English, asked by Anonymous, 9 months ago

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


Please answer the question that is in the attachment


no spams needed ​

Attachments:

Answers

Answered by Nereida
115

\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.

__________________


Anonymous: Good answer
Nereida: Thanks
Answered by Anonymous
81

\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.


Anonymous: Well done bro
Similar questions