CAT 2025 Lesson : Mensuration - Cylinder & Prism

A cylindrical jar contains water upto a height of 24 cm. If the water in this jar is transferred to another cylindrical jar whose base radius is double that of the earlier jar, then what is the height of the water (in cm) in the new jar?

Solution

Let the radius of the first and second jars be $\bm{x}$ cm and $\bm{2x}$ cm respectively. Let the height of the second jar be $y$ cm.

Volume of water in the first container $= \pi r^{2} h = \pi \times x^{2} \times 24 = 24 \pi x^{2}$

Volume of water in the second container $= \pi \times (2x)^{2} \times y = 4 \pi x^{2} y$

As the two volumes are equal,
$24 \pi x^{2} = 4 \pi x^{2} y$
⇒ $y = 6$ cm

Answer: 6

If the ratio of the diameters of 2 cylinders and the ratio of their volumes are 2 : 3 and 3 : 4 respectively, then what is the ratio of their heights?

Solution

As diameter is twice the radius, the ratio of radii will be the same as the ratio of the diameters.

Let the radii of the 2 cylinders be 2$r$ and 3$r$. Let $h_{1}$ and $h_{2}$ be their respective heights.

Ratio of volumes $= \dfrac {\pi (2r)^{2}h_{1}}{\pi (3r)^{2}h_{2}} = \dfrac{3}{4}$ ⇒ $\dfrac{4h_{1}}{9h_{2}} = \dfrac{3}{4}$

⇒ $\dfrac{h_{1}}{h_{2}} = \dfrac{27}{16}$

Answer: 27 : 16

A solid right triangular prism, with the base being an equilateral triangle, is melted and recast into a solid right hexagonal prism, where the base is a regular hexagon (with no material wastage). If the heights of the two right prisms are the same, then what is the ratio of the perimeters of the bases of the triangular prism and the hexagonal prism?
(1) $1 : \sqrt{2}$       (2) $\sqrt{3} : 1$        (3) $\sqrt{6} : 1$        (4) $\sqrt{3} : \sqrt{2}$

Solution

Let the length of the side of the equilateral triangle in the base of the first prism be $a_1$ and that of the regular hexagon in the second prism be $a_2$. Let the heights of the two prisms be $h$.

Volume of Equilateral triangular Prism $= \dfrac {\sqrt{3}}{4} \times a_{1}^{2} \times h$

Volume of Regular Hexagonal Prism $= 6 \times \dfrac {\sqrt{3}}{4} \times a_{2}^{2} \times h$

As one prism is recast into the other, the above volumes are equal.

$\dfrac {\sqrt{3}}{4} \times a_{1}^{2} \times h$ = $6 \times \dfrac {\sqrt{3}}{4} \times a_{2}^{2} \times h$

⇒ $\dfrac {a_{1}^{2}}{a_{2}^{2}} = 6$ ⇒ $\dfrac {a_{1}}{a_{2}} = \sqrt{6}$

Ratio of Perimeters $= \dfrac{3a_{1}}{6a_{2}} = \dfrac{1}{2} \times \dfrac{a_1}{a_2} = \dfrac{1}{2} \times \sqrt{6} = \dfrac{\sqrt{3}}{\sqrt{2}}$

Answer: (4) $\sqrt{3} : \sqrt{2}$