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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
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Correct option is D
V=43πR3;lnV=ln43π+lnR3Differentiating,dVV=3dRRError in the determination of the volume= 3 × 2% = 6%
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A physical quantity A is related to four observable a, b, c and d as . The percentage errors of measurement in a, b, c and d are 1 %, 3 %, 2 % and 2 % respectively. What is the percentage error in the quantity A ?
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