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

Question

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

Infinity Learn · Accepted Answer

Given, $$V1=(10.2$$±$$0.02)cm3and$$ $$V2=(6.4$$±$$0.01)cm3$$Δ$$V=$$±$$($$Δ$$V1+$$Δ$$V2)=$$±(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$$ product If Z $$=$$ AB, then maximum fractional error is Δ$$ZZ=$$±$$($$Δ$$AA+$$Δ$$BB)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 Δ$$ZZ=$$±$$($$Δ$$AA+$$Δ$$BB)Therefore$$, maximum fractional error in product of two $$($$ or $$more)$$ quantities is equal to sum of fractional errors in the individual quantities.

The volumes of two bodies are measured to be $V_{1} = (10.2 \pm 0.02) c m^{3} a n d V_{2} = (6.4 \pm 0.01) c m^{3} .$ The sum and difference in volumes with error limits is

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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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The volumes of two bodies are measured to be $V_{1} = (10.2 \pm 0.02) c m^{3} a n d V_{2} = (6.4 \pm 0.01) c m^{3} .$ The sum and difference in volumes with error limits is