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Question

A cone of height 16 cm and volume 2948 cm3 floats in a liquid of density 3.2 g cm3 such that its base lies below the surface of the liquid. If the area of cross section of conical region above the liquid surface is 6 cm2, find the density of the cone.
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Solution

H=16cm , volume V=2948cm3
V=2948=πR2H3
R2=3×2948πH....(1)
The area of cross section of conical region above the liquid surface is 6cm2.
The the radius of that section be r , 6=πr2
r2=6π ....(2)
Let the height of that cone be h , Now the triangle ABC and ADA2 are similar so Rr=Hh
Using the equation 1 and 2,
we get h=1.67cm
So the volume of cylinder immersed into the liquid V1=2948πr2h3=2944.66cm3
Weight of liquid =buoyancy force
V×ρcone×g=V1×ρl×g
2948×ρcone=2944.66×3.2
ρcone=3.196cm3

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