Question

The volume of two bodies are measured to be $V_1=(10.20\pm 0.02){\text{cm}}^3$ and $V_2=(6.40\pm 0.01){\text{cm}}^3$.
The difference in volume with appropriate significant figures is

• A. $(3.80\pm 0.04){\text{cm}}^3$
• B. $(3.80\pm 0.01){\text{cm}}^3$
• C. $(3.80\pm 0.03){\text{cm}}^3$
• D. $(3.80\pm 0.02){\text{cm}}^3$

Answer 1

The correct option is C $(3.80\pm 0.03){\text{cm}}^3$

Given,
$V_1=(10.20\pm 0.02){\text{cm}}^3=V_{1_{mean}}\pm \Delta V_{1_{mean}}$
$V_2=(6.40\pm 0.01){\text{cm}}^3=V_{2_{mean}}\pm \Delta V_{2_{mean}}$
So, $V_1-V_2=(V_{1_{mean}}-V_{2_{mean}})\pm (\Delta V_{1_{mean}}+\Delta V_{2_{mean}})$
[In addition or subtraction, the mean absolute error is added]
$\Rightarrow V_1-V_2=\left[\right(10.20-6.40)\pm (0.02+0.01\left)\right]{\text{cm}}^3$
$\Rightarrow V_1-V_2=(3.80\pm 0.03){\text{cm}}^3$