(1) The volume of water in the container V = SH = 4×10-3m2× 0.18m = 0.72×10-3m3.

∑ρ= mV

∴ The mass of water in the container is m = ρ V =1.0×103kg/m3× 0.72x010-3m3 = 0.72kg. 。

(2) The pressure of water at the bottom of the container is p = rhogh =1.0×103kg/m3× 9.8n/kg× 0.18m =1.765x103pa.

∫P = Fs

∴ pressure f = PS =1.765×103 pa× 4×10-3m2 = 7.056n. 。

(3)①∫ container height H = 0.2m, and container water height H = 0. 18m.

When the water depth reaches 0.2m (that is, when the water depth increases by 0.02m), the bottom pressure reaches the maximum.

∴ Volume of water discharged when the solid cube is gently immersed in the water in the container: v row = s△ h = 4×10-3m2× 0.02m = 0.8×10-4m3.

∫ The volume of solid cube V = 1× 10-4m3.

∴V line < v object; ?

This stone floats on the water.

Then f float = g row = g object = ρ object v object g = ρ water v row g.

∴ rho substance = rho water v discharge v substance =1.0×103kg/m3× 0.8×10? 4m3 1× 10？ 4m3=0.8× 103kg/m3

② When the water depth reaches 0.2m, the bottom pressure reaches the maximum.

p max = rhogh = 1.0× 103kg/m3×9.8n/kg×0.2m = 1960 pa。

Answer: (1) The mass of water in the container is 0.72kg. 。

(2) The water pressure at the bottom of the container is 7.056 Newton.

(3) When the density of solid block is at least 0.8× 103kg/m3, the pressure of water on the bottom of the container reaches the maximum; The maximum pressure is 1960Pa. ..