Correct answer is a) (0.25 ± 0.08) m

Here,

$\mathrm{A}=2.5 \mathrm{~ms}^{-1} \pm 0.5 \mathrm{~ms}^{-1}, \mathrm{~B}=0.10 \mathrm{~s} \pm 0.01 \mathrm{~s}$

$\mathrm{AB}=\left(2.5 \mathrm{~ms}^{-1}\right)(0.10 \mathrm{~s})=0.25 \mathrm{~m}$ $\frac{\Delta \mathrm{AB}}{\mathrm{AB}}=\left(\frac{\Delta \mathrm{A}}{\mathrm{A}}+\frac{\Delta \mathrm{B}}{\mathrm{B}}\right)=\left(\frac{0.5}{2.5}+\frac{0.01}{0.10}\right)=0.3$

$\Delta \mathrm{AB}=0.3 \times 0.25 \mathrm{~m}=0.075 \mathrm{~m}=0.08 \mathrm{~m}$

The value of $\mathrm{AB}(0.25 \pm 0.8) \mathrm{m}$