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CBSE Class 9 Mathematics Question 16 of 44

Surface Areas and Volumes — Question 8

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Question

Question 4

The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Answer

Given,

Radius of the balloon before pumping air (r1) = 7 cm

Radius of the balloon after pumping air (r2) = 14 cm

The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases. NCERT Class 9 Mathematics CBSE Solutions.

Initial surface area = 4πr12

Surface area after pumping air into ballon = 4πr22

Ratio = Initial surface areaSurface area after pumping air into ballon\dfrac{\text{Initial surface area}}{\text{Surface area after pumping air into ballon}}

=4πr124πr22=r12r22=(r1r2)2=(714)2=(12)2=14=1:4.= \dfrac{4πr_1^2}{4πr_2^2} \\[1em] = \dfrac{r_1^2}{r_2^2} \\[1em] = \Big(\dfrac{r_1}{r_2}\Big)^2 \\[1em] = \Big(\dfrac{7}{14}\Big)^2 \\[1em] = \Big(\dfrac{1}{2}\Big)^2 \\[1em] = \dfrac{1}{4} \\[1em] = 1 : 4.

Hence, the ratio of the surface area of the balloons = 1 : 4.