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(C) For a balanced meter bridge,the condition is given by $\frac{X}{R} = \frac{\ell}{100 - \ell}$,where $X$ is the unknown resistance and $R$ is the s

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$60 \pm 0.25 \ \Omega$

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(C) For a balanced meter bridge,the condition is given by $\frac{X}{R} = \frac{\ell}{100 - \ell}$,where $X$ is the unknown resistance and $R$ is the standard resistance.Given $R = 90 \ \Omega$ and $\ell = 40.0 \ cm$,we have:$X = R \frac{\ell}{100 - \ell} = 90 	imes \frac{40}{60} = 60 \ \Omega$.To find the error $\Delta X$,we use the logarithmic differentiation of the formula $X = R \frac{\ell}{100 - \ell}$:$\frac{\Delta X}{X} = \frac{\Delta \ell}{\ell} + \frac{\Delta \ell}{100 - \ell}$,where $\Delta \ell = 1 \ mm = 0.1 \ cm$.Substituting the values:$\frac{\Delta X}{60} = \frac{0.1}{40} + \frac{0.1}{60} = 0.1 \left( \frac{3 + 2}{120} ight) = 0.1 \left( \frac{5}{120} ight) = \frac{0.5}{120} = \frac{1}{240}$.$\Delta X = 60 	imes \frac{1}{240} = 0.25 \ \Omega$.Thus,the unknown resistance is $X = (60 \pm 0.25) \ \Omega$.

During an experiment with a meter bridge,the galvanometer shows a null point when the jockey is pressed at $40.0 \ cm$ using a standard resistance of $90 \ \Omega$,as shown in the figure. The least count of the scale used in the meter bridge is $1 \ mm$. The unknown resistance is:

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