Question

solve this ..........​

Answer 1

Question:

Twenty seven solid iron spheres each of radius 'r' and surface area S are melted to form a sphere with radius R and surface area S'. Find the

1) radius R of the new sphere

2) ratio of S and S'

Answer:

1. Radius of R of the new sphere is 3r.

2. Ratio of S and S' is 1 : 9.

Step-by-step-explanation:

We have given that, 27 solid iron spheres having radius 'r' and surface area S are melted.

And a new solid iron sphere was made having a radius of R and surface area S'.

We have to find the radius R of the new sphere.

As the new sphere is formed by melting 27 spheres, they both must have the same volume.

We know that,

$\pink{\sf\:Volume\:of\:sphere\:=\:\dfrac{4\:\pi\:r^3}{3}}\sf\:\:\:-\:-\:[\:Formula\:]$

Now,

$\sf\:Volume\:of\:new\:sphere\:=\:27\:\times\:Volume\:of\:original\:sphere\\\\\\\implies\sf\:\dfrac{\cancel{4}\:\cancel{\pi}\:R^3}{\cancel3}\:=\:27\:\times\:\dfrac{\cancel{4}\:\cancel{\pi}\:r^3}{\cancel3}\\\\\\\implies\sf\:R^3\:=\:27\:r^3\\\\\\\implies\sf\:R\:=\:\sf\:\sqrt[3]{\sf\:27}\:r\:\:\:-\:-\:[\:Taking\:cube\:roots\:]\\\\\\\implies\boxed{\red{\sf\:R\:=\:3r}}\sf\:\:-\:-\:-\:(\:1\:)$

$\rule{200}{1}$

Now,

We know that,

$\pink{\sf\:Surface\:area\:of\:sphere\:=\:4\:\pi\:r^2}\sf\:\:\:-\:-\:[\:Formula\:]$

Now,

$\displaystyle\dfrac{\sf\:Surface\:area\:of\:original\:sphere}{\sf\:Surface\:area\:of\:new\:sphere}\:=\:\dfrac{\sf\:4\:\pi\:r^2}{\sf\:4\:\pi\:R^2}\\\\\\\implies\sf\:\dfrac{S}{S'}\:=\:\dfrac{4\:\pi\:r^2}{4\:\pi\:(\:3r\:)^2}\:\:\:-\:-\:[\:From\:(\:1\:)\:]\\\\\\\implies\sf\:\dfrac{S}{S'}\:=\:\dfrac{\cancel{4\:\pi}\:r^2}{\cancel{4\:\pi}\:9r^2}\\\\\\\implies\sf\:\dfrac{S}{S'}\:=\:\dfrac{\cancel{r^2}}{9\:\cancel{r^2}}\\\\\\\implies\sf\:\dfrac{S}{S'}\:=\:\dfrac{1}{9}\\\\\\\implies\boxed{\red{\sf\:S\::S'\:=\:1\::\:9}}$

Answer 2

Answer:

(i) R = 3r

(ii) Ratio of S and S' = 1:9

Step-by-step explanation:

SOME IMPORTANT THINGS :—

(i) Volume of sphere = 4/3πr³

(ii) Always take the value of π as 22/7 until and unless not mentioned in the question.

(iii) Surface Area of sphere = 4πr²

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(I) SMALL IRON SPHERE

Radius of small iron sphere = r

Volume of 1 small iron sphere = 4/3 πr³

Number of small iron sphere = 27

Volume of 1 small iron sphere = 4/3πr³

Volume of 27 small iron sphere =

$27 \times \frac{4}{3} \times\pi \times {r}^{3}$

Volume of 27 small iron sphere = 9 X 4 X π X r³

Volume of 27 small iron sphere = 36πr³

LARGE IRON SPHERE

Radius of large iron sphere = R

Volume of large iron sphere = 4/3 X π X R³

Volume of large iron sphere = 4/3πR³

Volume of 27 small iron sphere = Volume of large iron sphere

=> 36πr³ = 4/3 πR³

=> 36πr³ X 3/4π = R³

=> 27r³ = R³

=> 3√27r³ = R

=> 3√3 X 3 X 3 X r X r X r = R

=> 3r = R

=> R = 3r

Therefore the radius of large iron sphere is three times (thrice) of the small iron sphere.

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(II) SMALLER IRON SPHERE

Surface Area of smaller iron sphere (S) = 4πr²

LARGER IRON SPHERE

Surface Area of larger iron sphere (S') = 4πR²

Surface Area of larger iron sphere (S') = 4π(3r)² [R = 3r Proved Earlier]

Surface Area of larger iron sphere (S') = 4π9r²

Ratio of S and S' = Surface Area of smaller iron sphere (S)/Surface Area of larger iron sphere (S')

Ratio of S and S' = S/S'

Ratio of S and S' = 4πr²/4π9r²

Ratio of S and S' = 1/9

Ratio of S and S' = 1:9

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