptt. I, II, III 2012]

Ans. (i) $\frac{R}{X}=\frac{l}{100-l}$ When $R$ and $X$ are doublèd and interchanged, new balance pointis

$l^{\prime }=(100-l)$

[Note : even if the student does not mention the formula, award 1 mark

(ii) No change in the position of balance point. 1

[CBSE Marking Scheme, 2011, 12]

Detailed Answer :

(i) Using the formula to obtain balance point:

$$\begin{array}{rlrl}\frac{R}{X}& =\frac{rl}{r(100-l)}[\because r=\frac{\rho }{A}]\therefore & \frac{R}{X}& =\frac{l}{(100-l)}[\end{array}$$

If $R$ and $X$ gets doubled and their position interchanges, then the new balance length $l^{\prime }$ will be :

$$\begin{array}{rlr}\frac{2X}{2R}& =\frac{l^{\prime }}{(100-l^{\prime })}\text{ So, }\frac{X}{R}& =\frac{l^{\prime }}{(100-l^{\prime })}\end{array}$$

Hence, the new balance point position will be :

$$\frac{100-l}{l}=\frac{l^{\prime }}{100-l^{\prime }}$$

(ii) If the galvanometer and the battery gets interchanged, there will be no effect on the position of balance point since in null position; there will be no flow of current in the circuit.

[A] Q. 3. The diagram below shows a potentiometer set up. On touching the jockey near to the end $X$ of the potentiometer wire, the galvanometer pointer deflects to left. On touching the jockey near to end $Y$ of the potentiometer, the galvanometer pointer again deflects to left but now by a larger amount. Identify the fault in the circuit and explain, using appropriate equations or otherwise, how it leads to such a one-sided deflection.

A&E [CBSE S.Q.P. 2018-19]

S01. The positive of $E_1$ is not connected to terminal $X$. $1/2$

In loop PGJX, $E_1-V_G+E_{XN}=0$

$V_G=E_1+E_{XN}$

$V_G=E_1+kl$

So, $V_G$ (or deflection) will be maximum when $l$ is maximum i.e., when jockey is touched near and $Y$. Also, $V_G$ (or deflection) will be minimum when $l$ is minimum i.e., when jockey is touched near end $X$.

[CBSE Marking Scheme 2018]