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
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 =
Answered by Thiyagarajan K | 07 Feb, 2019, 02:59: PM
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