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If the error in measurement of radius of sphere is 2%what will be percentage error in volume?

Answer» Volume of sphere =4/3*22/7*r^3=3*2%=6%
% error in radius of sphere =2%. volume of sphere =4/3×πr^3∆r/r=3×4/3×∆r/r∆r/r×100=4[∆r/r×100]∆r/r×100=4×2∆r/r×100=8%


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