@FELINTONETO Hi :-)

I've drawn a diagram to show the columns given the information in the question:

I'm here until about 12:30 UK time. If you want to ping me we can go through the question.

@JohnRennie Hey man! Thanks for the diagram. :) So, the way i did was making the iquality of pressures in the botton. so.... Hb.Ub = Ud.(Hb-Hc-Ha) +Uc.(Hb-Ha-Hd)+Ua.(Hb-Hc-Hd) <=> Ub.Hb = Hb(Ud+Uc+Ua) +Hc(-Ud-Uc-Ua)+Ha(-Ud-Uc)+Hd(-Uc-Ua)...But It didnt help

Once I get Ub = Ud+Uc+Ua....But Ud+Uc+Ua = 0, than Ub = 0... so i think I cant use these equality to have some answer..

I've tried other way but..still dont get a solid answer.

@FELINTONETO hi, sorry I was away from the computer.

I haven't actually worked this through, but let's look at what we can say.

The pressure at the lowest dashed line has to be the same on either side, so that means:

$$ U_B g h_B = U_D g h_D + U_C g h_C + U_A g h_A $$

And taking out a factor of $g$ we get:

$$ U_B h_B = U_D h_D + U_C h_C + U_A h_A $$

We also know that:

$$ h_B = h_D + h_C + h_A $$

We can substitute for $h_B$ to get:

$$ U_B(h_D + h_C + h_A) = U_D h_D + U_C h_C + U_A h_A $$

and rearranging gives:

$$ \left(\frac{U_D}{U_B} - 1\right) h_D + \left(\frac{U_C}{U_B} - 1\right) h_C + \left(\frac{U_A}{U_B} - 1\right) h_A = 0 $$

@FELINTONETO OK so far?

OMG! @JohnRennie. And I losing my mind 3 days for just this little rearranging.. :D THanks a lot man! Letter B is the answer. This question appeared in a test Ive made. Thanks! I will be saving this conversation once I will make more questions later. :D

You're welcome :-)

We don't like homework questions on the main site, but in the chat rooms they are fine.

@FELINTONETO Multiple choice questions can usually be solved by logic and insight. We only have one level at which liquids are equal, so we cannot work out any equalities. This eliminates options C, D, E immediately. The fact that C2 is twice the diameter of C1 is a red herring, because it does not affect the pressure at any level.

Finally we see that the (weighted) average density of A+C+D in column C2 (depth $h_B$) must be the same as the density of B in column C1. We are told that A is denser than B therefore at least one of C or D must be less dense than B. We don't have enough clues to deduce that either C or D is denser than B. This might be true but it is not necessarily true. ANSWER : option (B).