Question

Some physical quantities are given in Column I and some possible SI units in which these quantities may be expressed are given in Column II. Match the physical quantities in Column I with the units in Column II.
$\begin{array}{cccc}& \text{Column I}& & \text{Column II}\\ i& G{M}_e{M}_S& \text{a.}& \left(kg\right)\left(metre{)}^2\right(second{)}^{-2}\\ ii& 3RT& \text{b.}& \left(farad\right)\left(volt{)}^2\right(kg{)}^{-1}\\ iii& F^2/q^2{B}^2& \text{c.}& \left(volt\right)\left(coulomb\right)\left(metre\right)\end{array}$

Where G is universal gravitational constant; $M_e$= mass of the earth; $M_s$= mass of sun; $R$ = universal gas constant; $T$ = absolute temperature; $F$ = force; $q$ = charge; $B$ = magnetic field $\left(M{A}^{-1}{T}^{-2}\right)$. $(1\text{volt}=\frac{1\text{joule}}{1\text{coulomb}};1\text{farad}=\frac{1\text{coulomb}}{1\text{volt}})$

• A. $(i–c;ii–a;iii–b)$
• B. $(i–c;ii–b;iii–a)$
• C. $(i–a;ii–c;iii–b)$
• D. $(i–a;ii–b;iii–c)$

Answer 1

The correct option is A $(i–c;ii–a;iii–b)$

$\left(farad\right)\left(volt{)}^2\right(kg{)}^{-1}=\frac{C-V^2}{V-kg}$
$\left(farad\right)\left(volt{)}^2\right(kg{)}^{-1}=\frac{C-J}{C-kg}$
$\left(farad\right)\left(volt{)}^2\right(kg{)}^{-1}=m^2/s^2.....\left(1\right)$

$\left(volt\right)\left(coulomb\right)\left(metre\right)=\frac{J-C-m}{C}$
$\left(volt\right)\left(coulomb\right)\left(metre\right)=J-m$
$\left(volt\right)\left(coulomb\right)\left(metre\right)=\frac{kg-m^3}{s^2}..\left(2\right)$

For $\left(i\right)$:
$G{M}_e{M}_S=F{r}^2=kg-m^3/s^2$
from $\left(2\right)$,
$i-c$

For $\left(ii\right)$:
$3RT=\frac{J-K}{mol-K}=kg-m^2/s^2$
$ii-a$

For $\left(iii\right)$: using dimensions
$F^2/\left(q^2{B}^2\right)=\frac{L^2{M}^2}{T^4}.\frac{1}{A^2{T}^2}.\frac{A^2{T}^4}{M^2}$
$F^2/\left(q^2{B}^2\right)=\frac{L^2}{T^2}=m^2/s^2$
from $\left(1\right)$,
$iii-b$