Total mass of sand and test tube m = gg =1n10n/kg = 0.1kg =100g,

In the liquid of 1g/cm3, v behavior 1=mρ=Ggρ= 100cm3,

Similarly, in 2g/cm3 liquid, V 2 row =50cm3,

Let the distance from the scale line of 2g/cm3 to the bottom of the test tube be h 1,

∫S = Vh

∴ 100cm3 10cm+h 1=50cm3h 1，

Solution: h 1= 10cm,

∴S=Vh=50cm3 10cm=5cm2，

In the liquid of 1.2g/cm3, row V 3 =100g1.2g/cm3 = 2503cm3,

In the liquid of 1.4g/cm3, row V 4 =100g1.4g/cm3 = 5007cm3,

The distance from the scale line of 1.2g/cm3 to the bottom of the test tube h2=V row 3S=503cm,

The distance from the scale line of 1.4g/cm3 to the bottom of the test tube h3=V row 4S= 1007cm,

∴ The distance between two tick marks △ h = 503cm-1007cm = 5021cm ≈ 2.4cm. 。

So the answer is: 2.4.