An Important Question of class 10 Based on Mensuration Chapter of M.L Aggarwal for ICSE BOARD.

The radius of a spherical balloon increases as air is jumped into it.

Find the ratio of the surface areas of the balloon in two cases.

This is the Question Number 08, Exercise 17.3 of M.L Aggarwal.

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

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Given radius of the spherical balloon, r = 7 cm

Radius of the spherical balloon after air is pumped, R = 14 cm

Surface area of the sphere = 4r^{2}Ratio of surface areas of the balloons = 4r^{2}/4R^{2}= r^{2}/R^{2}= 7^{2}/14^{2}= 1/4Hence the ratio of the surface areas of the spheres is 1:4.