CUET UG 2025 — General Test Geometry

Given:

• Ratio of radii (A : B) = 2 : 3
• Ratio of volumes (A : B) = 9 : 7

The volume of a cylinder is $\pi r^2 h$πr2h

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Let the radius of cylinder A be $r_A = 2k$rA​=2k and the radius of cylinder B be $r_B = 3k$rB​=3k

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The ratio of volumes can be written as:

$$\dfrac{V_A}{V_B} = \dfrac{9}{7}$$VB​VA​​=79​

$$\dfrac{\pi r_A^2 h_A}{\pi r_B^2 h_B} = \dfrac{9}{7}$$πrB2​hB​πrA2​hA​​=79​

$$\dfrac{r_A^2 h_A}{r_B^2 h_B} = \dfrac{9}{7}$$rB2​hB​rA2​hA​​=79​

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Rearranging:

$$\left(\dfrac{r_A}{r_B}\right)^2 \times \dfrac{h_A}{h_B} = \dfrac{9}{7}$$(rB​rA​​)2×hB​hA​​=79​

Substituting $\dfrac{r_A}{r_B} = \dfrac{2}{3}$rB​rA​​=32​:

$$\left(\dfrac{2}{3}\right)^2 \times \dfrac{h_A}{h_B} = \dfrac{9}{7}$$(32​)2×hB​hA​​=79​

$$\dfrac{4}{9} \times \dfrac{h_A}{h_B} = \dfrac{9}{7}$$94​×hB​hA​​=79​

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Solving for the height ratio:

$$\dfrac{h_A}{h_B} = \dfrac{9}{7} \times \dfrac{9}{4}$$hB​hA​​=79​×49​

$$\dfrac{h_A}{h_B} = \dfrac{81}{28}$$hB​hA​​=2881​

Therefore, the ratio of heights (A : B) = 81 : 28