A container in the shape of a frustum of a cone having diameters of its two circular faces as 35 cm and 30 cm and vertical height 14 cm,

We have,

Height, $h=14 \mathrm{~cm}$,

Radius of upper end, $R=\frac{35}{2}=17.5 \mathrm{~cm}$ and

Radius of lower end, $r=\frac{30}{2}=15 \mathrm{~cm}$

Now,

Volume of the container $=\frac{1}{3} \pi h\left(R^{2}+r^{2}+R r\right)$

$=\frac{1}{3} \times \frac{22}{7} \times 14 \times\left(17.5^{2}+15^{2}+17.5 \times 15\right)$

$=\frac{44}{3} \times(306.25+225+262.5)$

$=\frac{44}{3} \times 793.75$

$=\frac{34925}{3} \mathrm{~cm}^{3}$

So, the mass of the oil that is completely filled in the container $=\frac{34925}{3} \times 1.2$

$=13970 \mathrm{~kg}$

$=13.97 \mathrm{~kg}$

$\therefore$ The cost of the oil in the container $=40 \times 13.97=₹ 558.80$