When the uncertainty in the position of a moving particle is $0$,then the uncertainty in momentum $(p)$ is equal to:

The uncertainty in the position of a moving bullet of mass $10 \ g$ is $10^{-5} \ m$. Calculate the uncertainty in its velocity.

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 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 $9 \times 10^{-28} \ g$):

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.

The uncertainties in the velocities of two particles,$A$ and $B$ are $0.05 \ ms^{-1}$ and $0.02 \ ms^{-1}$ respectively. The mass of $B$ is five times that of the mass of $A$. What is the ratio of uncertainties $\frac{\Delta x_A}{\Delta x_B}$ in their positions?