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The volumes of two bodies are measured to be `V_1 = (10.2+-0.02) cm^3 and V_2 = (6.4 +- 0.01)cm^3`. Calculate sum and difference in volumes with error limits.A. `(16.6pm0.03)(3.8pm0.03)`B. `(0.03pm3.8)(0.03pm16.6)`C. `(16.6pm3.8)(16.6pm0.03)`D. `(3.8pm0.05)(16.6pm0.03)` |
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Answer» Correct Answer - A Given, `V_(1)=(10.2pm0.02)cm^(3)` and `V_(2)=(6.4pm0.01)cm^(3)` `DeltaV=pm(DeltaV_(1)+DeltaV_(2))` `=pm(0.02+0.01)cm^(3)=pm0.03cm^(3)` `V_(1)+V_(2)=(10.2+6.4)cm^(3)=16.6cm^(3)` and `V_(1)-V_(2)=(10.2-6.4)cm^(3)=3.8cm^(3)` Hence, sum of volumes `=(16.6pm0.03)cm^(3)` and difference of volumes `=(3.8pm0.03)cm^(3)` |
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