Question

a solid weight 2375N in air and 125N when totally immersed in liquid of density 0.9 gm per cm que calculate the fesity of solid and the fesity of next liquid in which the same solid would float with one fifth of it's volume expose above the level of water

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Answer 1

2250/9000=0.25 m3= volume  of block

237.5/0.25= 950kg/m3 or 0.95 g/cm3

for next

Vldlg= VBDBg

VL/vB= db/dl    (Vl/VB= fraction of water inside liquid)

4/5= 950/dl

dl=950x5/4

dl= 1187.5 kg/m3

Answer 2

Hence the Density of liquid is  dl = 1187.5 kg/m3

Explanation:

Given data:

• Weight of solid in air = 2375 N
• Weight of solid in liquid = 125 N
• Denisty of liquid = 0.9 g/cm^3

Solution:

Volume of block  = 2250 / 9000 = 0.25 m^3

237.5 / 0.25 = 950kg/m3

Vldlg = VBDBg

4/5 = 950/dl

dl = 950 x 5 / 4  

dl = 1187.5 kg/m3

Hence the Density of liquid is  dl = 1187.5 kg/m3