A ball weighing $10 \mathrm{~g}$ is moving with a velocity of $90 \mathrm{~ms}^{-1}$. If the uncertainty in its velocity is $5 \%$, then the uncertainty in its position is $\times 10^{-33} \mathrm{~m}$. (Rounded off to the nearest integer)
[Given :$h=6.63 \times 10^{-34} \mathrm{Js}$ ]
(1)
$m=10 g=10^{-2} \mathrm{Kg}$
$v=90 \mathrm{~m} / \mathrm{sec}$
$\Delta \mathrm{v}=\mathrm{v} \times 5 \%=90 \times \frac{5}{100}=4.5 \mathrm{~m} / \mathrm{sec}$
$\mathrm{m} \cdot \Delta \mathrm{v} \cdot \Delta \mathrm{x} \geq \frac{\mathrm{h}}{4 \pi}$
$10^{-2} \times 4.5 \times \Delta \mathrm{x} \geq \frac{6.63 \times 3 \times 10^{-34}}{4 \times \frac{22}{7}}$
$\Delta \mathrm{x} \geq \frac{6.63 \times 7 \times 2 \times 10^{-34}}{9 \times 4 \times 22 \times 10^{-2}}$
$\Delta \mathrm{x} \geq 1.17 \times 10^{-33}=\mathrm{x} \times 10^{-33}$
$\mathrm{x}=1.17 \simeq 1$