The figure shows experimental set up of a meter bridge. When the two unknown resistances X and Y are inserted, the null point D is obtained 40 cm from the end A. When a resistance of 10 Ω is connected in series with X, the null point shifts by 10 cm. Find the position of the null point when the 10 Ω resistance is instead connected in series with resistance ‘Y’. Determine the values of the resistances X and Y.

#### Solution

For a metre bridge:

`X/Y = l_1/(100 - l_1)`… (1)

Where, it is given that *l*_{1} = 40 cm

`X/Y = 40/(100 -40) = 2/3`… (2)

When 10 Ω resistance is added in series to X, null point shifts by 10 cm.

`(X+10)/Y = (40+10)/(100 - (40 +10))`

`X +10 = 50/50`

`(X+10)/Y = 1 or X+10 = Y ..... (3)`

Substituting the value of X from equation (2), we obtain

`2/3y+10 = Y`

`10 = Y -2/3 Y`

or

`Y/3 =10`

`Y =30 Omega`

Substituting the value of *Y* in equation (3), we obtain

*X* + 10 = 30

*X *= 20 Ω

Position of the null point when 10 Ω resistance is put in series with *Y*,

`20/(30+10) = l_i/(100 - l_i)`

`2000 - 20l_1 = 40l_1`

`60l_1 = 2000`

`l_1 = 2000/60`

`l_1 = 33.3 cm`