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

Correct option is 2(16.6±0.03)cm3 and (3.8±0.03)cm3

Questions

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

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(16.6±0.02)cm3 and (3.8±0.02)cm3

(16.6±0.03)cm3 and (3.8±0.03)cm3

(19±0.03)cm3 and (3.8±0.03)cm3

(16.6±0.03)cm3 and (8.3±0.03)cm3

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detailed solution

Correct option is B

V1=(10.2±0.02)cm3V2=(6.4±0.01)cm3ΔV=±ΔV1+ΔV2=±(0.02+0.01)cm3=±0.03 cm3V1+V2=(10.2+6.4)cm3=16.6 cm3V1-V2=(10.2-6.4)cm3=3.8 cm3Hence, sum of volumes =(16.6±0.03)cm3and difference of volumes =(3.8±0.03)cm3

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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. Calculate sum and difference in volumes with error limits.

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