If the error in the measurement of radius of a sphere in `2%` then the error in the determination of volume of the spahere will be
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The correct Answer is:
D
Here `V_(DE)=V_(D)-V_(E)=20-(-20)=40V` Work done`=qV_(DE)` `=-1xx10^(-6)xx40` `=-4xx10^(-5) J`
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