There are 3 sphere having r cm radius which are completely fit (vertic

1.	There are 3 sphere having r cm radius which are completely fit (vertical manner) in a cylinder. Those sphere cut out from the cylinder. What the ratio of total surface area of all three sphere to curved surface area of cylinder?1). 1 ∶ 12). 2 ∶ 33). 3 ∶ 14). 4 ∶ 5
Answer» Solution : It is given that all the three SPHERES FIT correctly into the cylinder so, the height of the cylinder will be equal to the sum of DIAMETER of the three spheres, ===> $h = 3 d$ ( 'h' is the height of the cylinder and 'd' is the diameter of the sphere.) We know that $d = 2r$ so, $h = 3×2r = 6r$ $h = 6r$ Total Surface Area of a Sphere $= 4 \PI r^2$ Total Surface Area of 3 Spheres $= 3× 4 \pi r^2 = 12 \pi r^2$ Curved Surface Area of Cylinder $= 2 \pi r h$ Total Surface Area of $3$ Spheres : Curved Surface Area of Cylinder $= 12 \pi r^2 ÷2 \pi r h$ $= 6 r ÷ h$ $= 6 r ÷ 6 r( h = 6r )$ $= 1 : 1$ *So, the CORRECT option is 1).1 : 1*

There are 3 sphere having r cm radius which are completely fit (vertical manner) in a cylinder. Those sphere cut out from the cylinder. What the ratio of total surface area of all three sphere to curved surface area of cylinder?1). 1 ∶ 12). 2 ∶ 33). 3 ∶ 14). 4 ∶ 5

Answer»

Solution :

It is given that all the three SPHERES FIT correctly into the cylinder so, the height of the cylinder will be equal to the sum of DIAMETER of the three spheres,

===> $h = 3 d$ ( 'h' is the height of the cylinder and 'd' is the diameter of the sphere.)

We know that $d = 2r$ so,

$h = 3×2r = 6r$

$h = 6r$

Total Surface Area of a Sphere $= 4 \PI r^2$

Total Surface Area of 3 Spheres $= 3× 4 \pi r^2 = 12 \pi r^2$

Curved Surface Area of Cylinder $= 2 \pi r h$

Total Surface Area of $3$ Spheres : Curved Surface Area of Cylinder

$= 12 \pi r^2 ÷2 \pi r h$

$= 6 r ÷ h$

$= 6 r ÷ 6 r( h = 6r )$

$= 1 : 1$

So, the CORRECT option is 1).1 : 1

Discussion

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