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# A meter bridge is set-up as shown, to determine an unknown resistance X using a standard 10$\Omega$ resistor. The galvanometer shows null when tapping the key is at the 52 cm mark. The end corrections are 1 cm and 2 cm respectively for the ends A and B. The determined value of ‘X’ is –A) 10.2$\Omega$B) 10.6$\Omega$C) 10.8$\Omega$D) 11.1$\Omega$

Last updated date: 13th Jun 2024
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Hint: We need to understand the working principle in the meter-bridge in order to find the unknown resistance connected to the gap as we can see in the figure. The relation between the balancing length is required to solve this problem easily.

\begin{align} & \dfrac{X}{Y}=\dfrac{l}{100-l} \\ & \Rightarrow \dfrac{X}{10\Omega }=\dfrac{52+1}{100-(52-2)} \\ & \Rightarrow \dfrac{X}{10\Omega }=\dfrac{53}{50} \\ & \Rightarrow X=\dfrac{53}{50}\times 10\Omega \\ & \therefore X=10.6\Omega \\ \end{align}