Meter Bridge Questions in English

(C) In a meter bridge,the balancing condition is given by $\frac{P}{Q} = \frac{l}{100-l}$,where $P$ and $Q$ are resistances in the two gaps and $l$ is the balance length.
Case $1$: $R_1$ is in series with $10 \,\Omega$ in one gap,and $R_2$ is in the other gap. The balance point is at $50 \, cm$.
$\frac{R_1 + 10}{R_2} = \frac{50}{100-50} = \frac{50}{50} = 1$
$\Rightarrow R_2 = R_1 + 10$ --- (Equation $1$)
Case $2$: The $10 \,\Omega$ resistance is removed,so only $R_1$ remains in the first gap. The balance point shifts to $40 \, cm$.
$\frac{R_1}{R_2} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$
$\Rightarrow 3R_1 = 2R_2$ --- (Equation $2$)
Substitute Equation $1$ into Equation $2$:
$3R_1 = 2(R_1 + 10)$
$3R_1 = 2R_1 + 20$
$R_1 = 20 \,\Omega$.

(A) In a meter bridge experiment,it is assumed that the resistance of the $L$-shaped copper strips is negligible,but in reality,they possess some small resistance.
This leads to an error known as the end error.
To eliminate this error,the known resistance (from the resistance box) and the unknown resistance are interchanged,and the mean of the two readings is taken to cancel out the effect of the end resistance.

(A) The meter bridge works on the principle of a Wheatstone bridge.
When the null point is at the center of the wire, the ratio of the resistances in the two gaps is equal to the ratio of the lengths of the wire segments.
Since the null point is at the center, the lengths are equal, i.e., $l_1 = l_2 = 50\, cm$.
The balance condition for a Wheatstone bridge is given by $\frac{P}{Q} = \frac{l_1}{l_2}$.
Given that $P = 10\, \Omega$ and $l_1 = l_2$, we have $\frac{10}{Q} = \frac{50}{50} = 1$.
Therefore, $Q = 10\, \Omega$.

(D) In a meter bridge,the balancing condition is given by the formula $\frac{X}{R} = \frac{l}{100 - l}$,where $X$ is the unknown resistance,$R$ is the known resistance,and $l$ is the balancing length from the left end.
Given: $R = 1 \, \Omega$ and $l = 20 \, cm$.
The total length of the wire is $100 \, cm$,so the remaining length is $100 - 20 = 80 \, cm$.
Substituting the values into the formula: $\frac{X}{1} = \frac{20}{80}$.
Simplifying the fraction: $\frac{X}{1} = \frac{1}{4}$.
Therefore,$X = 0.25 \, \Omega$.

(C) The resistance of the wire segments $AC$ and $CB$ are calculated as follows:
$R_{AC} = 0.1 \, \Omega/cm \times 40 \, cm = 4 \, \Omega$
$R_{CB} = 0.1 \, \Omega/cm \times 60 \, cm = 6 \, \Omega$
In the balanced condition of the Wheatstone bridge, the ratio of resistances is equal:
$\frac{X}{6 \, \Omega} = \frac{R_{AC}}{R_{CB}} = \frac{4 \, \Omega}{6 \, \Omega}$
$\frac{X}{6} = \frac{4}{6} \implies X = 4 \, \Omega$
The total resistance of the circuit is the sum of the two parallel branches:
Branch $1$: $X + 6 \, \Omega = 4 \, \Omega + 6 \, \Omega = 10 \, \Omega$
Branch $2$: $R_{AC} + R_{CB} = 4 \, \Omega + 6 \, \Omega = 10 \, \Omega$
Since these two branches are in parallel, the equivalent resistance $R_{eq}$ is:
$R_{eq} = \frac{10 \, \Omega \times 10 \, \Omega}{10 \, \Omega + 10 \, \Omega} = 5 \, \Omega$
The current drawn from the battery is $I = \frac{V}{R_{eq}} = \frac{5 \, V}{5 \, \Omega} = 1.0 \, A$.
Thus, $X = 4 \, \Omega$ and $I = 1.0 \, A$.

(D) The balance condition for a meter bridge is given by $\frac{R}{S} = \frac{l}{100-l}$, where $R$ and $S$ are resistances in the two gaps and $l$ is the balance length from the left end.
Initially, $R = 10 \, \Omega$ and $S = 30 \, \Omega$. Thus, $\frac{10}{30} = \frac{l_1}{100-l_1} \Rightarrow 100 - l_1 = 3l_1 \Rightarrow 4l_1 = 100 \Rightarrow l_1 = 25 \, cm$.
After interchanging, $R = 30 \, \Omega$ and $S = 10 \, \Omega$. Thus, $\frac{30}{10} = \frac{l_2}{100-l_2} \Rightarrow 3(100 - l_2) = l_2 \Rightarrow 300 - 3l_2 = l_2 \Rightarrow 4l_2 = 300 \Rightarrow l_2 = 75 \, cm$.
The shift in the balance point is $|l_2 - l_1| = |75 - 25| = 50 \, cm$.

(A) In a meter bridge,the balancing condition is given by $\frac{R_1}{R_2} = \frac{l}{100 - l}$.
For the first case,$\frac{X}{Y} = \frac{20}{100 - 20} = \frac{20}{80} = \frac{1}{4}$.
So,$\frac{X}{Y} = \frac{1}{4}$,which implies $Y = 4X$.
In the second case,we balance $4X$ against $Y$. Let the new null point be at $l \ cm$.
Then,$\frac{4X}{Y} = \frac{l}{100 - l}$.
Substituting $Y = 4X$ into the equation,we get $\frac{4X}{4X} = \frac{l}{100 - l}$.
$1 = \frac{l}{100 - l} \Rightarrow 100 - l = l \Rightarrow 2l = 100 \Rightarrow l = 50 \ cm$.

(C) Let $R$ be the smaller resistance and $S$ be the larger resistance connected in the two gaps of the meter bridge.
According to the principle of the meter bridge,the balance condition is given by $\frac{R}{S} = \frac{l}{100 - l}$.
Given $l = 20 \ cm$,we have $\frac{R}{S} = \frac{20}{100 - 20} = \frac{20}{80} = \frac{1}{4}$.
Therefore,$S = 4R$ ..... $(i)$
When a resistance of $15 \ \Omega$ is connected in series with the smaller resistance $R$,the new resistance becomes $(R + 15) \ \Omega$.
The new balance point is at $l' = 40 \ cm$.
Applying the balance condition again: $\frac{R + 15}{S} = \frac{40}{100 - 40} = \frac{40}{60} = \frac{2}{3}$.
Substituting $S = 4R$ from equation $(i)$ into this expression:
$\frac{R + 15}{4R} = \frac{2}{3}$
$3(R + 15) = 2(4R)$
$3R + 45 = 8R$
$5R = 45$
$R = 9 \ \Omega$.

(A) In a meter bridge,the balancing condition is given by the ratio of resistances: $\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}}$.
Here,$R_{AC}$ is the resistance of the wire segment $AC$ and $R_{CB}$ is the resistance of the wire segment $CB$.
The resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area.
Substituting this into the balancing condition: $\frac{R_1}{R_2} = \frac{\rho (x) / A}{\rho (100-x) / A} = \frac{x}{100-x}$.
Since the cross-sectional area $A$ cancels out from the numerator and the denominator,the balancing length $x$ depends only on the ratio of the resistors $R_1$ and $R_2$.
Therefore,changing the radius of the wire $AB$ does not affect the balancing length $x$.

(A) In a meter bridge,the null point condition is given by the Wheatstone bridge principle: $\frac{R_1}{R_2} = \frac{P}{Q}$,where $P$ and $Q$ are the resistances of the two segments of the wire $AB$ divided by the jockey.
Let the resistance per unit length of the wire be $\sigma$. If the length of the wire is $L = 100 \ cm$,then $P = \sigma x$ and $Q = \sigma (L - x)$.
The condition becomes $\frac{R_1}{R_2} = \frac{\sigma x}{\sigma (L - x)} = \frac{x}{L - x}$.
Since the resistance per unit length $\sigma = \frac{\rho}{A}$ (where $\rho$ is resistivity and $A$ is the cross-sectional area),$\sigma$ cancels out from the ratio.
Therefore,the null point position $x$ is independent of the radius (and thus the cross-sectional area) of the wire $AB$.
Hence,if the radius of the wire is doubled,the null point position remains unchanged at $x$.

(A) In a meter bridge,the balancing condition is given by $\frac{R_1}{R_2} = \frac{l_1}{100 - l_1}$.
Given the initial condition,$\frac{X}{Y} = \frac{20}{100 - 20} = \frac{20}{80} = \frac{1}{4}$.
So,$Y = 4X$.
Now,we replace $X$ with $4X$ and balance it against $Y$. Let the new null point be at distance $l$.
The new condition is $\frac{4X}{Y} = \frac{l}{100 - l}$.
Substituting $Y = 4X$ into the equation,we get $\frac{4X}{4X} = \frac{l}{100 - l}$.
$1 = \frac{l}{100 - l}$.
$100 - l = l$.
$2l = 100$.
$l = 50 \ cm$.

(A) In a meter bridge,the condition for a balanced bridge when the galvanometer shows zero deflection is given by the formula:
$\frac{R_1}{R_2} = \frac{l}{100 - l}$
Here,$R_1 = 55 \, \Omega$,$R_2 = R$,and the balancing length $l = 20 \, \text{cm}$.
Substituting these values into the formula:
$\frac{55}{R} = \frac{20}{100 - 20}$
$\frac{55}{R} = \frac{20}{80}$
$\frac{55}{R} = \frac{1}{4}$
$R = 55 \times 4 = 220 \, \Omega$
Therefore,the value of the unknown resistance $R$ is $220 \, \Omega$.

(A) For a meter bridge,the balancing condition is given by $\frac{X}{Y} = \frac{l}{100 - l}$.
Given $l = 20 \ cm$,we have $\frac{X}{Y} = \frac{20}{100 - 20} = \frac{20}{80} = \frac{1}{4}$.
Now,if $X$ is replaced by $4X$,let the new null point be $l'$.
The new condition is $\frac{4X}{Y} = \frac{l'}{100 - l'}$.
Substituting $\frac{X}{Y} = \frac{1}{4}$,we get $4 \times \frac{1}{4} = \frac{l'}{100 - l'}$.
$1 = \frac{l'}{100 - l'} \Rightarrow 100 - l' = l' \Rightarrow 2l' = 100$.
Therefore,$l' = 50 \ cm$.

(C) In a meter bridge,the balance condition is given by $\frac{R_1}{R_2} = \frac{l_1}{100 - l_1}$.
Substituting $R_1 = 10 \, \Omega$ and $R_2 = 30 \, \Omega$:
$\frac{10}{30} = \frac{l_1}{100 - l_1} \implies 100 - l_1 = 3l_1 \implies 4l_1 = 100 \implies l_1 = 25 \, \text{cm}$.
Now,interchanging the resistances,we get $R_1 = 30 \, \Omega$ and $R_2 = 10 \, \Omega$.
$\frac{30}{10} = \frac{l_2}{100 - l_2} \implies 3(100 - l_2) = l_2 \implies 300 - 3l_2 = l_2 \implies 4l_2 = 300 \implies l_2 = 75 \, \text{cm}$.
The shift in the null point is $|l_2 - l_1| = |75 - 25| = 50 \, \text{cm}$.

(B) In the first case,at the balance point:
$\frac{5}{R} = \frac{l_1}{100 - l_1}$ $....(i)$
In the second case,the resistance $R$ is shunted with an equal resistance $R$,so the equivalent resistance becomes $R' = \frac{R \times R}{R + R} = \frac{R}{2}$.
The new balance point is at $1.6\,l_1$. Thus,at the balance point:
$\frac{5}{R/2} = \frac{1.6\,l_1}{100 - 1.6\,l_1}$ $....(ii)$
Dividing equation $(i)$ by equation $(ii)$:
$\frac{5/R}{5/(R/2)} = \frac{l_1 / (100 - l_1)}{1.6\,l_1 / (100 - 1.6\,l_1)}$
$\frac{1}{2} = \frac{l_1}{100 - l_1} \times \frac{100 - 1.6\,l_1}{1.6\,l_1}$
$\frac{1}{2} = \frac{100 - 1.6\,l_1}{1.6(100 - l_1)}$
$0.8(100 - l_1) = 100 - 1.6\,l_1$
$80 - 0.8\,l_1 = 100 - 1.6\,l_1$
$0.8\,l_1 = 20$
$l_1 = 25\,cm$
Substituting $l_1 = 25\,cm$ into equation $(i)$:
$\frac{5}{R} = \frac{25}{100 - 25} = \frac{25}{75} = \frac{1}{3}$
$R = 15\,\Omega$.

(A) In a meter bridge,the Wheatstone bridge principle is used to find the unknown resistance.
According to the principle,the ratio of the resistances in the left gap $(P)$ and the right gap $(Q)$ is equal to the ratio of the lengths of the wire segments on the left $(l)$ and right $(100 - l)$ sides.
Given,the balance point is at $l = 35 \ cm$.
Therefore,the ratio is $\frac{P}{Q} = \frac{l}{100 - l}$.
Substituting the values: $\frac{P}{Q} = \frac{35}{100 - 35} = \frac{35}{65}$.
Simplifying the fraction by dividing both numerator and denominator by $5$,we get $\frac{7}{13}$.
Thus,the ratio is $7 : 13$.

(A) For a meter bridge,the balancing condition is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$,where $l$ is the distance from end $A$.
In the first case,$l = 40\,cm$:
$\frac{R_1}{R_2} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$
$R_1 = \frac{2}{3} R_2$ ... $(1)$
In the second case,$R_2$ is shunted by $10\, \Omega$,so the new resistance $R_2'$ is $\frac{10 R_2}{10 + R_2}$. The new balance point is $l' = 50\,cm$:
$\frac{R_1}{R_2'} = \frac{50}{100-50} = \frac{50}{50} = 1$
$R_1 = R_2' = \frac{10 R_2}{10 + R_2}$ ... $(2)$
Equating $(1)$ and $(2)$:
$\frac{2}{3} R_2 = \frac{10 R_2}{10 + R_2}$
$2(10 + R_2) = 30$
$20 + 2 R_2 = 30$
$2 R_2 = 10$
$R_2 = 5\, \Omega$
Substituting $R_2 = 5\, \Omega$ into $(1)$:
$R_1 = \frac{2}{3} \times 5 = \frac{10}{3}\, \Omega$
Thus,$R_1 = \frac{10}{3}\, \Omega$ and $R_2 = 5\, \Omega$.

(B) The balancing condition for a meter bridge is given by:
$\frac{X}{R} = \frac{l_1}{100 - l_1}$
For the initial case where $X = 12\,\Omega$ and $R = 18\,\Omega$:
$\frac{12}{18} = \frac{l_1}{100 - l_1}$
$\frac{2}{3} = \frac{l_1}{100 - l_1}$
$200 - 2l_1 = 3l_1$
$5l_1 = 200$
$l_1 = 40\, cm$
Now,$R$ is changed to $8\,\Omega$. Let the new balancing length be $l_2$:
$\frac{12}{8} = \frac{l_2}{100 - l_2}$
$\frac{3}{2} = \frac{l_2}{100 - l_2}$
$300 - 3l_2 = 2l_2$
$5l_2 = 300$
$l_2 = 60\, cm$
The distance through which the jockey $J$ must be moved is:
$\Delta l = l_2 - l_1 = 60\, cm - 40\, cm = 20\, cm$

(C) In a meter bridge,the balancing condition is given by the formula: $\frac{P}{Q} = \frac{l_1}{l_2}$,where $P$ and $Q$ are the resistances in the two gaps,and $l_1$ and $l_2$ are the corresponding lengths of the wire.
Given,$P = 55 \, \Omega$ and $l_1 = 20 \, \text{cm}$.
The total length of the meter bridge wire is $100 \, \text{cm}$,so $l_2 = 100 - 20 = 80 \, \text{cm}$.
Substituting these values into the balancing condition:
$\frac{55}{R} = \frac{20}{80}$
$\frac{55}{R} = \frac{1}{4}$
$R = 55 \times 4 = 220 \, \Omega$.

(B) Let the resistances be $R_1$ and $R_2$. Given $R_1 + R_2 = 1000 \, \Omega$.
In the first case,the balance point is at length $l$ from the left end:
$\frac{R_1}{R_2} = \frac{l}{100-l} \implies \frac{R_1}{1000-R_1} = \frac{l}{100-l} \quad ... (i)$
After interchanging the resistances,the balance point shifts to the left by $10 \, cm$,so the new balance length is $(l-10) \, cm$:
$\frac{R_2}{R_1} = \frac{l-10}{100-(l-10)} = \frac{l-10}{110-l} \quad ... (ii)$
From $(i)$,$\frac{R_2}{R_1} = \frac{100-l}{l}$. Substituting this into $(ii)$:
$\frac{100-l}{l} = \frac{l-10}{110-l}$
$(100-l)(110-l) = l(l-10)$
$11000 - 100l - 110l + l^2 = l^2 - 10l$
$11000 = 200l \implies l = 55 \, cm$.
Substituting $l = 55$ into $(i)$:
$\frac{R_1}{1000-R_1} = \frac{55}{100-55} = \frac{55}{45} = \frac{11}{9}$
$9R_1 = 11000 - 11R_1$
$20R_1 = 11000 \implies R_1 = 550 \, \Omega$.

(B) In a meter bridge,the balance condition is given by $\frac{R}{S} = \frac{l_1}{100 - l_1}$.
At the null point $D$,the potential at $B$ is equal to the potential at $D$,so no current flows through the galvanometer.
If the jockey is moved to the left of $D$,the resistance of the left segment decreases,making the potential at $D$ higher than at $B$. Thus,current flows from $D$ to $B$ (i.e.,to $B$ from the wire).
If the jockey is moved to the right of $D$,the potential at $D$ becomes lower than at $B$,so current flows from $B$ to $D$ (i.e.,from $B$ to the wire).
Therefore,option $B$ is correct.

(B) In a meter bridge experiment,the sensitivity is highest when the null point is near the center of the wire (i.e.,$l \approx 50\,cm$).
Looking at the observations:
Observation $1$: $l = 43\,cm$
Observation $2$: $l = 51\,cm$
Observation $3$: $l = 59\,cm$
Observation $4$: $l = 70\,cm$
The null point for observation $2$ $(l = 51\,cm)$ is closest to the center $(50\,cm)$.
Therefore,the value obtained in observation $2$ is the most accurate.
The value is $28.82\,\Omega$.

(A) Let the unknown resistance be $X$. In a meter bridge,the balance condition is given by $\frac{R}{X} = \frac{\ell}{100-\ell}$,where $\ell$ is the balance length in $cm$.
Initially,$\frac{2}{X} = \frac{\ell}{100-\ell} \implies X = \frac{2(100-\ell)}{\ell} \quad (1)$
Since $X > 2$,the balance point shifts when resistances are interchanged. The new balance condition is $\frac{X}{2} = \frac{\ell+20}{100-(\ell+20)} = \frac{\ell+20}{80-\ell} \quad (2)$
Substituting $X$ from $(1)$ into $(2)$:
$\frac{2(100-\ell)}{\ell} \cdot \frac{1}{2} = \frac{\ell+20}{80-\ell}$
$\frac{100-\ell}{\ell} = \frac{\ell+20}{80-\ell}$
$(100-\ell)(80-\ell) = \ell(\ell+20)$
$8000 - 100\ell - 80\ell + \ell^2 = \ell^2 + 20\ell$
$8000 = 200\ell \implies \ell = 40\,cm$
Now,substitute $\ell = 40$ into equation $(1)$:
$X = \frac{2(100-40)}{40} = \frac{2 \times 60}{40} = 3\,\Omega$.

(D) Let the end corrections at $A$ and $B$ be $\alpha$ and $\beta$ respectively. The total length of the wire is $100\,cm$.
Case $1$: Resistance $P = 15\,\Omega$ is in the left gap and $Q = 20\,\Omega$ is in the right gap. The null point is at $l_1 = 42\,cm$.
The balance condition is: $\frac{P}{Q} = \frac{l_1 + \alpha}{100 - l_1 + \beta}$
$\frac{15}{20} = \frac{42 + \alpha}{100 - 42 + \beta} \Rightarrow \frac{3}{4} = \frac{42 + \alpha}{58 + \beta}$
$3(58 + \beta) = 4(42 + \alpha) \Rightarrow 174 + 3\beta = 168 + 4\alpha \Rightarrow 4\alpha - 3\beta = 6$ ... $(i)$
Case $2$: Resistances are interchanged,so $P = 20\,\Omega$ and $Q = 15\,\Omega$. The null point is at $l_2 = 57\,cm$.
The balance condition is: $\frac{P}{Q} = \frac{l_2 + \alpha}{100 - l_2 + \beta}$
$\frac{20}{15} = \frac{57 + \alpha}{100 - 57 + \beta} \Rightarrow \frac{4}{3} = \frac{57 + \alpha}{43 + \beta}$
$4(43 + \beta) = 3(57 + \alpha) \Rightarrow 172 + 4\beta = 171 + 3\alpha \Rightarrow 3\alpha - 4\beta = 1$ ... $(ii)$
Solving equations $(i)$ and $(ii)$:
Multiply $(i)$ by $4$ and $(ii)$ by $3$:
$16\alpha - 12\beta = 24$
$9\alpha - 12\beta = 3$
Subtracting the two equations: $7\alpha = 21 \Rightarrow \alpha = 3\,cm$.
Substituting $\alpha = 3$ in $(i)$: $4(3) - 3\beta = 6 \Rightarrow 12 - 3\beta = 6 \Rightarrow 3\beta = 6 \Rightarrow \beta = 2\,cm$.
Thus,the end corrections are $3\,cm$ and $2\,cm$.

(A) Initially,the balance condition for the meter bridge is given by $\frac{R_1}{R_2} = \frac{\ell}{100 - \ell}$.
Given $R_1 = 2 \, \Omega$ and $R_2 = 3 \, \Omega$,we have $\frac{2}{3} = \frac{\ell}{100 - \ell}$.
$2(100 - \ell) = 3\ell \Rightarrow 200 - 2\ell = 3\ell \Rightarrow 5\ell = 200 \Rightarrow \ell = 40 \, cm$.
When the resistances are interchanged,the new resistances are $R_1' = 3 \, \Omega$ and $R_2' = 2 \, \Omega$.
The new balance condition is $\frac{R_1'}{R_2'} = \frac{\ell'}{100 - \ell'}$.
$\frac{3}{2} = \frac{\ell'}{100 - \ell'} \Rightarrow 3(100 - \ell') = 2\ell' \Rightarrow 300 - 3\ell' = 2\ell' \Rightarrow 5\ell' = 300 \Rightarrow \ell' = 60 \, cm$.
The shift in the position of the jockey is $|\ell' - \ell| = |60 - 40| = 20 \, cm$.

(B) For a balanced meter bridge,the condition is $\frac{X}{l_1} = \frac{Y}{100 - l_1}$.
Given $Y = 12.5 \, \Omega$ and $l_1 = 39.5 \, cm$,we have:
$\frac{X}{39.5} = \frac{12.5}{100 - 39.5} = \frac{12.5}{60.5}$.
$X = \frac{12.5 \times 39.5}{60.5} \approx 8.16 \, \Omega$.
When resistances $X$ and $Y$ are interchanged,the new balance condition is $\frac{Y}{l_2} = \frac{X}{100 - l_2}$.
Substituting $X = \frac{Y \times l_1}{100 - l_1}$,we get $\frac{Y}{l_2} = \frac{Y \times l_1}{(100 - l_1)(100 - l_2)}$.
This simplifies to $100 - l_2 = l_1 \times \frac{l_2}{100 - l_1}$,which implies $l_2 = 100 - l_1$.
Therefore,$l_2 = 100 - 39.5 = 60.5 \, cm$.

(D) In the balanced position of the meter bridge,the condition is given by $\frac{P}{Q} = \frac{l}{100-l}$,where $l$ is the distance of the null point from end $A$.
Initially,$P = 4\,\Omega$ and $l = 60\,cm$. Therefore,$100-l = 40\,cm$.
Substituting these values into the balance condition:
$\frac{4}{Q} = \frac{60}{40}$
$\frac{4}{Q} = \frac{3}{2}$
$Q = \frac{8}{3}\,\Omega$.
Now,an unknown resistance $R$ is connected in series with $P$,so the new resistance is $P' = P + R = 4 + R$. The new null point is at $l' = 80\,cm$,so $100-l' = 20\,cm$.
Using the balance condition again:
$\frac{4+R}{Q} = \frac{80}{20}$
$\frac{4+R}{8/3} = 4$
$4+R = 4 \times \frac{8}{3}$
$4+R = \frac{32}{3}$
$R = \frac{32}{3} - 4 = \frac{32-12}{3} = \frac{20}{3}\,\Omega$.
Thus,the value of the unknown resistance $R$ is $\frac{20}{3}\,\Omega$.

(C) In a meter bridge,the balance condition is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$.
Given $\frac{X}{Y} = \frac{40}{100-40} = \frac{40}{60} = \frac{2}{3}$.
So,$X = \frac{2}{3}Y$.
Now,we balance $3X$ against $Y$. Let the new null point be at $l' \ cm$.
Then,$\frac{3X}{Y} = \frac{l'}{100-l'}$.
Substituting $X = \frac{2}{3}Y$ into the equation:
$\frac{3(\frac{2}{3}Y)}{Y} = \frac{l'}{100-l'}$
$2 = \frac{l'}{100-l'}$
$2(100 - l') = l'$
$200 - 2l' = l'$
$3l' = 200$
$l' = \frac{200}{3} \approx 66.67 \ cm$.
Rounding to the nearest integer,the null point is close to $67 \ cm$.

(A) For a meter bridge to be sensitive,the material of the wire (typically Manganin or Constantan) must have high resistivity so that the resistance per unit length is significant,allowing for precise measurements.
Additionally,it must have a low temperature coefficient of resistance so that the resistance of the wire does not change significantly with small fluctuations in temperature during the experiment,ensuring accuracy and stability.

(B) Let $R_{AB} = 4\,\Omega$ be the resistance of the wire $AB$. Let $R_{AJ}$ be the resistance of the segment $AJ$. The potential difference across $AJ$ is given by $V_{AJ} = I \cdot R_{AJ}$,where $I$ is the current in the main circuit.
For the first case,the current in the main circuit is $I_1 = \frac{6}{R_{h1} + R_{AB}} = \frac{6}{2 + 4} = 1\,\text{A}$.
The potential drop across $AJ$ is $V_{AJ} = I_1 \cdot R_{AJ} = 1 \cdot R_{AJ} = \varepsilon_1 = 0.5\,\text{V}$.
Thus,$R_{AJ} = 0.5\,\Omega$.
For the second case,the current in the main circuit is $I_2 = \frac{6}{R_{h2} + R_{AB}} = \frac{6}{6 + 4} = 0.6\,\text{A}$.
Since the null point $J$ is the same,the resistance $R_{AJ}$ remains $0.5\,\Omega$.
The emf $\varepsilon_2$ is equal to the potential drop across $AJ$ in the second case:
$\varepsilon_2 = I_2 \cdot R_{AJ} = 0.6 \times 0.5 = 0.3\,\text{V}$.

(C) For a meter bridge,the balancing condition is $\frac{R_1}{R_2} = \frac{l}{100-l}$.
Given $l = 40\,cm$,so $\frac{R_1}{R_2} = \frac{40}{60} = \frac{2}{3} \implies R_2 = 1.5 R_1$ .....$(i)$
When $10\,\Omega$ is connected in series with $R_1$,the new resistance is $(R_1 + 10)\,\Omega$. The null point shifts by $10\,cm$. Since $R_1$ increases,the null point shifts towards $B$,so the new length $l' = 40 + 10 = 50\,cm$.
Thus,$\frac{R_1 + 10}{R_2} = \frac{50}{50} = 1 \implies R_1 + 10 = R_2$ .....$(ii)$
Substituting $(i)$ into $(ii)$: $R_1 + 10 = 1.5 R_1 \implies 0.5 R_1 = 10 \implies R_1 = 20\,\Omega$ and $R_2 = 30\,\Omega$.
Now,we need to connect a resistance $R$ in parallel with $(R_1 + 10) = 30\,\Omega$ such that the null point returns to $40\,cm$ (i.e.,ratio remains $\frac{2}{3}$).
Let the equivalent resistance be $R_{eq} = \frac{30R}{30+R}$.
Then $\frac{R_{eq}}{R_2} = \frac{2}{3} \implies \frac{R_{eq}}{30} = \frac{2}{3} \implies R_{eq} = 20\,\Omega$.
So,$\frac{30R}{30+R} = 20 \implies 30R = 600 + 20R \implies 10R = 600 \implies R = 60\,\Omega$.

(C) Given $\frac{dR}{d\ell} = \frac{k}{\sqrt{\ell}}$,where $k$ is a constant.
Integrating from $\ell = 0$ to $\ell = 1\, m$,the total resistance $R_{AB}$ of the wire is:
$R_{AB} = \int_{0}^{1} \frac{k}{\sqrt{\ell}} d\ell = k [2\sqrt{\ell}]_{0}^{1} = 2k$.
Let the length $AP = L$. The resistance of segment $AP$ is $R_{AP} = \int_{0}^{L} \frac{k}{\sqrt{\ell}} d\ell = k [2\sqrt{\ell}]_{0}^{L} = 2k\sqrt{L}$.
The resistance of segment $PB$ is $R_{PB} = \int_{L}^{1} \frac{k}{\sqrt{\ell}} d\ell = k [2\sqrt{\ell}]_{L}^{1} = 2k(1 - \sqrt{L})$.
For a balanced meter bridge with equal resistances $R'$ in the gaps,the condition is $\frac{R'}{R_{AP}} = \frac{R'}{R_{PB}}$,which implies $R_{AP} = R_{PB}$.
Therefore,$2k\sqrt{L} = 2k(1 - \sqrt{L})$.
$\sqrt{L} = 1 - \sqrt{L} \implies 2\sqrt{L} = 1 \implies \sqrt{L} = 0.5$.
$L = (0.5)^2 = 0.25\, m$.

(A) In a meter bridge, the unknown resistance $X$ is given by the formula $X = R \frac{l_2}{l_1} = R \frac{(100 - l)}{l}$, where $l$ is the balancing length from the left end.
For reading $1$: $X = 1000 \times \frac{(100 - 60)}{60} = 1000 \times \frac{40}{60} \approx 666.67 \, \Omega$.
For reading $2$: $X = 100 \times \frac{(100 - 13)}{13} = 100 \times \frac{87}{13} \approx 669.23 \, \Omega$.
For reading $3$: $X = 10 \times \frac{(100 - 1.5)}{1.5} = 10 \times \frac{98.5}{1.5} \approx 656.67 \, \Omega$.
For reading $4$: $X = 1 \times \frac{(100 - 1)}{1} = 1 \times 99 = 99 \, \Omega$.
Comparing the calculated values of $X$, the values for readings $1, 2,$ and $3$ are consistent (around $660 \, \Omega$), whereas the value for reading $4$ is significantly different. Thus, reading $4$ is inconsistent.

(B) In a meter bridge,the balance condition is given by $\frac{P}{Q} = \frac{l_1}{100 - l_1}$.
Given $P = 5\,\Omega$ and $Q = R\,\Omega$,we have:
$\frac{5}{R} = \frac{l_1}{100 - l_1} \quad \dots(1)$
When $R$ is shunted with an equal resistance $R$,the equivalent resistance becomes $R' = \frac{R \times R}{R + R} = \frac{R}{2}$.
The new balance point is at $1.6\,l_1$. Thus:
$\frac{5}{R/2} = \frac{1.6\,l_1}{100 - 1.6\,l_1} \Rightarrow \frac{10}{R} = \frac{1.6\,l_1}{100 - 1.6\,l_1} \quad \dots(2)$
Dividing equation $(2)$ by equation $(1)$:
$\frac{10/R}{5/R} = \frac{1.6\,l_1}{100 - 1.6\,l_1} \times \frac{100 - l_1}{l_1}$
$2 = \frac{1.6(100 - l_1)}{100 - 1.6\,l_1}$
$200 - 3.2\,l_1 = 160 - 1.6\,l_1$
$40 = 1.6\,l_1 \Rightarrow l_1 = 25\,cm$.
Substituting $l_1 = 25$ in equation $(1)$:
$\frac{5}{R} = \frac{25}{100 - 25} = \frac{25}{75} = \frac{1}{3}$
$R = 15\,\Omega$.

(A) In a metre bridge experiment,the null point condition is given by the Wheatstone bridge principle:
$\frac{R_1}{R_2} = \frac{R_{AC}}{R_{CB}}$
Where $R_{AC}$ is the resistance of the wire segment $AC$ and $R_{CB}$ is the resistance of the wire segment $CB$.
Since the resistance of a wire is given by $R = \rho \frac{L}{A}$,where $\rho$ is the resistivity,$L$ is the length,and $A$ is the cross-sectional area $(\pi r^2)$,
$\frac{R_{AC}}{R_{CB}} = \frac{\rho (x) / A}{\rho (100-x) / A} = \frac{x}{100-x}$
Thus,the condition for null deflection is $\frac{R_1}{R_2} = \frac{x}{100-x}$.
This equation shows that the null point position $x$ depends only on the values of the resistors $R_1$ and $R_2$ and the total length of the wire. It is independent of the cross-sectional area (and thus the radius) of the wire $AB$,provided the wire is uniform.
Therefore,if the radius of the wire $AB$ is doubled,the null position $x$ will remain unchanged.

(D) In a meter bridge,the condition for zero deflection in the galvanometer is given by the Wheatstone bridge principle:
$\frac{P}{Q} = \frac{R}{S}$
Here,$P = 15 \, \Omega$ and $Q = 10 \, \Omega$.
The resistance of the wire segments $AB$ and $BC$ are proportional to their lengths $\ell$ and $(100 - \ell)$ respectively.
Thus,$\frac{15}{10} = \frac{\ell}{100 - \ell}$
$\Rightarrow 1.5 = \frac{\ell}{100 - \ell}$
$\Rightarrow 1.5(100 - \ell) = \ell$
$\Rightarrow 150 - 1.5\ell = \ell$
$\Rightarrow 150 = 2.5\ell$
$\Rightarrow \ell = \frac{150}{2.5} = 60 \, cm$.
Therefore,the length $AB$ is $60 \, cm$.

(C) For the given meter bridge,the balancing condition is given by $\frac{P}{Q} = \frac{l_1}{100 - l_1}$.
Given $l_1 = 55 \, cm$,so $100 - l_1 = 45 \, cm$.
Given $P = 3 \, \Omega$,we have $\frac{3}{Q} = \frac{55}{45} = \frac{11}{9}$.
Thus,$Q = 3 \times \frac{9}{11} = \frac{27}{11} \, \Omega$.
When an unknown resistance $x$ is connected in series with $P$,the new balancing length $l_1' = 75 \, cm$,so $100 - l_1' = 25 \, cm$.
The new balance condition is $\frac{P + x}{Q} = \frac{75}{25} = 3$.
Substituting the values,$3 + x = 3 \times Q = 3 \times \frac{27}{11} = \frac{81}{11}$.
Therefore,$x = \frac{81}{11} - 3 = \frac{81 - 33}{11} = \frac{48}{11} \, \Omega$.

(A) In a meter bridge,the balancing condition is given by $\frac{R}{S} = \frac{l}{100-l}$,where $l$ is the distance of the null point from $A$.
Initially,$l_1 = 25\, cm$. So,$\frac{R}{S} = \frac{25}{100-25} = \frac{25}{75} = \frac{1}{3}$. Thus,$S = 3R$.
When a resistance of $10\,\Omega$ is connected in parallel with $S$,the new resistance $S'$ is given by $\frac{1}{S'} = \frac{1}{S} + \frac{1}{10}$,so $S' = \frac{10S}{S+10}$.
The new null point is at the midpoint,so $l_2 = 50\, cm$. The new balancing condition is $\frac{R}{S'} = \frac{50}{100-50} = 1$,which means $R = S'$.
Substituting $S' = R$ into the equation $R = \frac{10S}{S+10}$,we get $R = \frac{10(3R)}{3R+10}$.
Dividing by $R$ (assuming $R \neq 0$),we get $1 = \frac{30}{3R+10}$.
$3R + 10 = 30$,so $3R = 20$,which gives $R = \frac{20}{3} \approx 6.67\,\Omega$.

(A) For a meter bridge,the balancing condition is given by $\frac{R_1}{R_2} = \frac{l}{100-l}$.
When $R_1$ and $R_2$ are interchanged,the new balancing length becomes $l' = l + 25$.
The new condition is $\frac{R_2}{R_1} = \frac{l+25}{100-(l+25)} = \frac{l+25}{75-l}$.
Taking the reciprocal of the first equation,we get $\frac{R_2}{R_1} = \frac{100-l}{l}$.
Equating the two expressions for $\frac{R_2}{R_1}$:
$\frac{100-l}{l} = \frac{l+25}{75-l}$.
$(100-l)(75-l) = l(l+25)$.
$7500 - 100l - 75l + l^2 = l^2 + 25l$.
$7500 - 175l = 25l$.
$200l = 7500 \implies l = 37.5 \, cm$.
Now,substitute $l = 37.5$ into the first equation:
$\frac{R_1}{R_2} = \frac{37.5}{100-37.5} = \frac{37.5}{62.5} = \frac{375}{625} = \frac{3}{5}$.

(D) According to the principle of the Wheatstone bridge,the condition for balance is independent of the positions of the cell and the galvanometer. This is known as the reciprocity theorem in electrical networks. If the galvanometer and the cell are interchanged,the bridge will still work,and the balance condition will remain the same,i.e.,$\frac{P}{Q} = \frac{l_{1}}{l_{2}}$.

(D) For a meter bridge,the balancing condition is given by $\frac{R}{S} = \frac{l}{100 - l}$.
Given $l = 25 \; cm$,we have $\frac{R}{S} = \frac{25}{100 - 25} = \frac{25}{75} = \frac{1}{3}$.
Thus,$S = 3R$.
The resistance of a wire is given by $R = \rho \frac{l}{A} = \rho \frac{l}{\pi (d/2)^2} = \frac{4 \rho l}{\pi d^2}$.
When the length is halved $(l^{\prime} = l/2)$ and the diameter is halved $(d^{\prime} = d/2)$,the new resistance $R^{\prime}$ becomes:
$R^{\prime} = \frac{4 \rho (l/2)}{\pi (d/2)^2} = \frac{4 \rho l / 2}{\pi d^2 / 4} = 2 \left( \frac{4 \rho l}{\pi d^2} \right) = 2R$.
Now,for the new balancing length $l^{\prime}$,the condition is $\frac{R^{\prime}}{S} = \frac{l^{\prime}}{100 - l^{\prime}}$.
Substituting $R^{\prime} = 2R$ and $S = 3R$:
$\frac{2R}{3R} = \frac{l^{\prime}}{100 - l^{\prime}}$
$\frac{2}{3} = \frac{l^{\prime}}{100 - l^{\prime}}$
$200 - 2l^{\prime} = 3l^{\prime}$
$5l^{\prime} = 200$
$l^{\prime} = 40 \; cm$.

(N/A) From the first balance point, we get:
$\frac{R}{S} = \frac{33.7}{100 - 33.7} = \frac{33.7}{66.3} \dots (i)$
After $S$ is connected in parallel with a resistance of $12 \; \Omega$, the equivalent resistance $S_{eq}$ is:
$S_{eq} = \frac{12S}{S + 12}$
The new balance condition gives:
$\frac{R}{S_{eq}} = \frac{51.9}{100 - 51.9} = \frac{51.9}{48.1}$
$\frac{R(S + 12)}{12S} = \frac{51.9}{48.1} \dots (ii)$
Substituting $\frac{R}{S} = \frac{33.7}{66.3}$ from $(i)$ into $(ii)$:
$\frac{33.7}{66.3} \cdot \frac{S + 12}{12} = \frac{51.9}{48.1}$
$\frac{S + 12}{S} = \frac{51.9}{48.1} \cdot \frac{66.3}{33.7} \cdot 12 \approx 2.15$
Solving for $S$, we get $S \approx 13.5 \; \Omega$.
Substituting $S$ back into $(i)$:
$R = S \cdot \frac{33.7}{66.3} = 13.5 \cdot 0.508 \approx 6.86 \; \Omega$.

(N/A) meter bridge with resistors $X$ and $Y$ is represented in the given figure.
$(a)$ Balance point from end $A$,$l_{1} = 39.5\; cm$.
Resistance of the resistor $Y = 12.5\; \Omega$.
Condition for the balance is given as:
$\frac{X}{Y} = \frac{l_{1}}{100 - l_{1}}$
$X = Y \times \frac{l_{1}}{100 - l_{1}} = 12.5 \times \frac{39.5}{100 - 39.5} = 12.5 \times \frac{39.5}{60.5} \approx 8.16\; \Omega$.
Therefore,the resistance of resistor $X$ is approximately $8.16\; \Omega$.
The connections between resistors in a Wheatstone or meter bridge are made of thick copper strips to minimize their resistance,which is not taken into consideration in the bridge formula.
$(b)$ If $X$ and $Y$ are interchanged,then $l_{1}$ and $100 - l_{1}$ get interchanged.
The balance point of the bridge will be at $100 - l_{1}$ from $A$.
$100 - l_{1} = 100 - 39.5 = 60.5\; cm$.
Therefore,the balance point is $60.5\; cm$ from $A$.
$(c)$ When the galvanometer and cell are interchanged at the balance point of the bridge,the galvanometer will show no deflection. Hence,no current would flow through the galvanometer.

(A) In a meter bridge circuit, the branch or arm that connects the source of electromotive force (battery) to the bridge is known as the $battery \text{ } arm$.
The branch or arm that connects the galvanometer between the null point on the wire and the junction of the two resistors is known as the $galvanometer \text{ } arm$.

(N/A) meter bridge consists of a wire of uniform cross-section and uniform resistivity with a length of $1 \ m$. This wire is stretched between two thick copper strips fixed on a wooden board.
The copper strips are $L$-shaped at the ends,providing terminals for connections. There is a central straight copper strip that provides a gap for connecting a galvanometer.
$A$ meter scale is fixed parallel to the wire to measure the balancing length.
$A$ battery,a key $(K_1)$,and a rheostat are connected in series between the two ends $A$ and $C$ of the wire to maintain a steady current.
$A$ galvanometer $(G)$ is connected between the central terminal $B$ and a jockey. The jockey can be moved along the wire to find the null point $D$ where the galvanometer shows zero deflection.

(N/A) meter bridge is based on the principle of the Wheatstone bridge.
$1$. Connect the unknown resistor $R$ in one gap and a known resistance $S$ from a resistance box in the other gap of the meter bridge.
$2$. $A$ jockey is slid along the wire $AC$ until the galvanometer shows zero deflection at a point $D$. This is called the null point.
$3$. Let the length of the wire $AD$ be $l$. Then the length of the wire $DC$ is $(100 - l) \text{ cm}$.
$4$. Let $\rho$ be the resistance per unit length of the wire. The resistance of segment $AD$ is $P = \rho l$ and the resistance of segment $DC$ is $Q = \rho(100 - l)$.
$5$. According to the Wheatstone bridge principle,at the balanced condition:
$\frac{R}{S} = \frac{P}{Q} = \frac{\rho l}{\rho(100 - l)}$
$6$. Simplifying this,we get:
$\frac{R}{S} = \frac{l}{100 - l}$
$7$. Therefore,the unknown resistance $R$ is given by:
$R = S \left( \frac{l}{100 - l} \right)$
$8$. By measuring the length $l$ and knowing $S$,the value of $R$ can be calculated. To minimize errors,the null point should be obtained near the center of the wire (between $40 \text{ cm}$ and $60 \text{ cm}$).

(N/A) In a meter bridge experiment,the positions of the known resistance $(R)$ and the unknown resistance $(S)$ are interchanged to eliminate the end errors.
End errors occur due to the resistance of the copper strips at the ends of the meter bridge wire and the contact resistance at the terminals.
By interchanging the resistances,we take two sets of readings.
Let the true value of the unknown resistance be $S$ and the end errors be $\alpha$ and $\beta$.
In the first case,the balance condition is $\frac{R}{S+\alpha} = \frac{l_1}{100-l_1}$.
In the second case,after interchanging,the condition becomes $\frac{S}{R+\beta} = \frac{l_2}{100-l_2}$.
By taking the average of the two measurements,the systematic errors caused by the end resistance are effectively cancelled out,leading to a more accurate result.

(B) In a meter bridge,the balance condition is given by the formula $\frac{X}{R} = \frac{l}{100 - l}$,where $X$ is the unknown resistance,$R$ is the known resistance,and $l$ is the balancing length from one end.
Given that the null point is at $l = 50 \ cm$,the remaining length is $100 - 50 = 50 \ cm$.
Substituting the values into the formula: $\frac{X}{10} = \frac{50}{50}$.
This simplifies to $\frac{X}{10} = 1$.
Therefore,$X = 10 \ \Omega$.

(C) In a meter bridge,the balancing condition is given by $\frac{R}{S} = \frac{\ell_1}{\ell_2}$,where $R$ is the resistance in the left gap and $S$ is the resistance in the right gap.
Given $S = 10\, \Omega$ and the ratio $\frac{\ell_1}{\ell_2} = \frac{3}{2}$.
Substituting these values: $\frac{R}{10} = \frac{3}{2} \implies R = 15\, \Omega$.
The total length of the resistance wire $R$ is $1.5\, m$.
Therefore,the length per unit resistance is $\frac{1.5\, m}{15\, \Omega} = 0.1\, m/\Omega$.
Converting this to the required form: $0.1\, m = 10 \times 10^{-2}\, m$.
Thus,the value is $10$.

(C) In a balanced Wheatstone bridge condition,the ratio of resistances in the arms is equal.
Given the resistances $R_1 = 12\, \Omega$ and $R_2 = 6\, \Omega$.
The wire $AB$ has a total length of $72\, cm$. Let the resistance per unit length of the wire be $\lambda$.
The resistance of segment $AP$ is $R_{AP} = \lambda x$ and the resistance of segment $PB$ is $R_{PB} = \lambda (72 - x)$.
For zero deflection in the galvanometer,the bridge is balanced:
$\frac{12}{x} = \frac{6}{72 - x}$
$12(72 - x) = 6x$
$864 - 12x = 6x$
$18x = 864$
$x = \frac{864}{18} = 48\, cm$.