A cube of side length lis floating in a liquid of density 1400 kg/m3 with half of the volume inside the liquid. Another liquid of density 1000 kg/mºis pour gently
in the beaker. Then answer the following question. (Consider liquid as immiscible)
The Density of material is
Answers
Explanation:
b=
3
2
a
Explanation:
since the cube floats between water and other material we can say that
weight of object = buoyancy force by water + buoyancy force by liquid
buoyancy force = weight of liquid displaced
weight = volume * density * g
volume = Area * Height
⇒ Weight = Area * Height * density * g
we have to find the height in water
let cube is having side a and b is the length up-to which it is immersed in water.
so (a-b) is the length up to witch the cube is immersed in water.
buoyancy force by water
= Area * Height * density * g
\begin{gathered}= a^{2} * b * density_{water} * g \\= bga^{2}d_{water}\end{gathered}
=a
2
∗b∗density
water
∗g
=bga
2
d
water
buoyancy force by water
= Area * Height * density * g
\begin{gathered}= a^{2} * (a-b) * density_{liquid} * g \\= (a-b)ga^{2}d_{liquid}\end{gathered}
=a
2
∗(a−b)∗density
liquid
∗g
=(a−b)ga
2
d
liquid
buoyancy force by water + buoyancy force by liquid
= [bga^{2}d_{water}] + [a^{2} * g * (a-b) * density_{liquid}]=[bga
2
d
water
]+[a
2
∗g∗(a−b)∗density
liquid
]
= a^{2}g[bd_{water} + (a-b)d_{liquid}]=a
2
g[bd
water
+(a−b)d
liquid
]
weight of object
Weight = Area * Height * density * g
\begin{gathered}= a^{2} * a * density_{cube} * g \\= ga^{3}d_{cube}\end{gathered}
=a
2
∗a∗density
cube
∗g
=ga
3
d
cube
weight of object = buoyancy force by water + buoyancy force by liquid
ga^{3}d_{cube}ga
3
d
cube
= a^{2}g[bd_{water} + (a-b)d_{liquid}]=a
2
g[bd
water
+(a−b)d
liquid
]
\begin{gathered}ad_{cube} = bd{water} + (a-b)d_{liquid}\\900a = 1000b + (a-b)700\\200a = 300b\\b = \frac{2}{3}a\end{gathered}
ad
cube
=bdwater+(a−b)d
liquid
900a=1000b+(a−b)700
200a=300b
b=
3
2
a