In a meter bridge, as shown in the figure, it is given that resistance $Y = 12.5 Ω$ and that the balance is obtained at a distance $39.5 c m$ from end $A$ (by jockey J). After interchanging the resistances $X$ and $Y$ , a new balance point is found at a distance $l_{2}$ from end $A$ . What are the values of $X$ and $l_{2}$ ?

19.15 Ω

and

39.5 c m

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8.16 Ω

and

60.5 c m

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19.15 Ω

and

60.5 c m

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8.16 Ω

and

39.5 c m

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Solution

The correct option is D $8.16 Ω$ and $60.5 c m$
$\frac{X}{Y}$ = $\frac{39.5}{60.5}$

$X = \frac{39.5}{60.5} \times 12.5$

$= 8.16 Ω$

$\frac{12.5}{8.16} = \frac{l_{2}}{100 - l_{2}}$

$1.53 (100 + l_{2}) = l_{2}$

$153 - 1.53 l_{2} = l_{2}$

$\frac{153}{2.53} = l_{2}$

$= 60.5 c m$

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