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An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is 1/4 of the radius of the original ball, how many such balls are made? Compare the surface area, of all the smaller balls combined together with that of the original ball.

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In this question we have to find the no. Of balls and compare the ratio of combined surface area of smaller balls with that of the original ball given tha an iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is 1/4 of the radius of the original ball.
Very important question of mensuration including very important concept of it.
need solution!
Rd sharma class 10 maths exercise-16.1 mensuration.

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  1. Let the radius of the big ball be x cm

    The, the radius of the small ball = x/4 cm

    And, let the number of balls = n

    Then according to the question, we have

    Volume of n small balls = Volume of the big ball

    n x 4/3 π(x/4)3 = 4/3 πx3

    n x (x3/ 64) = x3

    n = 64

    Therefore, the number of small balls = 64

    Next,

    Surface area of all small balls/ surface area of big ball = 64 x 4π(x/4)2/ 4π(x)2

    = 64/16 = 4/1

    Thus, the ratio of the surface area of the small balls to that of the original ball is 4:1

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