Question

If three cylinders of radius r and height h are placed vertically such that the curved surface of each cylinder touches the curved surfaces of the other two cylinders tangentially, then the volume of the air space left between the three cylinders is

(a) hr² (3 + π) 

(b) hr² \(\big(\sqrt3+\frac{π}{2}\big)\)

(c) hr² \(\big(\sqrt3-\frac{π}{2}\big)\)

(d) hr² \(\big(\sqrt3-π\big)\)

Answer 1

(c)  hr² \(\big(\sqrt3-\frac{π}{2}\big)\)

The bases of the three cylinders when placed as given are as shown in the figure : 

Let the radius of the base of each cylinder = r cm. 

We are required to find the volume of air. 

Space left between the cylinders = Area of shaded portion x height of cylinder 

Now, area of shaded portion 

= Area of ΔABC – Sum of areas of sectors of the three bases ΔABC, as can be seen is an equilateral triangle of side 2r. 

∴ Area of Δ ABC = \(\frac{\sqrt3}{4}\times(2r)^2=\sqrt3r^2\)

Area of (sector AEF + sector BED + sector CFD) (∴ sector angles ∠A = ∠B = ∠C = 60º) 

= 3 x \(\frac{\mathrm60^o}{\mathrm360^o}\timesπr^2=\frac{πr^2}{2}\)

∴ Required volume = \(\big(\sqrt3r^2-\frac{π}{2}r^2\big)h=\big(\sqrt3-\frac{π}{2}\big)r^2h.\)