1)a mettalic spherical shell of internal and external diameters 8 cm and 12 cm is melted and recast into the - Maths - Some Applications of Trigonometry

Question

1)a mettalic spherical shell of internal and external diameters
8 cm and 12 cm is melted and recast into the form of of a cone of
base diameter 8 cm find the height of the cone
2)a mettallic spere of radius 15 cm is melted and recast into
small cones,each of radius 5 cm and eight 10 cm find the no of
cones
3)the curved surface of a cylinder is 264m2 and its
volume is 924 m2 find ratio of its diameter to its
height

Ankush Jain · Accepted Answer

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1. Given internal radius (r1) of sphere = 8 cm

internal radius (r2) of sphere = 12 cm

Diameter of cone = 8 cm

∴ Radius of cone $= \frac{8}{2} = 4 cm$

Let h be the height of cone

Volume of spherical shell = Volume of cone

$\frac{4}{3} π (r_{2}^{3} - r_{1}^{3}) = \frac{1}{3} π r^{2} h \frac{4}{3} π [(12)^{3} - (8)^{3}] = \frac{1}{3} π (4)^{2} \times h 4 [1728 - 512] = 4^{2} \times h 4 \times 1216 = 4^{2} \times h \Rightarrow h = 304 cm$

Therefore, the height of cone is 304 cm

3. Given, curved surface area of cylinder = 264 m²

Volume of cylinder = 924 m²

$∴ \frac{Volume of cylinder}{Curved surface area of cylinder} = \frac{924}{264} \Rightarrow \frac{π r^{2} h}{2 π r h} = \frac{924}{264}, where r and h be the radius and height of cylinder respectively \Rightarrow \frac{r}{2} = \frac{924}{264} \Rightarrow r = 3.5 \times 2 = 7 cm$

⇒ Diameter = 2 × radius of cylinder = 2 × 7 = 14

Now, curved surface area of cylinder = 264 m²

$\Rightarrow 2 π r h = 264 \Rightarrow 2 \times \frac{22}{7} \times 7 \times h = 267 \Rightarrow h = \frac{264}{44} m h = 6 m \frac{Diameter of cylinder}{Height of cylinder} = \frac{14}{6} = \frac{7}{3} or 7:3$

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