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Solution - An Iron Spherical Ball Has Been Melted and Recast into Smaller Balls of Equal Size. If the Radius of Each of the Smaller Balls is 1/4 of the Radius of the Original Ball, How Many Such Balls Are Made? Compare the Surface Area, of All the Smaller Balls Combined Together With That of the Original Ball. - CBSE Class 10 - Mathematics

Question

An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is 1/4 of the radius of the original ball, how many such balls are made? Compare the surface area, of all the smaller balls combined together with that of the original ball.

Solution

Given that radius of each of smaller ball
=1/4 Radius of original ball.

Let radius of smaller ball be . r

Radius of bigger ball be 4r

Volume of big spherical ball =`4/3pir^3`    (∵ r = 4r)

`V_1 = 4/3pi(4r)^3`      ............(1)

Volume of each small ball = `4/3pir^3`

`v_2=4/3PIR^3`          .............(2)

Let no of balls be 'n'

`n = V_1/V_2`

⇒ `n=(4/3pi(4r)^3)/(4/3pi(r)^3)`

⇒ n = 43 = 64

∴ No of small balls = 64

Curved surface area of sphere = 4πr2

Surface area of big ball (S1) = 4π(4r)2    ............(3)

Surface area of each small ball(S1) = 4πr2

Total surface area of 64 small balls

`(S_2)=64xx4pir^2`        ..............(4)

By combining (3) and (4)

⇒ `S_2/3=4`

⇒ S2 = 4s

∴Total surface area of small balls is equal to 4 times surface area of big ball.

 

 

 

  Is there an error in this question or solution?

APPEARS IN

 R.D. Sharma 10 Mathematics (with solutions)
Chapter 16: Surface Areas and Volumes
Q: 29

Reference Material

Solution for question: An Iron Spherical Ball Has Been Melted and Recast into Smaller Balls of Equal Size. If the Radius of Each of the Smaller Balls is 1/4 of the Radius of the Original Ball, How Many Such Balls Are Made? Compare the Surface Area, of All the Smaller Balls Combined Together With That of the Original Ball. concept: Volume of a Combination of Solids. For the course CBSE
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