If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be
- 8%
- 2%
- 4%
- 6%
Answer
(4) 6%
Solution
V = 43 r3
VV 100 = 3 rr 100
Percentage error in volume = 3 2\% = 6\%
Answer: (4)
If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be
(4) 6%
V = 43 r3
VV 100 = 3 rr 100
Percentage error in volume = 3 2\% = 6\%
Answer: (4)
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