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Question

If the error in measuring the radius of a sphere is 2%, then the error in the measurement of volume is:

A
8%
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B
6%
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C
2%
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D
9%.
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Solution

The correct option is B 6%
Percentage error in radius is given as 2% i.e. Δrr×100=2 %
Volume of sphere V=4π3r3
Percentage error in volume ΔVV×100=3×Δrr×100=3×2=6 %

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