The volumes of two bodies are measured to be V1=(10.2±0.02)cm3 and V2=(6.4±0.01)cm3. The sum and difference in volumes with error limits is

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(16.6±0.03)cm3 and (3.8±0.03)cm3

(16.6±0.01)cm3 and (3.8±0.01)cm3

(16.2±0.03)cm3 and (3.6±0.03)cm3

(16.2±0.01)cm3 and (3.6±0.01)cm3

answer is A.

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Detailed Solution

Given, V1=(10.2±0.02)cm3and V2=(6.4±0.01)cm3ΔV=±(ΔV1+ΔV2)=±(0.02+0.01)cm3=±0.03 cm3V1+V2=(10.2+6.4)cm3=16.6 cm3and V1−V2=(10.2−6.4)cm3=3.8 cm3Hence, sum of volume =(16.6±0.03)cm3and difference of volume =(3.8±0.03)cm3In product If Z = AB, then maximum fractional error is ΔZZ=±(ΔAA+ΔBB)Therefore, maximum fractional error in product of two (or more) quantities is equal to sum of fractional errors in the individual quantities. In division If Z = A /B, then maximum fractional error is ΔZZ=±(ΔAA+ΔBB)Therefore, maximum fractional error in product of two ( or more) quantities is equal to sum of fractional errors in the individual quantities.

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