Question

a body of mass M and density D1 lies in a liquid of density D2 the minimum work done by the applied force to lift the body to a height is
ans is the picture
prove it​

Answer 1

Explanation:

I considered d1>d2 therefore the upthrust does not equal to the weight of the object.  

Since they are asking for the minimum work done we need to take the minimum force required to lift the object

Thus,

F = mg - U

But  

W =  F * d

Where F = mg - U and d = h

Therefore we can write,  

W = (mg - U) * h ------ 1

**Now we need to remove this U (upthrust) from our Work done equation. So write it in the form of Volume and densities

V is the volume of the object

U = Vd2g ------ 2

and mg = Vd1g ------ 3

**Divide equation 2 by 3  

2/3

U/mg = Vd2g/Vd1g

Simplify and U will get

U = mgd2/d1 ------ 4

Substitute equation 4 in equation 1

W = (mg - mgd2/d1) * h

when factorized

W = mgh (1-d2/d1)

I ma but rusty so please forgive me if there are any mistakes.  

Answer 2

Hey mate refer the attachment........