Computer the work which must be performed to slowly pump the water out of a hemispherical reservoir of radius `R=0.6m`.

Answer

`101.8"Kg"f-m`

Explanation

`W=int (dm) gx`
`W=intpi(R sin theta)^(2) dxrhog (R cos theta) " "……(1)`
since `x=R cos theta`
so, `dx =-R sin theta d theta " "…..(2)`
From `(1)` and `(2)`
`W = underset(pi//2)overset(0)intpi(Rsintheta)^(2)(-Rsintheta d theta)rhog(Rcos theta)`
`W=-piRrho^(4)gunderset(pi//2)overset(0)int sin^(3) theta cos theta d theta`
put `sin theta = t " so", cos theta d theta = dt`
`W = -piR rho^(4)g underset(pi//2)overset(0)intt^(3)dt " "W=(-piR^(4)rhog)/(4)[t^(4)]_(pi//2)^(0)`
`W=(piR^(4)g)/(4)[sin^(4)theta]_(pi//2)^(0) " "=W=(piR^(4)rhog)/(4)`
`W=(3.14xx(0.6)^(4)xx1000xxg)/(4)impliesW=101.8kgf-m`