The frequency $(v)$ of an oscillating liquid drop may depend upon radius $(r)$ of the drop, density $(\rho)$ of liquid and the surface tension $(s)$ of the liquid as : $v=r^{ a } \rho^{ b } s ^{ c }$. The values of $a , b$ and $c$ respectively are

If we use permittivity $ \varepsilon $, resistance $R$, gravitational constant $G$ and voltage $V$ as fundamental physical quantities, then

The force of interaction between two atoms is given by $F\, = \,\alpha \beta \,\exp \,\left( { – \frac{{{x^2}}}{{\alpha kt}}} \right);$ where $x$ is the distance, $k$ is the Boltzmann constant and $T$ is temperature and $\alpha $ and $\beta $ are two constants. The dimension of $\beta $ is

If ${E}, {L}, {m}$ and ${G}$ denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of ${P}$ in the formula ${P}={EL}^{2} {m}^{-5} {G}^{-2}$ are

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.