CUET UG 2025 — General Test Geometry

The volume formulas needed are:

Cylinder: $V = \pi r^2 h$V=πr2h

Cone: $V = \frac{1}{3}\pi r^2 h$V=31​πr2h

Sphere: $V = \frac{4}{3}\pi r^3$V=34​πr3

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For the cylinder with radius $r = 7$r=7 cm and height $h = 10$h=10 cm:

$$V_A = \pi r^2 h$$VA​=πr2h

$$V_A = \pi \times (7)^2 \times 10$$VA​=π×(7)2×10

$$V_A = \pi \times 49 \times 10$$VA​=π×49×10

$V_A = 490\pi$VA​=490π cm³

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For the cone with radius $r = 7$r=7 cm and height $h = 9$h=9 cm:

$$V_B = \frac{1}{3}\pi r^2 h$$VB​=31​πr2h

$$V_B = \frac{1}{3} \times \pi \times (7)^2 \times 9$$VB​=31​×π×(7)2×9

$$V_B = \frac{1}{3} \times \pi \times 49 \times 9$$VB​=31​×π×49×9

$$V_B = \frac{1}{3} \times \pi \times 441$$VB​=31​×π×441

$V_B = 147\pi$VB​=147π cm³

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For the sphere with radius $r = 6$r=6 cm:

$$V_C = \frac{4}{3}\pi r^3$$VC​=34​πr3

$$V_C = \frac{4}{3} \times \pi \times (6)^3$$VC​=34​×π×(6)3

$$V_C = \frac{4}{3} \times \pi \times 216$$VC​=34​×π×216

$$V_C = \frac{864}{3} \times \pi$$VC​=3864​×π

$V_C = 288\pi$VC​=288π cm³

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Comparing the volumes:

Cylinder (A): $490\pi$490π cm³

Sphere (C): $288\pi$288π cm³

Cone (B): $147\pi$147π cm³

The decreasing order is: (A), (C), (B)