$A$ golf ball of mass $m$ has a speed of $50 \ m \ s^{-1}$. If the speed can be measured within an accuracy of $2 \%$,the uncertainty in the position is

Given below are two statements $:$
Statement $(I):$ It is impossible to specify simultaneously with arbitrary precision,both the linear momentum and the position of a particle.
Statement $(II) :$ If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron,then the uncertainty in the measurement of velocity is $\geq \sqrt{\frac{h}{4\pi}} \times \frac{1}{m}$ which simplifies to $\geq \frac{1}{2m} \sqrt{\frac{h}{\pi}}$. In the light of the above statements,choose the correct answer from the options given below $:$

The moving bullet of a gun has $10 \ g$ mass and $10^{-5} \ m$ uncertainty in position. Find the uncertainty in velocity.

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

What is the uncertainty product $\Delta v \cdot \Delta x$ for a particle of mass $1 \ mg$? What does this imply?

The measurement of the electron position is associated with an uncertainty in momentum,which is equal to $1 \times 10^{-18} \ g \ cm \ s^{-1}$. The uncertainty in electron velocity is (mass of an electron is $9 \times 10^{-28} \ g$):