CUET UG 2025 — General Test Geometry

A flask in the shape of a right circular cone of height 36 cm is completely filled with tea. The tea is then poured into a cylindrical flask whose radius is two-third of the radius of the base of the cone.

Let the radius of the conical flask be $r$r and the radius of the cylindrical flask be $R = \frac{2r}{3}$R=32r​.

Let the height of tea in the cylindrical flask be $H$H.

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The volume of tea remains constant when poured from one flask to another.

Volume of cone $= \frac{1}{3}\pi r^2 h$=31​πr2h

$$= \frac{1}{3} \times \pi \times r^2 \times 36$$=31​×π×r2×36

$$= 12\pi r^2$$=12πr2

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Volume of cylinder $= \pi R^2 H$=πR2H

$$= \pi \times \left(\frac{2r}{3}\right)^2 \times H$$=π×(32r​)2×H

$$= \pi \times \frac{4r^2}{9} \times H$$=π×94r2​×H

$$= \frac{4\pi r^2 H}{9}$$=94πr2H​

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Since the volume of tea remains the same:

$$12\pi r^2 = \frac{4\pi r^2 H}{9}$$12πr2=94πr2H​

$$12 = \frac{4H}{9}$$12=94H​

$$108 = 4H$$108=4H

$$H = 27$$H=27

Therefore, the height of tea in the cylindrical flask is 27 cm.