Find out the quantity of dry material required for 50 m of cement plaster 1.20 cm thick and in 1:5

Given data:
Mix ratio = 1:5 , thickness of plaster = 1.2 cm = 0.012 m

Only one side of cement plaster is given i.e. 50m

Let us assume another side = 1 m

Therefore, the Area of cement plaster = ( 50*1) m²

Now, the wet volume of cement plaster  = area × thickness of cement plaster

= 50m² × 0.012m =0.6 m³

Adding 20% mortar for unevenness, joint filling, etc.

Therefore, total wet volume of cement plaster = 0.6 +( 20% ×0.6)

= 0.72 m³

Now, Dry volume of cement plaster = wet volume × 1.35

[ here 1.35 is the safety factor ]

= 0.72 × 1.35 = 0.972 m³

For cement:

Quantity of cement  = (total dry volume × cement ratio)/total ratio

= (0.972×1)/ (1+5)= 0.972/6 = 0.162m³

Now, the weight of cement = Volume of cement × density of cement

= 0.162mm³ × 1440Kg/mm³ = 233.28 Kg

As we know,

1 bag of cement = 50Kg

So,

233.28/50 = 4.66 = 5 bags

Here, 5 bags of dry cement is used.

For sand:

Quantity of sand  = ( Total dry volume × sand ratio) / total ratio

= ( 0.972 ×5) /(1+5) = 4.86/6 = 0.81m³

Now, the weight of sand = Volume of sand × density of sand

= 0.81mm³  × 1920 Kg/mm³ = 1555.2 Kg

[Density of cement and sand are considered from standards]

For water :

Quantity of water  = 20% of total dry material

= 20 % × ( wt. of cement + wt. of sand )

= 20 % × ( 233.28 + 1555.2 ) = 357.696 liters