CUET UG 2025 — General Test Geometry

A cone is melted and reformed into 3 cubes. The volume of the cone equals the sum of volumes of the three cubes.

Volume of a cone:

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

$$V_{cone} = \frac{1}{3} \times \pi \times (7)^2 \times 3$$Vcone​=31​×π×(7)2×3

$$V_{cone} = \frac{1}{3} \times \pi \times 49 \times 3$$Vcone​=31​×π×49×3

$$V_{cone} = 49\pi \text{ cm}^3$$Vcone​=49π cm3

Using $\pi = \frac{22}{7}$π=722​:

$$V_{cone} = 49 \times \frac{22}{7}$$Vcone​=49×722​

$$V_{cone} = 7 \times 22$$Vcone​=7×22

$$V_{cone} = 154 \text{ cm}^3$$Vcone​=154 cm3

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Volume of cube with side 3 cm:

$$V_1 = 3^3$$V1​=33

$$V_1 = 27 \text{ cm}^3$$V1​=27 cm3

Volume of cube with side 5 cm:

$$V_2 = 5^3$$V2​=53

$$V_2 = 125 \text{ cm}^3$$V2​=125 cm3

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The volume of the cone equals the sum of volumes of all three cubes:

$$V_{cone} = V_1 + V_2 + V_3$$Vcone​=V1​+V2​+V3​

$$154 = 27 + 125 + V_3$$154=27+125+V3​

$$154 = 152 + V_3$$154=152+V3​

$$V_3 = 2 \text{ cm}^3$$V3​=2 cm3

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For the third cube with volume 2 cm³:

$$(side)^3 = 2$$(side)3=2

$$side = \sqrt[3]{2}$$side=32​

$$side = 2^{1/3}$$side=21/3

Therefore, the side of the third cube is $\sqrt[3]{2}$32​ cm, which is approximately 1.26 cm or $\sqrt{2}$2​ cm (approximately 1.414 cm).