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

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

By Kepler’s third law, $T^2 \propto r^3 \Rightarrow T \propto r^{3/2}$

We know that $T$ is a function of $R$ and $g$.

Let $T \propto r^{3/2}R^ag^b \Rightarrow T = kr^{3/2}R^ag^b$ .....(i)

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

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

$[M^0L^0 T] = k[L^3]^{1/2}[L]^{a}[LT^{-2}]^b = k[L^{a + b + 3/2}T^{-2b}]$

On comparing the powers of same terms, we get

$a + b + 3/2 = 0$    ...(ii)

$-2b = 1 \Rightarrow b = -1/2$   ...(iii)

From Eq. (ii), we get $a-1 / 2+3 / 2=0 \Rightarrow a=-1$

Substituting the values of $a$ and $b$ in Eq. (i), we get

$T = kr^{3/2}R^{-1}g^{-1/2}$

$ T = \frac{k}{R}\sqrt{\frac{r^3}{g}} $

Note: When we are applying formulae, we should be careful about $r$ (radius of orbit) and $R$ (radius of planet).

(a) Oleic acid does not dissolve in water hence, it is dissolved in alcohol.

(b) Lycopodium powder spreads over the entire surface of water when it is sprinkled evenly. When a drop of prepared solution is dropped on water, oleic acid does not dissolve in water. Instead it spreads on the water surface pushing the lycopodium powder away to clear a circular area where the drop falls. We can therefore, measure the area over which oleic acid spreads.

(c) In each mL of solution prepared volume of oleic acid $=\frac{1}{20} \mathrm{~mL} \times \frac{1}{20}=\frac{1}{400} \mathrm{~mL}$

(d) Volume of $n$ drops of this solution of oleic acid can be calculated by means of a burette and measuring cylinder and measuring the number of drops.

(e) If $n$ drops of the solution make 1 mL , the volume of oleic acid in one drop will be $\frac{1}{(400) n} \mathrm{~mL}$.