Question

The volumes of two bodies are measured to be $V_1=(10.2\pm 0.02)c{m}^3$ and $V_2=(6.4\pm 0.01)c{m}^3.$ Calculate difference in volumes with error limits.

• A. and difference of volumes $=(3.8\pm 0.03)c{m}^3$
• B. $(3.8\pm 0.03)c{m}^4$
• C. $(3.4\pm 0.04)c{m}^5$
• D. $(3.5\pm 0.05)c{m}^6$

Answer 1

The correct option is A and difference of volumes $=(3.8\pm 0.03)c{m}^3$

$V_1=(10.2\pm 0.02)c{m}^3.$
and $V_2(6.4\pm 0.01)c{m}^3.$
$\Delta V=\pm (\Delta V_1+\Delta V_2)$
$=\pm (0.02+0.01)c{m}^3=\pm 0.03c{m}^3$
$V_1+V_2=(10.2+6.4)c{m}^3=16.6c{m}^3$
and $V_1-V_2=(10.2-6.4)c{m}^3=3.8c{m}^3$
Hence, sum of volumes $=(16.6\pm 0.03)c{m}^3$
and difference of volumes $=(3.8\pm 0.03)c{m}^3$