A meter bridge determines an unknown resistance X using a standard 5 Ω resistor. The galvanometer shows the null point at the 50 cm mark, the unknown resistor being on the side of end A . The end corr

Question

A meter bridge determines an unknown resistance $X$ using a standard $5 Ω$ resistor. The galvanometer shows the null point at the 50 cm mark, the unknown resistor being on the side of end $A$ . The end corrections are 1 cm and 2 cm for the ends $A$ and $B$ respectively. The determined value of $X$ is

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Answer:

A

Solution:

$\frac{X}{5} = \frac{50 + 1}{( 100 - 50 ) + 2}$

$= \frac{51}{52}$

$X = \frac{5 \times 51}{52} = 4.90 Ω$

TestHub · Accepted Answer

5 X ​ = ( 100 − 50 ) + 2 50 + 1 ​ = 52 51 ​ X = 52 5 × 51 ​ = 4.90 Ω

Physics - Current Electricity Question with Solution | TestHub

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