Solving Buoyancy & Upthrust: Understand C & D

[Suraj M]

Homework Statement

A solid is completely immersed in a liquid. The force exerted by the liquid on the solid will(more than one right)
(a)increase if it is pushed deeper inside the liquid.
(b)change if its orientation is changed
(c)decrease if it is taken partially out of the liquid
(d)be in the vertically upward direction

Homework Equations

U=Vd_lg

The Attempt at a Solution

Option C and D are true i got that, and the answers also say that the answer is C and D.
But why is Option A not true, the upthrust does increase when the solid is pushed down, right??
I just want to be sure. Thank you in advance

 

[Bystander]

But why is Option A not true, the upthrust does increase when the solid is pushed down, right??

How do you propose that the force should increase?

 

[Suraj M]

With depth there would be increase in pressure, as liquids have a higher compressibility factor than solids. Hence the density of the liquid would increase at a higher rate as compared to the solid. So the upthrust should increase, right?

 

[Bystander]

liquids have a higher compressibility factor than solids

Is this statement true for all possible liquid-solid pairs?

 

[Suraj M]

I'm not sure. By you're tone, i guess it isn't true for all.
Could you give an example of a pair where this wouldn't be true‽

 

[Bystander]

Mercury ( ~ 3 ppm/atm) and let's say hard rubber.

 

[Suraj M]

wouldn't rubber float in mercury?

 

[Bystander]

A solid is completely immersed in a liquid.

You could be using rubber coated uranium bricks; you're not constrained by liquid and solid densities in any way in the problem statement, you're merely calculating net buoyancy forces.

 

[Suraj M]

oh ok. thank you!