OpenStudy (anonymous):

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.

OpenStudy (nathan917):

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

OpenStudy (anonymous):

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

OpenStudy (nathan917):

ok.

OpenStudy (anonymous):

OpenStudy (nathan917):

Sure.

OpenStudy (nathan917):

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

OpenStudy (anonymous):

@Jack1 : Help me .

OpenStudy (jack1):

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

OpenStudy (anonymous):

Aah, okay.

OpenStudy (jhonyy9):

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

OpenStudy (jack1):

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

OpenStudy (jack1):

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

OpenStudy (jack1):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (jack1):

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

OpenStudy (jack1):

do itashi mapellete

OpenStudy (jack1):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

$s=\frac{ 15+9+12 }{2 }$=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