A meter bridge set up as shown to determine end correction at A and B . When a resistance of 15 Ω i

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3.Current Electricity

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A meter bridge set up as shown to determine end correction at $A$ and $B$. When a resistance of $15\ \Omega$ is used in left gap and of $20\ \Omega$ in right gap, then null point comes at a distance $42\ cm$ from $A$. When these resistances are interchanged null point comes at a distance $57\ cm$ from $A$. Values of end corrections are

A

$1\ cm$, $2\ cm$

B

$2\ cm$, $3\ cm$

C

$3\ cm$, $4\ cm$

D

$3\ cm$, $2\ cm$

Solution

Let and corrections at $\mathrm{A}$ and $\mathrm{B}\, \alpha$ and $\beta$

$\frac{15}{42+\alpha}=\frac{20}{58+\beta} \Rightarrow \frac{58+\beta}{42+\alpha}=\frac{4}{3}$

$\Rightarrow 3 \beta+174=168+4 \alpha$

$4 \alpha-3 \beta=6$ …………$(i)$

$\frac{20}{57+\alpha}=\frac{15}{43+\beta} \Rightarrow \frac{57+\alpha}{43+\beta}$

$=\frac{4}{3} \Rightarrow 171+3 \alpha=172+4 \beta$

$3 \alpha-4 \beta=1$ …………$(ii)$

from $(i)$ and $(ii)$

$\alpha=3 \mathrm{\,cm}, \beta=2 \mathrm{\,cm}$

Standard 12

Physics

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