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A meter bridge is set-up as shown, to determine an unknown resistance \mathrm{X} using a standard \mathrm{10\Omega} resistor. The galvanometer shows null point when tapping key is at \mathrm{ 52cm} mark. The end corrections are \mathrm{ 1 cm} and \mathrm{ 2cm} respectively for the ends A and B. The determined value of \mathrm{ X} is:

Option: 1

10.2\Omega


Option: 2

10.6\Omega


Option: 3

10.8\Omega


Option: 4

11.1\Omega


Answers (1)

best_answer

Applying the condition of balanced Wheatstone bridge, we get

\mathrm{\frac{X}{10 \Omega}=\frac{(52+1) \mathrm{cm}}{(100-52+2) \mathrm{cm}}=\frac{53}{50} }
\mathrm{ X=10 \Omega \times \frac{53}{50}=10.6 \Omega}
Hence Option 2 is correct.

Posted by

Sanket Gandhi

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