Question

A cylindrical bucket 32 cm high and 18 cm of radius of the base is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm.

Answer 1

As we know if a solid object turn into another or if a thing which take a shape of one object turn into another object then there volume will be equal.

Take pie = ¶

volume of cylinder = volume of cone

¶×rsquare×h= 1/3¶r square h

18×18×32×3/24=r square

18×18×8×3/6= r square

18×18×4×3/3=r square

√18×18×2×2=r

18×2=r

r =36 cm

h= 24 cm

slant height = √36×36+24×24

slant height = √1872

slant height=

√2×2×2×2×13×3×3

slant height = 12√13 cm

Answer 2

GIVEN :-

cylinder :

• hieght ( h1 ) = 32 cm

• radius ( r1 ) = 18 cm

cone :

• height ( h2 ) = 24 cm

TO FIND :-

• radius ( r2 ) and slant hieght ( s ) of heap ( cone )

SOLUTION :-

since in cylinder bucket is emptied to make a colonial heap

volume of cylinder = volume of cone

now volume of cylinder :-

hieght ( h1 ) = 32 cm

radius ( r1 ) = 18 cm

we know the formula that volume of cylinder :-

$\implies \boxed {\rm{volume \: = \pi {r}^{2} h}}$

$\implies \rm{\pi ({18})^{2}(32) }$

$\implies \rm{ \bold{ volume = \pi \times 18 \times 18 \times 32 }}$

now volume of cylinder :-

height ( h2 ) = 24 cm

let radius be r cm

and slant hieght be l cm

we know the formula of volume of cone :-

$\implies \boxed {\rm{volume \: = \dfrac{1}{3} \pi {r}^{2} h}}$

$\implies \rm{ \dfrac{1}{3} \pi {r}^{2} (24)}$

$\implies \rm{ \bold{ volume \: of \: cone = 8\pi {r}^{2} }}$

now volume of cylinder = volume of cone hence ,

$\implies \rm{ \pi \times 18 \times 18 \times 32= 8\pi {r}^{2} }$

$\implies \rm{ {r}^{2} = \dfrac{ 18 \times 18 \times 32}{8} }$

$\implies \rm{ {r}^{2} = 18 \times 18 \times 4}$

$\implies \rm{ {r}^{2} = {36}^{2} }$

$\implies \rm{ \bold{ {r} = {36 \: cm}} }$

now we have to find slant hieght :-

$\implies \boxed{ \rm{{l}^{2} = {h}^{2} + {r}^{2}}}$

$\implies \rm{{l}^{2} = {24}^{2} + {36}^{2}}$

$\implies \rm{{l}^{2} = 1872}$

$\implies \rm{l = \sqrt{1872} }$

$\implies \rm{ \bold{l = 12 \sqrt{13} \: cm }}$

HENCE,

$\implies \boxed{ \boxed{ \rm{l = 12 \sqrt{13} \: cm \: \: and \: \: \: r \: = 36cm }} }$