The effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles. Justify the statement with the help of a suitable example.

Statement justification with example

According to Heisenberg uncertainty principle:

$\Delta x\Delta p\ge \frac{h}{4\pi }$

And, $\Delta p=m\Delta v$.

Thus, $\Delta x,m\Delta v\ge \frac{h}{4\pi }$

Macroscopic particles are visible to naked eyes while microscopic particles includes atoms and molecules that are not visible to naked eyes

For Example, If the uncertainty principle is applied to a macroscopic object of mass one milligram $\left({10}^{-6}kg\right)$, then

$\Delta v\Delta x=\frac{h}{4\pi \times m}$

$\Delta v\Delta x=\frac{6.626\times {10}^{-34}kg{m}^2{s}^{-}}{4\times 3.14\times {10}^{-6}kg}$

$\Delta v\Delta x=0.52\times {10}^{-28}{m}^2{s}^{-1}$

While, this same value for product of uncertainties (velocity and momentum ) for a microscopic particle ( i.e. electron with mass of $(9.11\times 10-31Kg)$ is

$\Delta v\Delta x=\frac{6.626\times {10}^{-34}kg{m}^2{s}^{-}}{4\times 3.14\times 9.11\times {10}^{-31}kg}$

$\Delta v\Delta x=0.57\times {10}^{-4}{m}^2{s}^{-1}$

The value of $\Delta v\Delta x$ (product of uncertainity) obtained for macroscopic particle is extremely small in comparison to microscopic particle. and thus it is insignificant for the uncertainty principle to apply to macroscopic particle

Conclusion:

The effect of uncertainty principle is significant only for motion of microscopic particles and is negligible for the macroscopic particles.