Two solids A and B float in water.It is observed that A floats with 1/2 of its body immersed in water and B floats with 1/4 of its volume above the water level .What is the ratio of their densities of A to that of B?
WITH EXPLANATION PLS!!!
Answers
Answer:
=(1/2)×(4/3)=(2/3).Ans..
Explanation:
From the general concept of buoyancy for floating of bodies in water,it can be concluded regarding the volumetric analysis about the immersed percentage of net volume as V' in relation to total one with V,governed by the equation.→V'×1=V×d,where V' & V are in cm³ ,d is the relative density of the body. So V'/V=d.So for body A (dA, the density of A)=VA'/VA & for the body B,dB =VB'/VB,VA & VB are the volumes of the two bodies A&B respectively with VA'&VB' are their respective volume of immersion in water.As per question
VA'/VA=(1/2) and {(VB-VB')/VB}=(1/4)=>
1-(VB'/VB)=(1/4)=>(VB'/VB)={1-(1/4)}=(3/4)
So dA= (1/2)& dB=(3/4).Their ratio ={(1/2)/(3/4)}
=(1/2)×(4/3)=(2/3).Ans..
The answer is 2:3......