Einstein’s mass-energy relation emerging out of his famous theory of relativity relates mass $(m)$ to energy $(E)$ as $E = mc^2$, where $c$ is speed of light in vacuum. At the nuclear level, the magnitudes of energy are very small. The energy at nuclear level is usually measured in $MeV$, where $1\,MeV = 1.6\times 10^{-13}\,J$ ; the masses are measured i unified atomicm mass unit (u) where, $1\,u = 1.67 \times 10^{-27}\, kg$
$(a)$ Show that the energy equivalent of $1\,u$ is $ 931.5\, MeV$.
$(b)$ A student writes the relation as $1\,u = 931.5\, MeV$. The teacher points out that the relation is dimensionally incorrect. Write the correct relation.
$(a)$ We know that,
$1 \mathrm{amu}= 1 u=1.67 \times 10^{-27} \mathrm{~kg}$
$\text { Applying } \mathrm{E}=m c^{2}$
$\text { Energy } =\mathrm{E}=\left(1.67 \times 10^{-27}\right)\left(3 \times 10^{8}\right)^{2} \mathrm{~J}$
$ 1.67 \times 9 \times 10^{-11} \mathrm{~J}$
$\mathrm{E} =\frac{1.67 \times 9 \times 10^{-11}}{1.6 \times 10^{-13}} \mathrm{MeV}$
$=939.4 \mathrm{MeV} \approx 931.5 \mathrm{MeV}$
$(b)$ The dimensionally correct relation is, $1 \mathrm{amu} \times c^{2}=1 u \times c^{2}=931.5 \mathrm{MeV}$
In the relation $P = \frac{\alpha }{\beta }{e^{ - \frac{{\alpha Z}}{{k\theta }}}}$ $P$ is pressure, $Z$ is the distance, $k$ is Boltzmann constant and $\theta$ is the temperature. The dimensional formula of $\beta$ will be
Frequency is the function of density $(\rho )$, length $(a)$ and surface tension $(T)$. Then its value is
If Surface tension $(S)$, Moment of Inertia $(I)$ and Planck’s constant $(h)$, were to be taken as the fundamental units, the dimensional formula for linear momentum would be