Four wires of equal length 2 m are arranged as shown in the figure. Wires, 2, 3 and 4 are of equal cross-sectional area and wire 1 is of half the cross section of these wires. By how much distance pointer at point 5 must be moved to get null point? 

For getting null point

$\frac{R_1}{R_3}=\frac{R_2}{R_4}as{R}_2=R_4$

Let the pointer at point 5 is moved to left by distance x to get null point as shown in the figure. If resistance per unit length of wire 3 is r then that of wire 1 will be 2r. 

$(2-x)2r=x\times 2r+2\times r$

$or4x=2$

$orx=\frac{2}{4}=0.5m$