We know that dimension of energy $=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]$

Let $M_1, L_1, T_1$ and $M_2, L_2, T_2$ are units of mass, length and time in given two systems.

$$ \begin{array}{ll} & \therefore M_1=1 \mathrm{~kg}, L_1=1 \mathrm{~m}, T_1=1 \mathrm{~s} \\\\ & M_2=\alpha \mathrm{kg}, L_2=\beta \mathrm{m}, \mathrm{T}_2=\gamma \mathrm{s} \end{array} $$

The magnitude of a physical quantity remains the same, whatever be the system of units of its measurement i.e.,

$$ \begin{aligned} n_1 u_1 & =n_2 u_2 \\ n_2 & =n_1 \frac{u_1}{u_2}=n_1 \frac{\left[\mathrm{M}_1 L_1^2 \mathrm{~T}_1^{-2}\right]}{\left[\mathrm{M}_2 L_2^2 T_2^{-2}\right]}=5\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right] \times\left[\frac{L_1}{\mathrm{~L}_2}\right]^2 \times\left[\frac{\mathrm{T}_1}{T_2}\right]^{-2} \\\\ & =5\left[\frac{1}{\alpha} \mathrm{kg}\right] \times\left[\frac{1}{\beta} \mathrm{m}\right]^2 \times\left[\frac{1}{\gamma} \mathrm{s}\right]^{-2} \\\\ & =5 \times \frac{1}{\alpha} \times \frac{1}{\beta^2} \times \frac{1}{\gamma^{-2}} \\ n_2 & =\frac{5 \gamma^2}{\alpha \beta^2} \end{aligned} $$

Thus, new unit of energy will be $\frac{\gamma^2}{\alpha \beta^2}$.

The volume of a liquid flowing out per second of a pipe is given by $V=\frac{\pi}{8} \frac{p r^4}{\eta l}$

Dimension of $$ V=\frac{\text { Dimension of volume }}{\text { Dimension of time }}=\frac{\left[\mathrm{L}^3\right]}{[T]}=\left[\mathrm{L}^3 \mathrm{~T}^{-1}\right] $$

$(\because V$ is the volume of liquid flowing out per second $)$

Dimension of $p=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]$

Dimension of $\eta=\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right]$

Dimension of $l=[\mathrm{L}]$

Dimension of $r=[\mathrm{L}]$

Dimensions of LHS, $[V]=\frac{\left[L^3\right]}{[T]}=\left[L^3 \mathrm{~T}^{-1}\right]$

Dimensions of RHS, $\frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right] \times\left[\mathrm{L}^4\right]}{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-1}\right] \times[\mathrm{L}]}=\left[\mathrm{L}^3 \mathrm{~T}^{-1}\right]$

As dimensions of LHS is equal to the dimensions of RHS. Therefore, equation is correct dimensionally.

Given, physical quantity is $X = a^2 b^3 c^{5/2} d^{-2}$.

Maximum percentage error in $X$ is

$$\frac{\Delta X}{X} \times 100 = \pm \left[ 2 \left(\frac{\Delta a}{a} \times 100 \right) + 3 \left(\frac{\Delta b}{b} \times 100 \right) + \frac{5}{2} \left(\frac{\Delta c}{c} \times 100 \right) + 2 \left(\frac{\Delta d}{d} \times 100 \right) \right]$$

$= \pm \left[2(1) + 3(2) + \frac{5}{2}(3) + 2(4) \right]$ %

$= \pm \left[2 + 6 + 7.5 + 8 \right] = \pm 23.5 \%$

Thus, percentage error in quantity $X$ = ± 23.5%.

Mean absolute error in $X$ = $\pm 0.235 \pm 0.24$ (rounding-off up to two significant digits).

The calculated value of $x$ should be rounded off to two significant digits.

$\therefore X$ = 2.8

Given, expression is

$P=E L^2 m^{-5} G^{-2}$

where $E$ is energy

$$[E] = [ML^2 T^{-2}]$$

$m$ is mass

$$[m] = [M]$$

$L$ is angular momentum

$$[L] = [ML^2 T^{-1}]$$

$G$ is gravitational constant

$$[G] = [M^{-1}L^3 T^{-2}]$$

Substituting dimensions of each term in the given expression,

$\begin{aligned} & {[P]=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right] \times\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^2 \times[\mathrm{M}]^{-5} \times\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^{-2}} \\\\ & =\left[\mathrm{M}^{1+2-5+2} \mathrm{~L}^{2+4-6} \mathrm{~T}^{-2-2+4}\right]=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^0\right] \end{aligned}$

Therefore, $P$ is a dimensionless quantity.

We know that, dimensions of

$$ \text { (h) }=\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right] $$

Dimensions of

(c) $=\left[\mathrm{LT}^{-1}\right]$

Dimensions of gravitational constant $(G)=\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]$

(i) Let $$ m \propto c^x h^y G^z $$

$$ \Rightarrow m=k c^x h^y G^z $$   ....(i)

where, $k$ is a dimensionless constant of proportionality.

Substituting dimensions of each term in Eq. (i), we get

$$ \begin{aligned} {\left[\mathrm{ML}^0 \mathrm{~T}^0\right] } & =\left[\mathrm{LT}^{-1}\right]^x \times\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^y\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^z \\\\ & =\left[\mathrm{M}^{y-z} \mathrm{~L}^{x+2 y+3 z} \mathrm{~T}^{-x-y-2 z}\right] \end{aligned} $$

Comparing powers of same terms on both sides, we get

$$ \begin{array}{r} y-z=1 \,\,\,\, ....(ii) \\ x+2 y+3 z=0 \,\,\,\, ....(iii) \\ -x-y-2 z=0 \,\,\,\, ....(iv) \end{array} $$

Adding Eqs. (ii), (iii) and (iv), we get

$$ 2 y=1 \Rightarrow y=\frac{1}{2} $$

Substituting value of $y$ in Eq. (ii), we get

$$ z=-\frac{1}{2} $$

From Eq. (iv)

$$ x=-y-2 z $$

Substituting values of $y$ and $z$, we get

$$ x=-\frac{1}{2}-2\left(-\frac{1}{2}\right)=\frac{1}{2} $$

Putting values of $x, y$ and $z$ in Eq. (i), we get

$$ \begin{aligned} & m=k c^{1 / 2} h^{1 / 2} G^{-1 / 2} \\\\ & \Rightarrow \quad m=k \sqrt{\frac{c h}{G}} \\ & \end{aligned} $$

$$ \begin{aligned} & \text{Let} \quad L \propto C^x h^y G^z \\\\ & \Rightarrow L=k c^x h^y G^z \,\,\,\, ....(v) \end{aligned} $$

where, $k$ is a dimensionless constant.

Substituting dimensions of each term in Eq. (v), we get

$$ \begin{aligned} {\left[\mathrm{M}^0 \mathrm{LT}^0\right] } & =\left[\mathrm{LT}^{-1}\right]^x \times\left[\mathrm{ML}^2 \mathrm{~T}^{-1}\right]^y \times\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]^z \\\\ & =\left[\mathrm{M}^{y-z} \mathrm{~L}^{x+2 y+3 z} \mathrm{~T}^{-x-y-2 z}\right] \end{aligned} $$

On comparing powers of same terms, we get

$$ \begin{array}{r} y-z=0 \,\,\,\,....(vi)\\ x+2 y+3 z=1 \,\,\,\,....(vii)\\ -x-y-2 z=0 \,\,\,\,....(viii) \end{array} $$

Adding Eqs. (vi), (vii) and (viii), we get

$$ \begin{array}{r} 2 y=1 \\\Rightarrow y=\frac{1}{2} \end{array} $$

Substituting value of $y$ in Eq. (vi), we get

$$ z=\frac{1}{2} $$

From Eq. (viii),

$$ x=-y-2 z $$

Substituting values of $y$ and $z$, we get

$$ x=-\frac{1}{2}-2\left(\frac{1}{2}\right)=-\frac{3}{2} $$

Putting values of $x, y$ and $z$ in Eq. (v), we get

$$ \begin{aligned} & L=k c^{-3 / 2} h^{1 / 2} G^{1 / 2} \\\\ & L=k \sqrt{\frac{h G}{c^3}} \end{aligned} $$

(iii) Let $T \propto C^x h^y G^z$

$$ \Rightarrow \quad T=k c^x h^y G^z $$ ....(ix)

where, $k$ is a dimensionless constant.

Substituting dimensions of each term in Eq. (ix), we get

$$ \begin{aligned} {\left[M^0 L^0 T\right.} & =\left[L T^{-1}\right]^x \times\left[M^2 T^{-1}\right]^y \times\left[M^{-1} L^3 T^{-2}\right]^z \\\\ & =\left[M^{y-z} L^{x+2 y+3 z} T^{-x-y-2 z}\right] \end{aligned} $$

On comparing powers of same terms, we get

$$ \begin{array}{r} y-z=0 \,\,\,\,....(x) \\ x+2 y+3 z=0 \,\,\,\,....(xi) \\ -x-y-2 z=1 \,\,\,\,....(xii) \end{array} $$

Adding Eqs. (x), (xi) and (xii), we get

$$ \begin{aligned} 2 y & =1 \\ \Rightarrow y & =\frac{1}{2} \end{aligned} $$

Substituting value of $y$ in Eq. (x), we get

$$ z=y=\frac{1}{2} $$

From Eq. (xii),

$$ x=-y-2 z-1 $$

Substituting values of $y$ and $z$, we get

$$ x=-\frac{1}{2}-2\left(\frac{1}{2}\right)-1=-\frac{5}{2} $$

Putting values of $x, y$ and $z$ in Eq. (ix), we get

$$ \begin{aligned} T & =k c^{-5 / 2} h^{1 / 2} G^{1 / 2} \\\\ T & =k \sqrt{\frac{h G}{c^5}} \end{aligned} $$