Question

$\huge\mathsf\pink{Gummie\:public}$


Please answer the question that is in the attachment


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Answer 1

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

1. 3 r1
2. 1:9

$\huge\star{\green{\underline{\mathfrak{Explanation :-}}}}$

Given :-

• The radius of the 27 solid iron spheres = r
• The surface area of these small spheres = s
• The surface area of the big sphere formed by melting the small ones = S

To find :-

1. The radius of the big sphere formed
2. The ratio between the surface areas of small sphere and the big sphere

Solution :-

1)

We know that,

The volume of all the 27 small spheres will be equal to the volume of the big sphere formed.

So,

$27(\cancel{\dfrac{4}{3} \pi} {(r1)}^{3} ) = \cancel{ \dfrac{4}{3} \pi }{(r2)}^{3}$

$27 {(r1)}^{3} = {(r2)}^{3}$

$(r2) = \sqrt[3]{27(r1) {}^{3} }$

$(r2) = 3(r1)$

So, the radius is 3×r1.

2)

s = 4 pi (r1)^2

S = 4 pi (r2)^2 = 4 pi (3 r1)^2 = 36 pi r1^2

So, ratio :-

$\cancel \dfrac{4\pi(r1) {}^{2} }{36\pi {(r1)}^{2} }$

$= \dfrac{1}{9}$

So, the ratio is 1:9.

__________________

Answer 2

$\bold{\Huge{\underline{\boxed{\sf{\green{ANSWER\::}}}}}}$

$\bold{\Large{\underline{\sf{\red{Given\::}}}}}$

Twenty-seven solid iron spheres, each of radius 'r' & surface area S are melted to form a sphere with surface area S.

$\bold{\Large{\underline{\sf{\pink{To\:find\::}}}}}$

• Radius r' of the new sphere.
• Ratio of S & S'.

$\bold{\Large{\underline{\sf{\purple{Explanation\::}}}}}$

We know that formula of the volume of sphere:

$\bold{\frac{4}{3} \pi r^{3} }$   [cubic units]

$\bold{\huge{\underline{\sf{\orange{First\:Case\::}}}}}$

• Volume of a solid Iron sphere= $\bold{\frac{4}{3} \pi r^{3} }$

Therefore,

→ Volume of twenty-seven solid iron sphere= $\bold{\cancel{27}*\frac{4}{\cancel{3}} \pi r^{3} }$

→ Volume of 27 solid Iron sphere= 36πr³

&

Volume of new sphere= 36πr³

Let the new radius of sphere be r'

Volume of the new sphere= $\bold{\frac{4}{3} \pi r'^{3} }$

Now,

→ $\bold{\frac{4}{3}\cancel{ \pi} r'^{3} =36\cancel{\pi} r3^{2} }$

→ $\frac{4}{3} r'^{3} =36r^{3}$

→ r'³ = $\bold{\frac{\cancel{36}r^{3} *3}{\cancel{4}} }$

→ r'³ = 27r³

→ r'³ = $\bold{(27r^{3} )^{\frac{1}{3} } }$

→ r'³ = $\bold{(3*3*3r^{3} )^{\frac{1}{3} } }$

→ r'³ = 3r

$\bold{\huge{\underline{\sf{\orange{Second\:Case\::}}}}}$

We know that formula of the surface area of sphere: 4πr²  [sq.units]

We have,

• S= 4πr²
• S'= 4π(3r)²

∴ $\bold{\frac{S}{S'} }=\bold{\frac{4\pi r^{2} }{4\pi (3r)^{2} } }$

→ $\bold{\frac{S}{S'} }=\bold{\frac{\cancel{4\pi }r^{2} }{\cancel{4\pi} (3r)^{2} } }$

→ $\bold{\frac{S}{S'} }=\bold{\frac{r^{2} }{(3)r^{2} } }$

→ $\bold{\frac{S}{S'} }=\bold{\frac{r^{2} }{(9)r^{2} } }$

→ $\bold{\frac{S}{S'} }=\bold{\frac{\cancel{r^{2}} }{(9)\cancel{r^{2} }} }$

→ $\bold{\frac{S}{S'} =\frac{1}{9} }$

→ S:S' = 1:9

Thus,

Radius r' of the new sphere is 3r &

Ratio of S & S' is 1:9.