Question

Match the following two columns.

Column I

Column II

Answer 1

For (A)
We know that $F_g=G{M}_e{M}_5/r^2$
Dimension of $F_g$ is $\left[ML{T}^{-2}\right]$
Dimension of $r^2$ is$\left[L^2\right]$
Therefore, putting in eq. we get
$\left[ML{T}^{-2}\right]=dim\left(G{M}_e{M}_5\right)/\left[L^2\right]$
$dim\left(G{M}_e{M}_5\right)=\left[M{L}^3{T}^{-2}\right]$
therefore, $A\underset{}{\overset{}{\to }}2$

Now we know that, for a gas average kinetic energy is given by
$KE=\frac{3nRT}{2}$
$KE=\frac{3wRT}{2M}$
Therefore, $dim\left(KE\right)=dim\left(\frac{3wRT}{2M}\right)$
$dim\left(\frac{3RT}{2M}\right)=dim\left(\frac{KE}{w}\right)$
$dim\left(\frac{3RT}{2M}\right)=\frac{\left[M{L}^2{T}^{-2}\right]}{\left[M\right]}$
$dim\left(\frac{3RT}{2M}\right)=\left[L^2{T}^{-2}\right]$
Therefore, $B\underset{}{\overset{}{\to }}3$

We know that magnetic force is given by,
$F=qvB\sin \theta$
$F=qvB$ when $\sin \theta =1$
$F^2=(qvB{)}^2$
$v^2=\frac{F^2}{(qB{)}^2}$
$dim\left(v^2\right)=dim\left(\frac{F^2}{(qB{)}^2}\right)$
$dim\left(\frac{F^2}{(qB{)}^2}\right)=\left[L^2{T}^{-2}\right]$
therefore, $C\stackrel{}{\to }3$

We know that gravitational potentialon earth's surface is given by
$V_g=\frac{G{M}_e}{R_e}$
$\frac{U_g}{m}=\frac{G{M}_e}{R_e}$
$dim\left(\frac{G{M}_e}{R_e}\right)=dim\left(\frac{U_g}{m}\right)$
$dim\left(\frac{G{M}_e}{R_e}\right)=\left[L^2{T}^{-2}\right]$
therefore, $D\stackrel{}{\to }3$