Question

A cone of height 16cm and volume 2948 cm3 floats in a liquid of density 3.2g/cm3 such that its base lies below the surface of the liquid

Answer 1

Explanation:

H=16cm , volume V=2948cm^3V=2948cm

3

V=2948=\dfrac{\pi R^2H}{3}V=2948=

3

πR

2

H

R^2=\dfrac{3\times 2948}{\pi H}R

2

=

πH

3×2948

....(1)

The area of cross section of conical region above the liquid surface is 6cm^26cm

2

.

The the radius of that section be r , 6=\pi r^26=πr

2

r^2=\dfrac{6}{\pi}r

2

=

π

6

....(2)

Let the height of that cone be h , Now the triangle ABCABC and ADA_2ADA

2

are similar so \dfrac{R}{r}=\dfrac{H}{h}

r

R

=

h

H

Using the equation 1 and 2,

we get h=1.67 cmh=1.67cm

So the volume of cylinder immersed into the liquid V_1=2948-\dfrac{\pi r^2h}{3}=2944.66cm^3V

1

=2948−

3

πr

2

h

=2944.66cm

3

Weight of liquid =buoyancy force

V\times \rho_{cone}\times g=V_1\times \rho_{l}\times gV×ρ

cone

×g=V

1

×ρ

l

×g

2948\times \rho_{cone}=2944.66\times 3.22948×ρ

cone

=2944.66×3.2

\rho_{cone}=3.196cm^{-3}ρ

cone

=3.196cm

−3