A cylindrical wooden block of density half the density of water is floating in water in a cylindrical container. The cross section of the wooden block. And its height are A and h respectively. The cross sectional area of the container is 2A. The wooden block is pushed vertically so that it gradually gets immersed in water. Calculate the amount of work done in pushing the block. Density of water `=rho_(0)`.

Options

A. `(h_1 -h_2) grho`
B. `(h_1 -h_2) gArho`
C. `1/2 (h_1-h_2)^2 gArho`
D. `1/4 (h_1-h_2)^2 gArho`

Answer

D

Explanation

`hrarr` Common height
Conservation of mass –
`rhoA_(1) h_1 +rhoA_(2) h_2 = rhoh(A_1+A_2)`
`implies h = ((h_1+h_2)/(2)) " "` [As `A_1=A_2 =A` ]
As `(h_1/2) rarr` initial C.G. `" "` [centre of gravity]
Similarly, `(h_2/2) rarr` initial C.G. `" "` [centre of gravity]
The initial potential energy
`U_i=(h_(1)Arho)g""(h_1)/2+(h_(2)Arho)gh_2/2=rhogA ((h_1^2+h_2^2))/(2)`
when vessels are connected –
`h/2 rarrC` Centre of gravity
i.e. `(((h_1+h_2))/4)` [ as `h=(h_1+h_2)//2`]
Potential energy –
`U_f = [((h_1+h_2)/2)Arho]g((h_1+h_2)/4)=Arhog[((h_1+h_2)^(2))/(4)]`
Work done by gravity –
`W=U_i-U_f=1/4 .rhogA[2h_1^2+2h_2^2-h_1^2-h_2^2-2h_1h_2]`
`W=1/4 rhogA(h_1-h_2)^2`