Bulk modulus $K = \rho \frac{dP}{d\rho}$.

• Rearranging for change in density $d\rho$: $d\rho = \frac{\rho dP}{K}$
• Given $dP = P$, the increase is $\frac{\rho P}{K}$.

Stress is defined as $\frac{\text{Force}}{\text{Area}}$ [cite: 457].

• Force (Weight) $W \propto \text{Volume} \times \text{density} \propto L^3$
• Area $A \propto L^2$
• $\text{Stress} \propto \frac{L^3}{L^2} \propto L$
• Since $L$ increases by $9$, the stress increases by a factor of $9$[cite: 457].

Let $k = \dfrac{e^2}{4\pi\varepsilon_0}$. Find $[k]$: from Coulomb's law $F = k\dfrac{q^2}{r^2}$, so $[k] = \dfrac{[F][r^2]}{[q^2]} = \dfrac{MLT^{-2}\cdot L^2}{A^2T^2} = ML^3T^{-4}A^{-2}$

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

$\left[Gk\right] = M^{-1}L^3T^{-2} \cdot ML^3T^{-4}A^{-2} = L^6T^{-6}A^{-2}$

$\left[\sqrt{Gk}\right] = L^3T^{-3}A^{-1}$

$\left[\dfrac{1}{c^2}\sqrt{Gk}\right] = \dfrac{L^3T^{-3}A^{-1}}{L^2T^{-2}} = LT^{-1}A^{-1}$...

The correct combination (known as the classical electron radius analogue) is $\dfrac{1}{c^2}\left[G\dfrac{e^2}{4\pi\varepsilon_0}\right]^{1/2}$, which gives dimension $[L]$.

$LC = \frac{0.5}{100} = 0.005 \text{ mm}$.

Zero Error: Since the zero is 3 divisions below the mean line, it is a positive zero error. Zero Error $= +3 \times 0.005 = +0.015 \text{ mm}$.

Observed Reading: $5.5 + (48 \times 0.005) = 5.5 + 0.240 = 5.740 \text{ mm}$.

Actual Thickness: Observed Reading $-$ Zero Error $= 5.740 - 0.015 = 5.725 \text{ mm}$.

Matching logic:

• Acetaldehyde (CH$_3$CHO): Aliphatic aldehyde — gives silver mirror with Tollen's reagent (P).
• Benzaldehyde (PhCHO): Aromatic aldehyde — forms addition compound with NaHSO$_3$ (R) (bisulphite addition).
• Cyclohexanol: Secondary alcohol — reacts with Lucas reagent (Q) (ZnCl$_2$/HCl) to form turbid chloride (within minutes).
• Phenol: Gives characteristic violet colour with FeCl$_3$ (S).

Correct: (A)→(P), (B)→(R), (C)→(Q), (D)→(S)