When two resistances X and Y are put in the left hand and right hand gaps in a wheatstone meter bridge,  the null point is at 60cm. If X is shunted by a resistance equal to half of itself then find the shift in the null point.

26.7 cm

Arrangement is shown in the figure.         

$ \frac{X}{Y} = \frac{60}{40} = \frac{3}{2}$

When X is shunted then resistance in the left gap becomes

$ X'= \frac{X.X/2}{X+X/2} = \frac{X}{3}$

$ Now , \frac{X/3}{Y} = \frac{l}{100 - l} = \frac{1}{2}$

$ l = \frac{100}{3} cm$

$ Shift = 60 cm - 33.3 cm = 26.7 cm$