Discussion Forum

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

Answer:

Explanation:

 For balanced meter bridge

    $\frac{X}{R}=\frac{l}{(100-l)}$   ( where , R=90 Ω)

  $\therefore$     $\frac{X}{90}=\frac{40}{(100-40)}$

  $\therefore$     $X=60$

  $X=R\frac{l}{(100-l)}$

 $\frac{\triangle X}{X}=\frac{\triangle l}{l}+\frac{\triangle l}{100-l}$

                        $=\frac{0.1}{40}+\frac{0.1}{60}$

  $\triangle X=0.25$

   So,   X= $60\pm 0.25$ Ω