find the percentage of reduction in surface area
Q.6. Three solid metal spheres of radii 3 cm, 4 cm and 5 cm respectively are melted together. The obtained metal is recast into a single solid sphere. The percentage reduction in the surface area of the solid sphere is
A. 24 %
B. 26 %
C. 28 %
D. 30 %

Dear Student,

Please find below the solution to the asked query:

We know :  Surface area of sphere  =  4 π r2

So,

Surface area of sphere with radius 3 cm = 4 π × ( 3 )2 =  36 π cm2   ,

Surface area of sphere with radius 4 cm = 4 π × ( 4 )2 =  64 π cm2  

And

Surface area of sphere with radius 5 cm = 4 π × ( 5 )2 =  100 π cm2  

Then,

Total surface area of all three spheres = 36 π + 64 π+ 100 π  =  200 π cm2 

We know volume of sphere = 43× πr3  

As given : Three solid metal spheres of radii 3 cm, 4 cm and 5 cm respectively are melted together. The obtained metal is recast into a single solid sphere. So

Volume of all three sphere =  Volume of new sphere

We assume radius of new sphere =  r  , So

43× π × ( 3 )3   + 43× π × ( 4 )3   + 43× π × ( 5 )3   = 43× πr3  ,

43× π [ ( 3 )3   +  ( 4 )3   + ( 5 )3  ] = 43× πr3  ,

[ 27 +  64  + 125 ] = r3  ,

[216 ] = r3  ,

r = 6  cm

Then,

Surface area of new sphere = 4 π × ( 6 )2 =  144 π cm2  

Change in surface area =  200 π -  144 π =  56π cm2

Thus ,

Percentage reduction in the surface area of the solid sphere = 56 π200 π × 100 = 56200 × 100= 562 = 28 %
Therefore,

Option ( C )                                                          ( Ans )

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C.28%
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howww ???
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