2 solid spheres of radius 2 m, 1 solid cylinder of radius 2 m, height 4 meter and 3 solid cones of r

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22 Mar 2017 Radius of Sphere

2 solid spheres of radius 2 m, 1 solid cylinder of radius 2 m, height 4 meter and 3 solid cones of radius 2 m, height 6 m are melted to form a bigger solid sphere. What is the radius of this bigger solid sphere in meters?
Please note that symbol ^ denotes power of

OPtion

1) 4.17
2) 4.12
3) 4.06
4) 4.01
5) 3.87
6) 3.81
7) 3.73
8) 3.66
9) 3.58
10)3.51

Solution

Answer: 3.58m
Volume of 2 smaller sphere's = 2*(4/3)*Pi*r^3
Volume of one Cylinder = Pi*r^2*h1
Volume of 3 Cones = 3*(1/3)*Pi*r^2*h2
Volume of Larger Sphere = (4/3)*Pi*R^3
Here r=h=2m, h1=4m, h2=6m and R=?
Volume of Larger Sphere = Volume of 2 smaller sphere's + Volume of one Cylinder +Volume of 3 Cones
(4/3)*Pi*R^3 = 2*(4/3)*Pi*r^3 + P1*r^2*h1 + 3*(1/3)*Pi*r^2*h2
=> 4/3*R^3 = (8/3)*8 + 16 + 24
=> R^3/3 = 16/3 + 4 + 6 = (16+12+18)/3 = 46/3
=> R^3 = 46
R = 46^(1/3) =3.58 m

Option 9)

Correct Option: 9 Best Solution (2)

Payal   7 years AGO

volume of sphere =4/3 *(pi)*r^3
volume of cylinder = (pi)* r^2 *h
volume of cone = (pi) r^2*h/3

let radius of new sphere be r'
therefore we have
Volume of new sphere = sum of volume of all items

4/3*(pi)*r'^3= 2*(volume of sphere)+volume of cylinder+3(volume of cone)

4/3*(pi)*r'^3=2*4/3*(pi)*2^3+(pi)*2^2*4+3*(pi)*262*6/3
solving this
r'^3=46
r=3.58

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DIVYANSHU SHEKHAR   7 years AGO

Volume of sphere=2*4/3*pi*r^3=67.01cubic met.
Volume of solid cylinder=pi*r^2*h=50.264 cubic met.
Volume of solid cone=3*1/3*pi*r^2*h=75.396 cubic met.
Total volume=192.67 cubic metre
=>4/3*pi*r^3=192.67
=>r=3.58m

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