Question

The diameters of internal and external surfaces of a hollow spherical shell are 10 cm and 6 cm respectively. If it is melted and recast into a solid cylinder of length \(2\frac{2}3\) of cm, find the diameter of the cylinder.

Answer 1

Diameter of internal surface = 10 cm 

∴ Radius of internal surface = \(\frac{10}2\) cm = 5 cm 

Diameter of external surface = 6 cm 

∴ Radius of external surface =\(\frac{6}2\) cm = 3 cm 

Volume of spherical shell hollow = \((\frac{4}{3})π(R^3 - r^3)\)

= \((\frac{4}{3})\timesπ\times(5^3 - 3^3)\)

Height of solid cylinder = \(\frac{8}3\) cm 

Let the radius of the solid cylinder be ‘r’ cm 

Volume of the solid cylinder = πr²h

= \(\frac{22}{7}\times{r}\times{r}\times{\frac{8}{3}}\) cm³

Volume of the solid cylinder = Volume of spherical shell hollow

⇒ π x r x r x \(\frac{8}3\) = \((\frac{4}{3})\timesπ\times(5^3 - 3^3)\)

⇒ \(r^2\times\frac{8}3\) = \((\frac{4}{3})\times(125 - 27)\)

⇒ \(r^2\times\frac{8}3\) = \(\frac{4}3\times{98}\)

⇒ r² = \(\frac{4}3\times{98}\) x \(\frac{3}{8}\)

⇒ r² = 49

⇒ r = \(7\)

Diameter of cylinder = \(14\) cm

Answer 2

Diameters= 10, 6 cm. So Radius = 5, 3 cm

So total volume = Volume of external sphere - volume of internal sphere

= (4π/3)(b^3 - a^3) {a,b are radius}

= (4π/3)(125-27)

=392π/3

Now, as volume should be conserved, the volume of the cylinder is also 392π/3

So, πr^2h = 392π/3

8r^2/3 = 392/3

r^2 = 49

r = 7

2r = d = 14

So the ans is 14

Hope it helps

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