Question

The surface area of a solid metallic sphere is 616 cm^2. It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained?

Answer 1

$\bold{question -}$

The surface area of a solid metallic sphere is 616 cm^2. It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained?

$\mathfrak{\underline{\underline{answer :}}}$

[π = 22/7]

4 ×22/7× r²= 616 (total surface area of sphere)

r^2=616*7/88

r^2=7*7

r=7

volume of large sphere =4/3* 22/7* 7*7*7

volume of smaller sphere =4/3 *22/7* 3.5/2* 3.5/2*3.5/2

number of small sphere=volume of large sphere/volume of smaller sphere

                                      =4/3*22/7*7*7*7/4/3*22/7*3.5/2*3.5/2* 3.5/2

=7*7*7*2*2*2*10*10*10/35*35*35

 =8*2*2*2/1*1*1

 =8*8/1

=64

So, total smaller balls are 64

Answer 2

$\huge\mathfrak\red{\underline{Answer:64\:spheres}}$

$\rule{100}2$

$\textbf{\underline{\underline{Step-By-Step\:Explanation:-}}}$

It is given that the surface area of a solid metallic sphere is 616 cm².

Then, it is melted and recast into smaller spheres of diameter 3.5 cm. From this, it is clear that the small spheres of diameter 3.5 cm would have the same surface area.

We know that,

→ Surface area of sphere = 4πr²

$\sf {4}\pi {r}^2 = {616}$

$\sf\implies {r}^2 =\dfrac{616\times 7}{4\times 22}$

$\sf\implies {r}^2 ={49}$

$\sf\implies {r} =\sqrt{49}$

$\sf\therefore {r} = {7}$

Hence,

Radius of the given sphere is 7 cm.

Also,

→ Volume of sphere $\sf =\dfrac{4}{3} \times\pi\times {r}^3$

So, Volume of bigger sphere :-

$\sf =\dfrac{4}{3}\times\dfrac{22}{7}\times {7}^3$

→ Volume of smaller sphere

Radius = 3.5/2

$\sf =\dfrac{4}{3}\pi\times {(3.5/2)}^3$

Finally,

Number of smaller spheres

$\sf=\dfrac{Volume\:of\:bigger\:sphere}{Volume\:of\:smaller\:sphere}$

$\sf =\dfrac{\dfrac{4}{3}\pi\times {7}^3}{\dfrac{4}{3}\pi\times{3.5/2}^3}$

$\sf = 64$

Therefore,

64 such spheres can be obtained.

$\rule{200}{2}$