Two nuclei of mass number $3$ combine with another nucleus of mass number $4$ to yield a nucleus of mass number $10$. If the binding energy per nucleon for the mass numbers $3$, $4$, and $10$ are $5.6 \text{ MeV}$, $7.4 \text{ MeV}$, and $6.1 \text{ MeV}$, respectively, then in the process, $\Delta Mc^2 = . . . . . . \text{ MeV}$.

Given below are two statements:
Statement $I$: For all elements, greater the mass of the nucleus, greater is the binding energy per nucleon.
Statement $II$: For all elements, nuclei with less binding energy per nucleon transform to nuclei with greater binding energy per nucleon.
In the light of the above statements, choose the correct answer from the options given below:

The dependence of binding energy per nucleon,$B_N$ on the mass number,$A$,is represented by

One microgram of matter converted into energy will give:

Assuming the experimental mass of $^{12}_{6}C$ as $12 \text{ u}$, the mass defect of $^{12}_{6}C$ atom is . . . . . . $\text{u}$. (Mass of proton $= 1.00727 \text{ u}$, mass of neutron $= 1.00866 \text{ u}$).

$A$ given coin has a mass of $3.0\; g$. Calculate the nuclear energy that would be required to separate all the neutrons and protons from each other. For simplicity, assume that the coin is entirely made of $_{29}^{63} Cu$ atoms (of mass $62.92960\; u$).