Meter Bridge Questions in English

(B) Let $X'$ be the equivalent resistance of $X$ and $S$ connected in parallel.
$X' = \frac{X \cdot S}{X + S} = \frac{5S}{5 + S}$.
In a meter bridge,the balancing condition is given by $\frac{X'}{Y} = \frac{l_1}{l_2}$,where $l_1 = 0.625 \ m$ and $l_2 = 1 - 0.625 = 0.375 \ m$.
Substituting the values:
$\frac{5S / (5 + S)}{2} = \frac{0.625}{0.375}$
$\frac{5S}{2(5 + S)} = \frac{625}{375} = \frac{5}{3}$
$15S = 10(5 + S)$
$15S = 50 + 10S$
$5S = 50$
$S = 10 \ \Omega$.

(D) Let the resistance in the left gap be $R_L$ and the resistance in the right gap be $R_R$.
In the first case,the right gap has two equal resistors $r$ in series,so $R_R = r + r = 2r$.
The balancing point is at $l_1 = 50 \ cm$.
Using the meter bridge formula: $\frac{R_L}{R_R} = \frac{l_1}{100 - l_1} \implies \frac{R_L}{2r} = \frac{50}{50} = 1 \implies R_L = 2r$.
In the second case,one resistor $r$ is removed from the right gap,so the new right resistance is $R_R' = r$.
This removed resistor $r$ is connected in parallel to $R_L$.
The new left resistance is $R_L' = \frac{R_L \cdot r}{R_L + r} = \frac{2r \cdot r}{2r + r} = \frac{2r^2}{3r} = \frac{2}{3}r$.
Let the new balancing point be $l_2$.
Then $\frac{R_L'}{R_R'} = \frac{l_2}{100 - l_2} \implies \frac{(2/3)r}{r} = \frac{l_2}{100 - l_2} \implies \frac{2}{3} = \frac{l_2}{100 - l_2}$.
$2(100 - l_2) = 3l_2 \implies 200 - 2l_2 = 3l_2 \implies 5l_2 = 200 \implies l_2 = 40 \ cm$.

(D) In a meter bridge,the principle is given by the formula $\frac{R}{S} = \frac{l_1}{100 - l_1}$,where $R$ is the resistance in the left gap and $S$ is the resistance in the right gap.
Given the ratio $\frac{R}{S} = \frac{2}{3}$.
Substituting the values into the formula: $\frac{2}{3} = \frac{l_1}{100 - l_1}$.
Cross-multiplying gives: $2(100 - l_1) = 3l_1$.
$200 - 2l_1 = 3l_1$.
$200 = 5l_1$.
$l_1 = \frac{200}{5} = 40 \,cm$.
Therefore,the balance length from the left end is $40 \,cm$.

(C) Let the initial balancing length be $l_1$ from the left end. The resistances in the gaps are $R_1 = 3 \Omega$ and $R_2 = 9 \Omega$.
In the initial balanced condition: $\frac{R_1}{R_2} = \frac{l_1}{100 - l_1} \Rightarrow \frac{3}{9} = \frac{l_1}{100 - l_1} \Rightarrow \frac{1}{3} = \frac{l_1}{100 - l_1}$.
$100 - l_1 = 3l_1 \Rightarrow 4l_1 = 100 \Rightarrow l_1 = 25 \text{ cm}$.
When a shunt $S$ is connected in parallel to the $9 \Omega$ resistor,the new resistance in the right gap becomes $R_2' = \frac{9S}{9+S}$.
The balancing point shifts by $25 \text{ cm}$. Since $R_2' < R_2$,the balancing point must shift towards the left. Thus,the new balancing length $l_2 = 25 - 25 = 0$ is not possible,so the shift must be towards the right,$l_2 = 25 + 25 = 50 \text{ cm}$.
For the new balance condition: $\frac{R_1}{R_2'} = \frac{l_2}{100 - l_2} \Rightarrow \frac{3}{R_2'} = \frac{50}{100 - 50} = 1$.
Therefore,$R_2' = 3 \Omega$.
Substituting $R_2' = \frac{9S}{9+S} = 3 \Rightarrow 9S = 27 + 3S \Rightarrow 6S = 27 \Rightarrow S = 4.5 \Omega$.

(A) Case $I$: For a balanced meter bridge,the ratio of resistances is equal to the ratio of the lengths of the wire segments.
$\frac{R}{S} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$
$S = \frac{3R}{2} \quad \dots (i)$
Case $II$: When a resistance of $10 \Omega$ is connected in parallel with $S$,the new resistance $S'$ is given by:
$S' = \frac{10S}{10+S}$
The new null point is at $90 \text{ cm}$.
$\frac{R}{S'} = \frac{90}{100-90} = \frac{90}{10} = 9$
$\frac{R(10+S)}{10S} = 9 \Rightarrow R(10+S) = 90S \quad \dots (ii)$
Substitute $S = \frac{3R}{2}$ from Eq. $(i)$ into Eq. $(ii)$:
$R(10 + \frac{3R}{2}) = 90(\frac{3R}{2})$
$10 + \frac{3R}{2} = 135$
$\frac{3R}{2} = 125$
$R = \frac{250}{3} \approx 83.33 \Omega$
Now,find $S$ using Eq. $(i)$:
$S = \frac{3}{2} \times \frac{250}{3} = 125 \Omega$
Thus,$R = 83.33 \Omega$ and $S = 125 \Omega$.

(B) In a meter bridge,the unknown resistance $S$ is connected in the left gap and a known resistance $R$ is connected in the right gap.
According to the principle of the Wheatstone bridge,at the balanced condition:
$\frac{S}{R} = \frac{l_1}{l_2}$
where $l_1$ is the balancing length from the left end and $l_2 = (100 - l_1)$ is the remaining length.
Given: $l_1 = 25 \ cm$,$R = 1 \ \Omega$.
Therefore,$l_2 = 100 - 25 = 75 \ cm$.
Substituting the values:
$\frac{S}{1} = \frac{25}{75}$
$S = \frac{1}{3} \ \Omega \approx 0.33 \ \Omega$.

(B) Let $X'$ be the equivalent resistance of $X$ and $S$ connected in parallel.
$X' = \frac{X \cdot S}{X + S} = \frac{5S}{5 + S}$.
In a meter bridge, the balancing condition is given by $\frac{X'}{Y} = \frac{l_1}{l_2}$, where $l_1 = 0.625 \, m$ and $l_2 = 1 - 0.625 = 0.375 \, m$.
Substituting the values: $\frac{5S / (5 + S)}{2} = \frac{0.625}{0.375}$.
$\frac{5S}{2(5 + S)} = \frac{625}{375} = \frac{5}{3}$.
$15S = 10(5 + S)$.
$15S = 50 + 10S$.
$5S = 50$.
$S = 10 \, \Omega$.

(C) In a metre bridge,the condition for the balance point is given by the formula: $\frac{P}{Q} = \frac{l}{100-l}$,where $P$ is the resistance in the left gap,$Q$ is the resistance in the right gap,and $l$ is the balance length from the left end.
Given the ratio $\frac{P}{Q} = \frac{2}{3}$.
Substituting the values into the formula:
$\frac{2}{3} = \frac{l}{100-l}$
Cross-multiplying gives:
$2(100-l) = 3l$
$200 - 2l = 3l$
$200 = 5l$
$l = \frac{200}{5} = 40 \ cm$.
Therefore,the balance point from the left is $40 \ cm$.

(C) For a meter bridge,the balance condition is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$,where $R_1$ is the resistance in the left gap and $R_2$ is the resistance in the right gap.
Case $I$: $P$ and $Q$ are in series,so $R_2 = P + Q$.
$\frac{30}{P+Q} = \frac{37.5}{100-37.5} = \frac{37.5}{62.5} = 0.6$
$P+Q = \frac{30}{0.6} = 50 \Omega$ ... $(i)$
Case $II$: $P$ and $Q$ are in parallel,so $R_2 = \frac{PQ}{P+Q}$.
$\frac{30}{\frac{PQ}{P+Q}} = \frac{71.4}{100-71.4} = \frac{71.4}{28.6} \approx 2.5$
$\frac{30(P+Q)}{PQ} = 2.5$
Since $P+Q = 50$,we have $\frac{30 \times 50}{PQ} = 2.5$
$PQ = \frac{1500}{2.5} = 600 \Omega^2$ ... $(ii)$
From $(i)$ and $(ii)$,we have a quadratic equation $x^2 - 50x + 600 = 0$.
$(x-30)(x-20) = 0$.
Thus,the values are $30 \Omega$ and $20 \Omega$.

(A) The meter bridge works on the principle of a balanced Wheatstone bridge.
Let $R_L$ be the resistance in the left gap and $R_R$ be the resistance in the right gap.
The balance point is at $l = 37.5 \ cm$ from the left.
The length of the wire in the right gap is $100 - l = 100 - 37.5 = 62.5 \ cm$.
According to the Wheatstone bridge principle,$\frac{R_L}{R_R} = \frac{l}{100-l}$.
Substituting the values,$\frac{R_L}{R_R} = \frac{37.5}{62.5} = \frac{3}{5}$.
The question asks for the ratio of the right gap resistance to the left gap resistance,which is $\frac{R_R}{R_L}$.
Therefore,$\frac{R_R}{R_L} = \frac{62.5}{37.5} = \frac{5}{3}$.

(C) In case $I$,the resistances are $R_1 = 2 \ \Omega$ and $R_2 = 3 \ \Omega$. The null point is at $l$. Using the meter bridge formula: $\frac{R_1}{R_2} = \frac{l}{100-l} \implies \frac{2}{3} = \frac{l}{100-l}$.
Solving this,$200 - 2l = 3l \implies 5l = 200 \implies l = 40 \ cm$.
In case $II$,the resistance in the right gap becomes $R_2' = \frac{3x}{3+x}$ because $x$ is connected in parallel to $3 \ \Omega$. The null point shifts by $10 \ cm$ to the right,so the new position is $l' = 40 + 10 = 50 \ cm$.
Using the formula: $\frac{R_1}{R_2'} = \frac{l'}{100-l'} \implies \frac{2}{\frac{3x}{3+x}} = \frac{50}{100-50} = \frac{50}{50} = 1$.
Therefore,$\frac{2(3+x)}{3x} = 1 \implies 6 + 2x = 3x \implies x = 6 \ \Omega$.

(B) Initially,for a meter bridge,the balance condition is given by $\frac{P}{Q} = \frac{l}{100-l}$.
Given $P = 2 \Omega$ and $Q = 3 \Omega$,we have $\frac{2}{3} = \frac{l}{100-l}$.
$2(100-l) = 3l \Rightarrow 200 - 2l = 3l \Rightarrow 5l = 200 \Rightarrow l = 40 \text{ cm}$.
When an unknown resistor $R$ is connected in parallel with $3 \Omega$,the new resistance $Q'$ becomes $\frac{3R}{3+R}$.
The null point shifts by $22.5 \text{ cm}$ toward $Y$,so the new length $l' = 40 + 22.5 = 62.5 \text{ cm}$.
The new balance condition is $\frac{2}{Q'} = \frac{62.5}{100-62.5} = \frac{62.5}{37.5} = \frac{5}{3}$.
Substituting $Q'$,we get $\frac{2}{\frac{3R}{3+R}} = \frac{5}{3} \Rightarrow \frac{2(3+R)}{3R} = \frac{5}{3}$.
$6(3+R) = 15R \Rightarrow 18 + 6R = 15R \Rightarrow 9R = 18 \Rightarrow R = 2 \Omega$.

(C) For a meter bridge,the balancing condition is given by $\frac{R_{1}}{R_{2}} = \frac{l}{100-l}$.
Initially,the null point is at $40 \text{ cm}$,so $l = 40 \text{ cm}$:
$\frac{R_{1}}{R_{2}} = \frac{40}{60} = \frac{2}{3} \implies R_{1} = \frac{2}{3}R_{2} \quad ... (1)$
When a $16 \ \Omega$ resistance is connected in parallel to $R_{2}$,the new equivalent resistance $R_{2}'$ is $\frac{R_{2} \times 16}{R_{2} + 16}$.
The new null point is at $50 \text{ cm}$,so $l = 50 \text{ cm}$:
$\frac{R_{1}}{R_{2}'} = \frac{50}{50} = 1 \implies R_{1} = R_{2}' = \frac{16R_{2}}{R_{2} + 16} \quad ... (2)$
Equating $(1)$ and $(2)$:
$\frac{2}{3}R_{2} = \frac{16R_{2}}{R_{2} + 16}$
$\frac{1}{3} = \frac{8}{R_{2} + 16} \implies R_{2} + 16 = 24 \implies R_{2} = 8 \ \Omega$.
Substituting $R_{2} = 8 \ \Omega$ into equation $(1)$:
$R_{1} = \frac{2}{3} \times 8 = \frac{16}{3} \ \Omega$.

(D) In a Wheatstone bridge (of which the metre bridge is a form),the galvanometer and the cell (battery) are conjugate branches.
According to the reciprocity theorem,interchanging the positions of the battery and the galvanometer does not change the condition of the bridge balance.
If it was balanced,it remains balanced; if it was unbalanced,the deflection might change in magnitude or direction,but it will still show a balance point.
Therefore,both deflections can be observed depending on the jockey position,and null deflection will still occur at the balance point.
Thus,option $D$ is correct.