8.0 Metre Bridge Or Slide wire bridge

A known resistance $R$ is taken out of resistance box and jockey is moved till the deflection in galvanometer is zero. $$\frac{P}{Q} = \frac{R}{S}$$

Here,

$P$= resistance of $l$ length of wire,

$Q$= resistance of $100-l$ length of wire,

$r$ = resistance of wire per unit length,

Then,

$P$ = $r.l$ and $Q =r.\left( {100 - l} \right)$

Therefore, $$\frac{P}{Q} = \frac{{rl}}{{r\left( {100 - l} \right)}} = \frac{l}{{100 - l}}$$

Thus ,$$\frac{R}{S} = \frac{l}{{100 - l}}$$

Hence, $$S = \left( {\frac{{100 - l}}{l}} \right)R$$

Question 24. In a meter bridge, the length of the wire is 100 $cm$. At what position will the balance point be obtained if the two resistors are in ratio 2:3.

Solution: For a balanced meter bridge, $$\frac{R}{S} = \frac{l}{{100 - l}}$$

Given, $$\frac{R}{S} = \frac{2}{3}$$

or, $$\frac{l}{{100 - l}} = \frac{2}{3} \Rightarrow 200 - 2l = 3l$$

Hence, $$l = \frac{{200}}{5} = 40cm$$

Question 25. In the metre bridge experiment, null point is obtained at 20$cm$ from one end of the wire when resistance $X$ is balanced against another resistance $Y$. If $X<Y$, then where will be the new position of the null point from the same end is obtained, if one decides to balance a resistance of $4X$ against it?

For balanced metre bridge,$$\frac{R}{S} = \frac{l}{{100 - l}}$$

In first case, $R=X$ , $S=Y$ and $l=20cm$

Thus, $$\frac{X}{Y} = \frac{{20}}{{100 - 20}} = \frac{{20}}{{80}} = \frac{1}{4}$$

In second case, $R = 4X$, $S = Y$

Thus, $$\frac{{4X}}{Y} = \frac{{l'}}{{100 - l'}} \Rightarrow 4 \times \frac{1}{4} = \frac{{l'}}{{100 - l'}}$$

$$ \Rightarrow l' = 100 - l' \Rightarrow l' = 50cm$$

Question 26. Resistance in the two gaps of a metre bridge are $10\Omega $ and $30\Omega $ respectively. If the resistance are interchanged, calculate the shift in balance point.

Solution: For balanced metre bridge,$$\frac{R}{S} = \frac{l}{{100 - l}}$$

In first case,$$\frac{R}{S} = \frac{l}{{100 - l}} = \frac{{10}}{{30}} = \frac{1}{3}$$

$$ \Rightarrow 3l = 100 - l \Rightarrow l = 25cm$$

In second case,$$\frac{R}{S} = \frac{l'}{{100 - l'}} = \frac{{30}}{{10}} = \frac{3}{1}$$

$$ \Rightarrow l' = 300 - 3l' \Rightarrow l' = 75cm$$

Shift in the balance point, $$l' - l = 75 - 25 = 50cm$$

Question 27. With certain resistance in the left gap of a slide wire bridge, the balance point is obtained when a resistance of $10\Omega $ is taken out from a resistance box. On increasing the resistance of the resistance box by $12.5\Omega $, the balance point shift by 20 $cm$. Find the unknown resistance.

Solution: In first case, let $l$ be the length from zero end at which balance point is obtained. Then,$$\frac{R}{{10}} = \frac{l}{{100 - l}}$$

In second case , resistance is increased by $12.5\Omega $ and balance point shift by 20cm.

$$\frac{R}{{22.5}} = \frac{{l - 20}}{{100 - \left( {l - 20} \right)}} = \frac{{l - 20}}{{120 - l}}$$

Dividing above equations, we get, $$\frac{{22.5}}{{10}} = \frac{l}{{100 - l}} \times \frac{{120 - l}}{{l - 20}}$$

$$ \Rightarrow {l^2} - 120l + 3600$$

On solving, we get $$l = 60$$

hence, Unknown resistance is $$\frac{X}{{10}} = \frac{{60}}{{100 - 60}} \Rightarrow X = \frac{{60 \times 10}}{{40}} = 15\Omega $$

Question 28. An unknown resistance ${R_1}$ is connected in series with resistance of $10\Omega $. This combination is connected to one gap in a metre bridge , while another gap is connected to another resistance ${R_2}$. The balance point is obtained at 50 $cm$. Now when the $10\Omega $ is removed, the balance point shifts to 40 $cm$. Find the value of ${R_1}$.

Solution: For balanced metre bridge,$$\frac{R}{S} = \frac{l}{{100 - l}}$$

In first case, $R = {R_1} + 10$, S=${R_2}$ and $l$=50cm

Thus, $$\frac{{{R_1} + 10}}{{{R_2}}} = \frac{{50}}{{100 - 50}} \Rightarrow {R_1} + 10 = {R_2}$$

In second case, R = ${R_1}$ , S=${R_2}$ and $l$=40cm

$$\frac{{{R_1}}}{{{R_2}}} = \frac{{40}}{{100 - 40}} = \frac{2}{3}$$

$$ \Rightarrow 3{R_1} = 2{R_2}$$

Using above relation in ${R_1} + 10 = {R_2}$ and on solving , we get$${R_1} = 20\Omega $$