Two steel sheets each of length a₁ and breadth a₂ are used to prepare the surface of two right circular cylinders - one having volume V₁ and height a₂ and the other having volume V₂ and height a₁. Then :

1. V₁ = V₂

2. a₁ V₁ = a₂ V₂

3. a₂ V₁ = a₁ V₂

4. $\dfrac{\text{V}$1}{\text{a}2} = \dfrac{\text{V}2}{\text{a}1}

For cylinder 1,

Height of cylinder, h = a₂

Radius of cylinder be r cm

Circumference of base = 2πr = a₁

$\Rightarrow 2 \times π \times \text{r} = \text{a}$1 \\[1em] \Rightarrow \text{r} = \dfrac{\text{a}1}{2π}

Volume of cylinder 1, V₁ = πr²h

$= π \times \Big(\dfrac{\text{a}$1}{2π}\Big)^2 \times \text{a}2 \\[1em] = π \times \dfrac{\text{a}1^2}{4π^2} \times \text{a}2 \\[1em] = \dfrac{\text{a}1^2 \text{a}2 }{4π}

For cylinder 2,

Height of cylinder, H = a₁

Radius of cylinder be R

Circumference of base = 2πR = a₂

$\Rightarrow 2 \times π \times \text{R} = \text{a}$2 \\[1em] \Rightarrow \text{R} = \dfrac{\text{a}2}{2π} \\[1em] \Rightarrow \text{R} = \dfrac{\text{a}_2}{2π}

Volume of cylinder 2, V₂ = πR²H

$= π \times \Big(\dfrac{\text{a}$2}{2π}\Big)^2 \times \text{a}1 \\[1em] = π \times \dfrac{\text{a}2^2}{4π^2} \times \text{a}1 \\[1em] = \dfrac{\text{a}2^2 \text{a}1}{4π}

Ratio of the volumes of the two cylinders:

$\Rightarrow \dfrac{\text{Volume of cylinder 1}}{\text{Volume of cylinder 2}} = \dfrac{\dfrac{\text{a}$1^2 \text{a}2 }{4π}}{\dfrac{\text{a}2^2 \text{a}1}{4π}} \\[1em] \Rightarrow \dfrac{\text{V}1}{\text{V}2} = \dfrac{\text{a}1^2 \text{a}2 }{4π} \times \dfrac{4π}{\text{a}2^2 \text{a}1} \\[1em] \Rightarrow \dfrac{\text{V}1}{\text{V}2} = \dfrac{\text{a}1}{\text{a}2} \\[1em] \Rightarrow \text{V}1 \text{a}2 = \text{V}2 \text{a}1

Hence, option 3 is the correct option.