Correct answer is (D)

$\begin{array}{rl}& v={\lambda }^{\mathrm{a}}{\mathrm{g}}^{\mathrm{b}}{\rho }^{\mathrm{c}}\\ \\ & \text{using dimension formula}\\ \\ & \Rightarrow \left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-1}\right]={\left[{\mathrm{L}}^1\right]}^{\mathrm{a}}{\left[{\mathrm{L}}^1{\mathrm{T}}^{-2}\right]}^{\mathrm{b}}{\left[{\mathrm{M}}^1{\mathrm{L}}^{-3}\right]}^{\mathrm{c}}\\ \\ & \Rightarrow \left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^{-1}\right]=\left[{\mathrm{M}}^{\mathrm{c}}{\mathrm{L}}^{\mathrm{a}+\mathrm{b}-\mathrm{c}}{\mathrm{T}}^{-2\mathrm{b}}\right]\\ \\ & \therefore \mathrm{c}=0,\mathrm{a}+\mathrm{b}-3\mathrm{c}=1,-2\mathrm{b}=-1\Rightarrow \mathrm{b}=\frac{1}{2}\\ \\ & \text{ Now }\mathrm{a}+\mathrm{b}-3\mathrm{c}=1\\ \\ & \Rightarrow \mathrm{a}+\frac{1}{2}-0=1\\ \\ & \Rightarrow \mathrm{a}=\frac{1}{2}\\ \\ & \therefore \mathrm{a}=\frac{1}{2},\mathrm{b}=\frac{1}{2},\mathrm{c}=0\end{array}$

Correct answer is (B)

Given that, $[X+\frac{a}{Y^2}][Y-b]=\mathrm{R}T$

$\therefore$ $X$ and $\frac{a}{Y^2}$ have the same dimensions and $Y$ and $b$ have the same dimensions, let's analyze the dimensions of $\frac{a}{b}$.

Since $X$ represents pressure, it has dimensions of $[M{L}^{-1}{T}^{-2}]$.

Since $X$ and $\frac{a}{Y^2}$ have the same dimensions, we have:

$\left[\frac{a}{Y^2}\right]=[M{L}^{-1}{T}^{-2}]$

Then, the dimensions of $a$ are:

$[a]=[M{L}^{-1}{T}^{-2}][Y^2]=[M{L}^5{T}^{-2}]$

Now, since $Y$ and $b$ have the same dimensions and $Y$ represents volume, we have:

$[b]=[L^3]$

Now, let's find the dimensions of the ratio $\frac{a}{b}$:

$\frac{[a]}{[b]}=\frac{[M{L}^5{T}^{-2}]}{[L^3]}=[M{L}^2{T}^{-2}]$

Indeed, the dimensions of $\frac{a}{b}$ are $[M{L}^2{T}^{-2}]$, which corresponds to the dimensions of energy.

Correct answer is (D)

We know that force (F) has dimensions of ${\mathrm{MLT}}^{-2}$, velocity (V) has dimensions of ${\mathrm{LT}}^{-1}$, and time (T) has dimensions of $\mathrm{T}$. To express density ($\rho$), which has dimensions of ${\mathrm{ML}}^{-3}$, in terms of F, V, and T, we need to find the exponents of F, V, and T.

Let the dimensions of density be expressed as:

${[\mathrm{M}]}^a{[\mathrm{L}]}^b{[\mathrm{T}]}^c$

Substituting the dimensions of F, V, and T:

${[\mathrm{F}]}^a{[\mathrm{V}]}^b{[\mathrm{T}]}^c=({\mathrm{MLT}}^{-2}{)}^a({\mathrm{LT}}^{-1}{)}^b(\mathrm{T}{)}^c$

Since density has dimensions of ${\mathrm{ML}}^{-3}$, we can set up the following equations:

$a=1$ (for the mass term M)

$a+b=-3$ (for the length term L)

$-2a-b+c=0$ (for the time term T)

Solving these equations, we get:

$a=1$

$b=-4$

$c=-2$

Thus, the dimensional formula of density in terms of F, V, and T is:

${\mathrm{FV}}^{-4}{\mathrm{T}}^{-2}$

Correct answer is (C)

$\begin{array}{rl}& [\mathrm{M}]=[\mathrm{G}{]}^{\mathrm{x}}[\mathrm{h}{]}^{\mathrm{y}}[\mathrm{c}{]}^{\mathrm{z}}\\ \\ & [\mathrm{M}]={\left[{\mathrm{M}}^{-1}{\mathrm{L}}^3{\mathrm{T}}^{-2}\right]}^x{\left[{\mathrm{ML}}^2{\mathrm{T}}^{-1}\right]}^y{\left[{\mathrm{LT}}^{-1}\right]}^z\\ \\ & \left[{\mathrm{M}}^1{\mathrm{L}}^0{\mathrm{T}}^0\right]=\left[{\mathrm{M}}^{-x+y}\right]\left[{\mathrm{L}}^{3x+2y+z}\right]\left[{\mathrm{T}}^{-2x-y-z}\right]\\ \\ & \mathrm{y}-\mathrm{x}=1......(1)\\ \\ & 3\mathrm{x}+2\mathrm{y}+\mathrm{z}=0.......(2)\\ \\ & -2\mathrm{x}-\mathrm{y}-\mathrm{z}=0........(3)\end{array}$

On solving, $\mathrm{x}=-\frac{1}{2},\mathrm{y}=\frac{1}{2},\mathrm{z}=\frac{1}{2}$

So $\mathrm{m}=\sqrt{\frac{\mathrm{hc}}{\mathrm{G}}}$