. If the % error in calculating the radius of a sphere is 2%, what will be the percentage error in calculating the volume? *
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As error in radius is 2 percent
u knw that volume of sphere =4/3πr3
in errors whether its on numerator side or denomitor it has to add
powers are simply multiplied
as it r3 it will b 3×R ie 3×2= 6 percent
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