Question

Two spheres of copper, of radii 1 cm and 2 cm, respectively, are melted inside a cylinder of radius 1 cm. Find the altitude of the cylinder.

Answer 1

vol of sphere 1 = (4/3)*pi*1^3
vol of sphere 2 = (4/3)*pi*2^3
total volume of melted material = (4/3)*pi*9 ( sum of the above)
vol of the moulded cone = (1/3)*pi*r^2*h, where r = radius and h is height. Given r=1 So we have
(4/3)*pi*9 = (1/3)*pi*h; cancelling pi on both sides, we have
h = 36 cm

Answer 2

But your answer is not in choices.
The choices are:
a. 2
b.3
c.4
d.12

Answer 3

ok.

Answer 4

@nathan917 : Can you please help me? :(

Answer 5

Sure.

Answer 6

I Think its 12. Im Not sure What Do you think @jhonyy9

Answer 7

@Jack1 : Help me .

Answer 8

i'm out champ, sorry but i think the answer is 36

Answer 9

Aah, okay.

Answer 10

@Jack1 this is ok ,but how you have got it ?

Answer 11

the above maths by nathan, the question was also done earlier today by nurali, same way and same answer, i cant fault it, only thing i can think of that would make a difference (but not by a factor of 3) is if it was an oblique cylinder

Answer 12

but there's no mention of what angle it's sitting at, and again, it wouldn't affect it by a factor of 3 or more, so... yeah

Answer 13

hang on...
vol of sphere 1 = (4/3)*pi*1^3
vol of sphere 2 = (4/3)*pi*2^3
total volume of melted material = (4/3)*pi*9 ( sum of the above)

%%%*** vol of the moulded cone = (1/3)*pi*r^2*h, where r = radius and h is height. Given r=1 So we have %%%*** this seems wrong, it's a cylinder, not a cone, so area just = pi r^2 h

... so:
(4/3)*pi*9 = *pi*h; cancelling pi on both sides, we have
h = 12 cm

Answer 14

its 12
see the volume of the cylinder will be addition of the volumes of the respective spheres

Answer 15

Thank you for helping me. This is a great help for me :D
Arigatou ^_^

Answer 16

sorry, *so VOLUME is just = pi r^2 h

Answer 17

do itashi mapellete

Answer 18

what's with the autocorrect on swearwords @Mertsj @ all mods
do itashimapellete

Answer 19

\[volume of spheres=\frac{ 4 }{ 3 }\pi \left( 1 \right)^{3}+\frac{ 4 }{ 3 }\left( 2 \right)^{3}\]
\[V=\frac{ 4 }{3 }\pi \left( 1+8 \right)=12\pi \]
Let h be the altitude of copper in the cylinder.
\[Volume of copper \in the cylinder=\pi \left( 1 \right)^{2}h=hpi\]
\[h \pi=12\pi ,h=12cm\]

Answer 20

\[s=\frac{ 15+9+12 }{2 }\]=18
\[Area A=\sqrt{s \left( s-a \right)\left( s-b \right)\left( s-c \right)}\]
\[A=\sqrt{18\left( 18-15 \right)\left( 18-9 \right)\left( 18-12 \right)}\]
\[A=\sqrt{18*3*9*6}=\sqrt{54*54}=54\]
\[9^{2}+12^{2}=81+144=225=15^{2}\]
therefore it is a right angled triangle.
A=1/2 *9*12=9*6=54