A 10 Ω box is in the right gap of a meter bridge, and the unknown X is in the left…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A $10\ \Omega$ box is in the right gap of a meter bridge, and the unknown $X$ is in the left. The first balance is at $38.0\ \text{cm}$ . After $X$ and the box are interchanged, the balance is at $63.0\ \text{cm}$ . The two should add to $100\ \text{cm}$ , but they do not. Find $X$ from each reading and take the average.

Need a reminder? The key ideaShow

At balance, $\dfrac{X}{R} = \dfrac{l}{100 - l}$ , with $l$ measured from the end where $X$ is joined. A resistor $S$ in parallel with $R$ makes the right arm $\dfrac{RS}{R + S}$ . For a wire, $\rho = \dfrac{\pi d^2 X}{4L}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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