(1)

$T\propto S^{\alpha }{A}^{\beta }{\rho }^{\gamma }{R}^{\delta }$

$\left[T\right]=$${\left[M{T}^{-2}\right]}^{\alpha }$${\left[L^2\right]}^{\beta }$${\left[M{L}^{-3}\right]}^{{}^{\gamma }}$${\left[L\right]}^{\delta }$

$\alpha +\gamma =0,$$2\beta -3\gamma +\delta =0,$$-2\alpha =1$

$\therefore$    $\alpha =-\frac{1}{2}, \gamma =\frac{1}{2}$

Now, $2\beta -\frac{3}{2}+\delta =0$

         $\Rightarrow 2\beta +\delta =\frac{3}{2}$

By checking with options 

       $\beta =-1, \delta =\frac{7}{2}\Rightarrow \text{Option (a) holds}$

So, $v\propto \sqrt{\Delta P}\propto \sqrt{\frac{S}{r}}\propto r^{-\frac{1}{2}}$

So, $a=-\frac{1}{2}$

(2)

$F=\alpha t^2+\beta t$

Dimension of $\alpha =\frac{\left[ML{T}^{-2}\right]}{\left[T^2\right]}=\left[ML{T}^{-4}\right]$

Dimension of $\beta =\frac{\left[ML{T}^{-2}\right]}{\left[T\right]}=\left[ML{T}^{-3}\right]$

Option a): $\frac{\beta t}{\alpha }=\frac{\left[ML{T}^{-3}\right]\left[T\right]}{\left[ML{T}^{-4}\right]}=\left[M^0{L}^0{T}^2\right]$

Option b): $\frac{\alpha t}{\beta }=\frac{\left[ML{T}^{-4}\right]\left[T\right]}{\left[ML{T}^{-3}\right]}=\left[M^0{L}^0{T}^0\right]$

Option c): $\alpha \beta t=\left[ML{T}^{-4}\right]\left[ML{T}^{-3}\right]\left[T\right]=\left[M^2{L}^2{T}^{-6}\right]$

Option d): $\frac{\alpha \beta }{t}=\frac{\left[ML{T}^{-4}\right]\left[ML{T}^{-3}\right]}{\left[T\right]}=\left[M^2{L}^2{T}^{-8}\right]$

Hence, option (b) is correct.

(3)

Let for energy, $\left[E\right]=F^{\alpha }{A}^{\beta }{T}^{\gamma }$

or $\left[M^1{L}^2{T}^{-2}\right]$$=$${\left[ML{T}^{-2}\right]}^{\alpha }$${\left[L{T}^{-2}\right]}^{\beta }$${\left[T\right]}^{\gamma }$

$M^1{L}^2{T}^{-2}=M^{\alpha }{L}^{\alpha +\beta }{T}^{-2\alpha -2\beta +\gamma }$

Comparing from both sides, $\alpha =1$

$\alpha +\beta =2\Rightarrow \beta =1$

$-2\alpha -2\beta +\gamma =-2\Rightarrow \gamma =2 \therefore \text{Energy}=\left[F\right]\left[A\right]\left[T^2\right]$

(4)

Dimensions of $\frac{e^2}{4\pi {\epsilon }_0}=[F\times d^2]=\left[M{L}^3{T}^{-2}\right]$

Dimensions of $G=\left[M^{-1}{L}^3{T}^{-2}\right],$ Dimensions of $c=\left[L{T}^{-1}\right]$

$l\propto {\left(\frac{{\displaystyle e^2}}{{\displaystyle 4\pi {\epsilon }_0}}\right)}^p{G}^q{c}^r \therefore \left[L^1\right]=$${\left[M{L}^3{T}^{-2}\right]}^{{}^p}$${\left[M^{-1}{L}^3{T}^{-2}\right]}^q$${\left[L{T}^{-1}\right]}^r$

On comparing both sides and solving, we get: $p=\frac{1}{2}, q=\frac{1}{2}, \text{and} r=-2 \therefore l\propto \frac{1}{c^2}{\left[\frac{{\displaystyle G{e}^2}}{{\displaystyle 4\pi {\epsilon }_0}}\right]}^{1/2}$

(1)

According to question, $l\propto h^p{c}^q{G}^r$

$l=k$ $h^p{c}^q{G}^r$                                                     ...(i)

Writing dimensions of physical quantities on both sides,

$\left[{\text{M}}^0{\text{LT}}^0\right]=$${\left[{\text{ML}}^2{\text{T}}^{-1}\right]}^p$${\left[{\text{LT}}^{-1}\right]}^{{}^q}$${\left[{\text{M}}^{-1}{\text{L}}^3{\text{T}}^{-2}\right]}^r$

Applying the principle of homogeneity of dimensions, we get 

$p-r=0$     ...(ii),            $2p+q+3r=1$         ...(iii),            $-p-q-2r=0$        ...(iv)

Solving eqns. (ii), (iii) and (iv), we get $p=r=\frac{1}{2}, q=-\frac{3}{2}$

From eqn. (i), we get $l=K\frac{\sqrt{hG}}{c^{3/2}}$

(3)

$\left[v_c\right]=\left[{\eta }^x{\rho }^y{r}^z\right]$                                  (given)                               ...(i)

Writing the dimensions of various quantities in eqn. (i), we get $\left[M^{0}L{T}^{-1}\right]=$${\left[M{L}^{-1}{T}^{-1}\right]}^x$${\left[M{L}^{-3}{T}^0\right]}^y$${\left[M^{0}L{T}^0\right]}^z$

                                                                                                                     $=\left[M^{x+y}{L}^{-x-3y+z}{T}^{-x}\right]$

Applying the principle of homogeneity of dimensions, we get $x+y=0;$ $-x-3y+z=1;$ $-x=-1$

On solving, we get $x=1, y=-1, z=-1$

(4)

Let mass $m\propto F^a{V}^b{T}^c$ or $m=k{F}^a{V}^b{T}^c$                   ...(i)

where $k$ is a dimensionless constant and $a,b$ and $c$ are the exponents.

Writing dimensions on both sides, we get

$\left[M{L}^0{T}^0\right]=$${\left[ML{T}^{-2}\right]}^a$${\left[L{T}^{-1}\right]}^b$${\left[T\right]}^c$

$\left[M{L}^0{T}^0\right]=\left[M^a{L}^{a+b}{T}^{-2a-b+c}\right]$

Applying the principle of homogeneity of dimensions, we get

$a=1$    ...(ii),     $a+b=0$    ...(iii),     $-2a-b+c=0$           ...(iv)

Solving eqns. (ii), (iii) and (iv), we get $a=1,b=-1,c=1$

From eqn. (i), $\left[m\right]=\left[F{V}^{-1}T\right]$