An electron moves with a speed of $300 \ ms^{-1}$ with an uncertainty of $0.01\%$. Find the uncertainty in its position.

The uncertainty in position $(\Delta x)$ is calculated using Heisenberg's uncertainty principle: $\Delta x \cdot m \Delta v \geq \frac{h}{4 \pi}$.
Given:
Speed $(v) = 300 \ ms^{-1}$
Uncertainty in speed $(\Delta v) = 0.01\% \text{ of } 300 \ ms^{-1} = \frac{0.01}{100} \times 300 = 3 \times 10^{-2} \ ms^{-1}$.
Mass of electron $(m) = 9.1 \times 10^{-31} \ kg$.
Planck's constant $(h) = 6.626 \times 10^{-34} \ J \ s$.
Substituting the values:
$\Delta x = \frac{h}{4 \pi m \Delta v}$
$\Delta x = \frac{6.626 \times 10^{-34}}{4 \times 3.1416 \times 9.1 \times 10^{-31} \times 3 \times 10^{-2}}$
$\Delta x \approx 1.93 \times 10^{-3} \ m$.