A wooden block floating in a bucket of water has (4/5) of its volume submerged. When certain amount of an oil poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is :

Question

A wooden block floating in a bucket of water has (4/5) of its volume submerged. When certain amount of an oil poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is : (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Mg=((4V/5))ρω g or ((M/V))=(4 ρω/5) or ρ=(4 ρω/5) When block floats fully in water and oil, then Mg=Fb1+Fb2 (ρ V)g=((V/2))ρ textoil g+(V/2)ρ ω g or Upsilon textoil =(3/5) ρω=0.6ρω

Q. A wooden block floating in a bucket of water has 54​ of its volume submerged. When certain amount of an oil poured into the bucket, it is found that the block is just under the oil surface with half of its volume under water and half in oil. The density of oil relative to that of water is :

Mg=(54V​)ρωg or (VM​)=54ρω​ or ρ=54ρω​ When block floats fully in water and oil, then Mg=Fb1​​+Fb2​​ (ρV)g=(2V​)ρoil​g+2V​ρωg or Υoil​=53​ρω=0.6ρω

$M g = (\frac{4 V}{5}) ρ ω g$
or $(\frac{M}{V}) = \frac{4 ρ ω}{5}$
or $ρ = \frac{4 ρ ω}{5}$
When block floats fully in water and oil, then
$M g = F_{b_{1}} + F_{b_{2}}$
$(ρ V) g = (\frac{V}{2}) ρ_{oil} g + \frac{V}{2} ρ ω g$
or $Υ_{oil} = \frac{3}{5} ρ ω = 0.6 ρ ω$