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
Application Videos
Concept Videos
CBSE 10 - Maths
Asked by dagardisha39 | 28 Apr, 2024, 12:58: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by raopaidi | 25 Apr, 2024, 04:06: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by aditikaldate7 | 21 Apr, 2024, 03:16: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by pathaksuman622 | 21 Apr, 2024, 11:56: AM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by yalagondanikhil | 16 Apr, 2024, 12:25: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by rrajansinghakb199 | 08 Apr, 2024, 05:12: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by nagalaxmidurgarao937 | 08 Apr, 2024, 01:42: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by mraniruddha03 | 02 Apr, 2024, 06:44: PM
ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by muttenenimalleswarrao | 29 Mar, 2024, 08:32: PM
ANSWERED BY EXPERT