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CBSE Class 10 Answered

A spherical rubber ball of radius 14 cm is cut by a knife at a distance of “x” cm from its centre, into 2 different pieces. What should be the value of “x” such that the cumulative surface area of the newly formed pieces is 3/28 more than the rubber ball’s original surface area? 
Asked by rushabh1234 | 07 Feb, 2019, 10:44: AM
answered-by-expert Expert Answer
It is assumed that given rubber ball is solid ball.
 
After cutting at a distance x from centre, the cross section circular areas with diameter AB on
both the cut portions are excess surface area.
 
excess surface area = 2×π×(142-x2 ) = (3/28)×4π×142 
 
Above expression can be simplified as  x2 = 154 or  x = begin mathsize 12px style square root of 154 space equals space 12.41 space c m end style
Answered by Thiyagarajan K | 07 Feb, 2019, 02:59: PM
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