Mercury's natural state is where the atoms are close to each other but are still free to pass by each other. In which state(s) could mercury naturally exist?
Liquid is the answer
Answer:
c. 2 MeV.
Explanation:
The computation of the binding energy is shown below
![= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV](https://tex.z-dn.net/?f=%3D%20%5BZm_p%20%2B%20%28A%20-%20Z%29m_n%20-%20N%5Dc%5E2%5C%5C%5C%5C%3D%5B%281%29%20%281.007825u%29%20%2B%20%282%20-%201%20%29%20%28%201.008665%20u%29%20-%202.014102%20u%5Dc%5E2%5C%5C%5C%5C%3D%20%280.002388u%29c%5E2%5C%5C%5C%5C%3D%20%28.002388%29%20%28931.5%20MeV%29%5C%5C%5C%5C%3D2.22%20MeV)
= 2 MeV
As 1 MeV = (1 u) c^2
hence, the binding energy is 2 MeV
Therefore the correct option is c.
We simply applied the above formula so that the correct binding energy could come
And, the same is to be considered
Shear Stress= Force along the tangent/ Area
= 2700/π*r^2
=2700/3.14*16
=54