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

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(B) In a meter bridge,the balance condition is given by $\frac{X}{R} = \frac{\ell_1 + \alpha}{\ell_2 + \beta}$,where $\alpha$ and $\beta$ are the end

Vedclass · Accepted Answer

$10.6$

Vedclass · Answer

(B) In a meter bridge,the balance condition is given by $\frac{X}{R} = \frac{\ell_1 + \alpha}{\ell_2 + \beta}$,where $\alpha$ and $\beta$ are the end corrections at ends $A$ and $B$ respectively.Given:Unknown resistance = $X$Standard resistance $R = 10 \, \Omega$Null point position $\ell_1 = 52 \, cm$Length of wire $\ell = 100 \, cm$,so $\ell_2 = 100 - 52 = 48 \, cm$End correction at $A$,$\alpha = 1 \, cm$End correction at $B$,$\beta = 2 \, cm$Substituting the values in the formula:$\frac{X}{10} = \frac{52 + 1}{48 + 2}$$\frac{X}{10} = \frac{53}{50}$$X = \frac{53 	imes 10}{50} = \frac{53}{5} = 10.6 \, \Omega$Therefore,the determined value of '$X$' is $10.6 \, \Omega$.

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