Option C ($15.38 per labour hour) is correct.

OAR uses budgeted figures: $40,000 ÷ 2,600 budgeted labour hours = $15.38 per labour hour.

Labour hours are the more appropriate basis as actual labour hours (2,900) exceed actual machine hours (2,700), indicating a labour-intensive department.

Lassaigne's test is used to detect Nitrogen, Halogens, and Sulfur. However, the compound must contain Carbon and Nitrogen (to form $CN^-$) or a Halogen (to form $X^-$) to give a positive result with $AgNO_3$.

• Aniline ($C_{6}H_{5}NH_{2}$) contains nitrogen, which forms $NaCN$. When reacted with $AgNO_3$, it doesn't give a halide precipitate like the other options which contain chlorine [cite: 3, 4].
• $C_{6}H_{5}Cl$, $CH_{3}Cl$, and $CH_{2}Cl_{2}$ all contain Chlorine and will form a white precipitate of $AgCl$[cite: 25, 26].

The textbook explains that photography was “developed in the 19th century through the artistic aspirations of two Frenchmen, Nicéphore Niépce and Louis-Jacques-Mandé Daguerre.”

• Joseph Nicéphore Niépce secured the world's earliest surviving photograph around 1826–1827 on a plate sensitised with bitumen, exposed for eight hours (this photograph is preserved at the University of Texas at Austin)
• From 1829 until Niépce's death in 1833, the two worked in partnership
• Louis Daguerre then developed the daguerreotype in 1837 — a process using copper plates lightly coated with sensitised silver and “developed” over mercury fumes, producing sharp, permanent images

The daguerreotype was the first commercially successful photographic process. Note: Henry Fox Talbot (Britain) independently developed the calotype process around the same time; George Eastman introduced roll film in 1888; Muybridge's 1877 experiments with sequential photographs of horses contributed to the development of cinema, not still photography.

If $\tan^{-1}x, \tan^{-1}y, \tan^{-1}z$ are in A.P., then $2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z$.

Also $x,y,z$ in A.P. means $2y=x+z$.

From $2\tan^{-1}y=\tan^{-1}x+\tan^{-1}z$: using the addition formula, $\tan(2\tan^{-1}y)=\tan(\tan^{-1}x+\tan^{-1}z)$, i.e., $\dfrac{2y}{1-y^2}=\dfrac{x+z}{1-xz}$.

Substituting $x+z=2y$: $\dfrac{2y}{1-y^2}=\dfrac{2y}{1-xz}$. So $1-y^2=1-xz$, giving $y^2=xz$.

Combined with $x+z=2y$ (A.P.) and $y^2=xz$ (G.P.) $\Rightarrow x=y=z$.