Question

The surface area of a solid sphere is increased by 21% without changing its shape. Find the percentage increase in its - 1) radius , 2) volume

Answer 1

Let the original and increased surface areas be A1 and A2.

A1­/A2 = (r1/r2)2

1.21A1­/A1 = (r2/r1)2
 
1.1 = r2/r1

∴ r2 = 1.1r1
 
∴ Increase percentage = (r2 – r1) / (r1) x 100 % = (1.1r1 – r1)/ (r1) x 100 % = 10

Hence, the percentage increase in diameter is 10 %
 
Let the original and increased volume be V1 and V2.

V1/V2 = (r1­/r2)3
 
V2/V1 = (11­/10)3
 
         = 1.331
 
∴ V2 = 1.331 V1
 
Increase % = (V2 – V1) / (V1) x 100 % = (1.331 V1 – V1) / V1 x 100 %
 
                    = 33.1%
 
Hence, the percentage increase in volume is 33.1 %

Answer 2

Answer:Let the original radius be 'r' and after changes the radius is 'R'

Now,

Surface Area of Sphere = 4πr²

4πr² = 100 (Original area)

4πR² = 121 (Area after increase of 21 %)

r²/R² = 100/121

r/R = 10/11

Now, increased value of radius in percentage.

11 - 10 = 1

= (1/10) × 100

= 10 %

So, radius will increase by 10 %

Now, Volume

Volume of Sphere = 4/3πr³

Ratio of volumes = 4/3πr³ : 4/3πR³

r³ : R³

(10)³ : (11)³

1000 : 1331

Increase in volume 

= 1331 - 1000

 = 331 

Now, increased value of volume in prcentage

(331/1000) ×100

= 33.1 %

So, the volume will be increased by 33.1 %

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