Fig. 4.11 shows a cube of each side 15 cm immersed in a tube containing water of density `10^(3) kg m^(-3)` such that its top surface is 20 cm below the free surface of water.

Calculate (i) the pressure at the top of cube,
(ii) the pressure at the bottom of cube,
(iii) the resultant pressure on cube.
(iv) the resultant thrust on cube.
Take atmospheric pressure = `10^(5)` Pa and `g=9.8 N kg^(-1)`.

Explanation

Given, atmospheric pressure `P_(0)=10^(5)Pa, g=9.8N kg^(-1)` , depth to top of the cube from the free surface of water `h_(1)=20 cm=0.2m` , depth of bottom of cube from the free surface of water `h_(2)=(20+15)cm=35=0.35 m`
(i) Pressure at the top surface of cube
`P_(1)=P_(0)+h_(1)rhog`
`=10^(5)+(0.20xx10^(3)xx9.8)`
`=1.0196xx10^(5)Pa`
(ii) Pressure at the bottom surface of cube
`P_(2)=P_(0)+h_(2)rhog`
`=10^(5)+(0.35xx10^(3)xx9.8)`
`=1.0343xx10^(5)Pa`
(iii) Resultant pressure on cube
`=P_(2)-P_(1)=1.0343xx10^(5)-1.0196xx10^(5)`
`=0.0147xx10^(5)Pa(or 1.47xx10^(3)Pa)`
(iv) Area of base of cube
`15cmxx15cm`
`=(15)/(100)mxx(15)/(100)m=225xx10^(-4)m^(2)`
`:.` Resultant pressure xx area of base
`=(1.47xx10^(3))xx(225xx10^(-4))`
`=33.074 N ("upwards")`
Note: The resultant pressure on cube will be upwards. The resultant thrust acting on cube in upward direction is called upthrust. Obviously this upthrust depends on the immersed volume (= area of base xx height) of the cube and not on its depth inside water. Here we can note that if cube is replaced by a lamina which has negligible thickness `( h to 0)`, pressure on the two sides of lamina will be equal, hence upthrust on it will be zero.