Two steel sheets each of length $a_{1}$ and breadth $a_{2}$ are used to prepare the surfaces of two right circular cylinders - one having volume $v_{1}$ and height a_2 and other having volume $v_{2}$ and height $a_{1}$ . Then, (a) $v_{1} = v_{2}$ (b) $a_{1} v_{1} = a_{1} v_{2}$

(c) $a_{1} v_{1} = a_{1} v_{2}$ (d) $\frac{v_{1}}{a_{1}} = \frac{v_{2}}{a_{2}}$

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Solution

Length of each sheet = $a_{1}$

and breadth = $a_{2}$

Volume of cylinder = $π r^{2} h$

In first case,

$v_{1}$ is volume and $a_{2}$ is the height

$a_{1}$ is the length i.e. circumference of the cylinder = $a_{1}$

$∴ R a d i u s = \frac{a_{1}}{2 π}$

$∴ V o l u m e v_{1} = {(\frac{a_{1}}{2 π})}^{2} a_{2}$

$= \frac{a_{1}^{2} a_{2}}{4 π^{2}} \Rightarrow 4 π^{2} v_{1} = a_{1}^{2} a_{2}$

Similarly, $v = \frac{a_{2}^{2} a_{1}}{4 π^{2}} \Rightarrow 4 π^{2} v_{2} = a_{1} a_{2}^{2}$

$\Rightarrow \frac{4 π^{2} v_{1}}{a_{1}} = a_{1} a_{2}$

and $\frac{4 π^{2} v_{2}}{a_{2}} = a_{1} a_{2}$

$∴ \frac{4 π^{2} v_{1}}{a_{1}} = \frac{4 π^{2} v_{2}}{a_{2}}$

$\frac{v_{1}}{a_{1}} = \frac{v_{2}}{a_{2}} \Rightarrow a_{2} v_{1} = a_{1} v_{2}$

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Q. Two steel sheets each of length a₁ and breadth a₂ are used to prepare the surfaces of two right circular cylinders ⇀ one having volume v₁ and height a₂ and other having volume v₂ and height a₁. Then,

(a) v₁ = v₂

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(c) a₂v₁ = a₁v₂

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a_{1}

and breadth

a_{2}

are used to prepare the surface of two right circular cylinders-one having volume

V_{1}

and height

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A liquid flows through a horizontal tube. The velocities of the liquid in the two sections, which have have areas of cross section $A_{1}$ and $A_{2}$ , are $v_{1}$ and $v_{2}$ respectively. The difference in the levels of the liquid in the two vertical tubes is $h$ . Then :