Two steel sheets each of length $a_{1}$ and breadth $a_{2}$ are used to prepare the surface of two right circular cylinders-one having volume $V_{1}$ and height $a_{2}$ and the other having volume $V_{2}$ and height $a_{1}$ . Then

V_{1} = V_{2}

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a_{1} V_{1} = a_{2} V_{2}

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a_{2} V_{1} = a_{1} V_{2}

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\frac{V {_{1}}^{2}}{a_{1}} = \frac{V {_{2}}^{2}}{a_{2}}

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Solution

The correct option is C $a_{2} V_{1} = a_{1} V_{2}$
Let the radius of base of the cylinders be r and R
$\Rightarrow a_{1} = 2 π r, a_{2} = 2 π R$
$\Rightarrow r = \frac{a_{1}}{2 π}, R = \frac{a_{2}}{2 π}$
Also
$v_{1} = π r^{2} a_{2} = π {(\frac{a_{1}}{2 π})}^{2} a_{2} = \frac{a_{1}^{2} a_{2}}{4 π}$
$v_{2} = π R^{2} a_{1} = π {(\frac{a_{2}}{2 π})}^{2} a_{1} = \frac{a_{2}^{2} a_{1}}{4 π}$
$\Rightarrow \frac{v_{1}}{v_{2}} = \frac{\frac{a_{1}^{2} a_{2}}{4 π}}{\frac{a_{2}^{2} a_{1}}{4 π}} = \frac{a_{1}}{a_{2}}$
$\Rightarrow \frac{v_{1}}{a_{1}} = \frac{v_{2}}{a_{2}} = a_{2} v_{1} = a_{1} v_{2}$

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