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.

Options

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)`

Answer

A

Explanation

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)`