Statement I : Correct

This statement is describing the uncertainty principle.

It states that it is impossible to measure both the position and the momentum of an object.

Statement II : Correct.

If the uncertainty in the measurement of position ($\Delta x$) and the uncertainty in the measurement of momentum ($\Delta p$) are equal for an electron, then the uncertainty in the measurement of velocity ($\Delta v$) is greater than or equal to $\sqrt{\frac{h}{\pi}}\frac{1}{2m}$

This is derived from the Heisenberg uncertainty principle, $$\Delta x\,.\,\Delta {p_x} \ge {h \over {4\pi }}$$

Given, $$\Delta x = \Delta p$$

Substitute this in $$\Delta x\,.\,\Delta p \ge {h \over {4\pi }}$$ as

$$\Delta p\,.\,\Delta p \ge {h \over {4\pi }}$$

$$\Delta {p^2} \ge {h \over {4\pi }}$$

$$\Delta {p^2} \ge {h \over {4\pi }}$$

$$\left\{ \matrix{ The\,formula\,for\,momentum\,(P)\,is\,P = mV \hfill \cr For\,uncerta{\mathop{\rm int}} y\,\Delta P = m\Delta v \hfill \cr} \right.$$

So, $$m\Delta v \ge \sqrt {{h \over {4\pi }}} $$

$$\Delta v \ge \sqrt {{h \over {4\pi }}} \times {1 \over m}$$

$$\Delta v \ge {1 \over 2} > \sqrt {{h \over \pi }} > {1 \over m}$$

$$\Delta v \ge \sqrt {{h \over \pi }} {1 \over {2m}}$$

Both the statements are true (correct). So answer is (2) Both statement I and statement II are true.