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Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the (i) radius r' of the new sphere, (ii) ratio of S and S'. - CBSE Class 9 - Mathematics

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Question

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

(i) radius r' of the new sphere, (ii) ratio of S and S'.

Solution

(i) Radius of 1 solid iron sphere = r

`"Volume of 1 solid iron sphere "=4/3pir^3`

`"Volume of 27 solid iron spheres "=27xx4/3pir^3`

27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this new sphere be r'.

`"Volume of new solid iron sphere "=4/3pi(r')^3`

`4/3pi(r')^3=27xx4/3pir^3`

`(r')^3=27r^3`

r' = 3r

 

(ii) Surface area of 1 solid iron sphere of radius r = 4πr2

Surface area of iron sphere of radius r' = 4π (r')2

= 4 π (3r)2 = 36 πr2

`S/(S')=(4pir^2)/(36pir^2)=1/9=1:9`

 

  Is there an error in this question or solution?

APPEARS IN

 NCERT Solution for Mathematics Class 9 (2018 to Current)
Chapter 13: Surface Area and Volumes
Ex. 13.80 | Q: 9 | Page no. 236

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Solution Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S'. Find the (i) radius r' of the new sphere, (ii) ratio of S and S'. Concept: Volume of a Sphere.
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