Question

$\huge\frak\red{ \underline {Question\:❓}}\ \textless \ br /\ \textgreater$

The surface area of a solid metallic sphere is 616 cm2. It is melted and recast into smaller sphere of diameter 3.5 cm. How many such sphere can be obtained ?​

Answer 1

Answer

Given:

The surface area of a solid metallic sphere is 616cm². It is melted and recast into smaller spheres of diameter 3.5cm.

To find:

The spheres are can be obtained.

Explanation:

We have,

The surface area of a solid sphere= 616cm²

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

→ 4× \frac{22}{7}

7

22

×r² = 616cm²

→ 88/7 × r² = 616

→ r² = \frac{616*7}{88}

88

616∗7

→ r² = \frac{\cancel{616}*7}{\cancel{88}}

88

616

∗7

→ r² = (7× 7)cm

→ r² = 49cm

→ r= √49cm

→ r= 7cm.

Therefore,

We know that volume of the sphere: \frac{4}{3} \pi r^{3}

3

4

πr

3

Volume of the big sphere= (\frac{4}{3} *\frac{22}{7} *7*7*7)cm^{3}(

3

4

∗

7

22

∗7∗7∗7)cm

3

Volume of the big sphere= (\frac{4}{3} *\frac{22}{\cancel{7}} *\cancel{7}*49)cm^{3}(

3

4

∗

7

22

∗

7

∗49)cm

3

Volume of the big sphere= (\frac{88*49}{3} )cm^{3}(

3

88∗49

)cm

3

Volume of the big sphere= \frac{4312}{3} cm^{3}

3

4312

cm

3

Volume of the big sphere= 1437.33cm³

&

Volume of the small sphere:

We have,

Diameter of small sphere,d= 3.5cm

Radius of small sphere,r= 3.5/2cm

→ (\frac{4}{3} *\frac{22}{7} *\frac{3.5}{2} *\frac{3.5}{2} *\frac{3.5}{2} )cm^{3}(

3

4

∗

7

22

∗

2

3.5

∗

2

3.5

∗

2

3.5

)cm

3

→ (\frac{\cancel{88}}{21} *\frac{42.875}{\cancel{8}} )cm^{3}(

21

88

∗

8

42.875

)cm

3

→ (\frac{11}{21} *42.875)cm^{3}(

21

11

∗42.875)cm

3

→ \cancel{(\frac{471.625}{21} )}cm^{3}

(

21

471.625

)

cm

3

→ 22.45cm³.

∴ The number of smaller sphere:

⇒ \frac{Volume\:of\:big\:sphere}{Volume\:of\:small\;sphere}

Volumeofsmallsphere

Volumeofbigsphere

⇒ \cancel{\frac{1437.33cm^{3} }{22.45cm^{3} }}

22.45cm

3

1437.33cm

3

⇒ 64.02 spheres

Thus,

The sphere can be obtained is 64 spheres . [approximately]

Answer 2

..Question..ㅤㅤㅤ

The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller sphere of diameter 3.5 cm. How many such sphere can be obtained ?

..Answer..

Given ,

ㅤㅤ➪ Surface area (sphere) = 616 cm²

ㅤㅤ➪ Diameter (small sphere) = 3.5 cm

☯︎ We Know That ,

°•° Surface area (sphere) = 4 π r²

•°• According to the Question

ㅤ➪ Surface area (sphere) = 616 cm²

ㅤ➪ 4 π r²ㅤ ㅤㅤ = 616 cm²

ㅤ➪ 4 × 22/7 × r² = 616

ㅤ➪ 22/7 × r²ㅤㅤ= 616 / 4

ㅤ➪ 22/7 × r²ㅤㅤ= 154

ㅤ➪ 22 × r²ㅤ ㅤ = 154 × 7

ㅤ➪ r²ㅤㅤㅤㅤㅤ = 1078

ㅤ➪ rㅤㅤㅤㅤㅤ = √ 1078

ㅤ➪ rㅤㅤㅤㅤㅤ = 32.8 ( approx )

°•° Radius of Sphere = 32.8 cm

•°• Diameter " ㅤ "ㅤ = 2 × 32.8

ㅤ ㅤ ㅤㅤㅤㅤㅤㅤㅤ = 65.6 cm

Now ,

☯︎ Diameter (small sphere) = 3.5 cm

•°• Surface area (small sphere) »

ㅤㅤㅤ➪ 4 π r²

ㅤㅤㅤ➪ 4 × 22/7 × (3.5)²

ㅤㅤㅤ➪ 4 × 22/7 × 3.5 × 3.5

ㅤㅤㅤ➪ 4 × 22 × 3.5 × 0.5

ㅤㅤㅤ➪ 4 × 22 × 1.75

ㅤㅤㅤ➪ 88 × 1.75

ㅤㅤㅤ➪ 154

•°• Surface area(small sphere)= 154 cm²

So , No. of small sphere obtained =

ㅤ➪ Area(sphere) ÷ Area(small sphere)

ㅤㅤ➪ 616 ÷ 154

ㅤㅤ➪ 4

߷ Total No. of small sphere = 4

_____________________

Give a look here ,

• Surface area ( sphere )

ㅤㅤㅤ➪ 4 π r²

• Volume of sphere

ㅤㅤㅤ➪ 4/3 π r³

• C.S.A ( cylinder )

ㅤㅤㅤ➪ 2 π r h

• Total Surface area ( cylinder )

ㅤㅤㅤ➪ 2 π r ( r + h )

• Surface area ( cuboid )

ㅤㅤㅤ➪ 2 ( lb + bh + hl )

• Surface area ( cube )

ㅤㅤㅤ➪ 6 s²

ㅤㅤㅤ꧁ ʙʀᴀɪɴʟʏ×ᴋɪᴋɪ ꧂