a rec.parallelpiped squ.base L=20,h=5cm fill water ,iron sphereof r=10cminserted in it.find rest water in it?

I presume you want to know the volume of the water remaining in the box.

The diameter of the sphere is 20 cm, which makes it just big enough to fit into the box. However, the sphere exceeds the box in height. Only 5 cm of the sphere is immersed.

Volume of the parallelopiped Vp = 20x20x5 = 2000 cm^3

Volume of sphere immersed in water Vs = 654.5 cm^3

The calculation of the volume of the sphere immersed in the water is a bit complicated. I can explain that personally, but it requires the user to understand integration.

Hence, volume of water left in box Vw= Vp - Vs = 1345.5 cm^3

Volome of the Pipe=Base x Height

=(20 x 20) x 5=2000 cu cm

Volume of the sphere= Pi x r^2

= 3.14 x 10^2

=314 cu cm

so the rest water=2000-314=1686 cu cm ans

Vp = 20^2*5 = 2000

Vs = Pi*10^2 = 100Pi

VH2O = 2000 - 100Pi