A meter bridge is set up as shown to determin | vedclass

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(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 =

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$3 \ cm, 2 \ cm$

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(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 for a meter bridge with end corrections 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 $l_2 = 57 \ cm$.$\frac{20}{15} = \frac{57 + \alpha}{100 - 57 + \beta} \Rightarrow \frac{4}{3} = \frac{57 + \alpha}{43 + \beta}$.$4(43 + \beta) = 3(57 + \alpha) \Rightarrow 172 + 4\beta = 171 + 3\alpha \Rightarrow 3\alpha - 4\beta = 1$  --- $(ii)$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: $7\alpha = 21 \Rightarrow \alpha = 3 \ cm$.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 $2 \ cm$.

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