A conical flask stands apex-down, with a full height of 12 cm and a top radius of 5 cm. Liquid is poured in until it reaches a depth of 9 cm, measured from ...

When liquid fills a cone that is apex-down, the surface of the liquid is a circle parallel to the base, so the small cone of liquid is similar to the whole flask. In similar cones every pair of corresponding lengths (radius, height, slant) is in the same ratio.

(a) The liquid is measured from the apex, so the height of the liquid cone (9 cm) compares directly with the full height (12 cm): the linear scale factor is \( \dfrac{9}{12} = \dfrac{3}{4} \) [1 mark]. Applying this scale factor to the full radius gives the liquid surface radius \( r = 5 \times \dfrac{3}{4} = 3.75 \) cm [1 mark].

(b) The volume of a cone is \( V = \dfrac{1}{3}\pi r^2 h \). Substituting the liquid's own radius and depth (not the flask's full dimensions): \( V = \dfrac{1}{3}\pi \times 3.75^2 \times 9 \) [1 mark], which simplifies to \( V = 42.1875\pi \ \text{cm}^3 \) [1 mark].

(c) Evaluating with \( \pi \) gives \( 42.1875\pi = 132.54\ldots \). Rounding to 3 significant figures (the first three non-zero digits, not 3 decimal places) gives \( V = 133 \ \text{cm}^3 \) [1 mark].

The key idea to remember: because liquid depth is measured from the apex, the depth-to-height ratio is exactly the linear scale factor between the liquid cone and the flask, so there is no need to find the missing top part separately.