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For a cricket tournament involving $8$ countries, a special trophy, as shown below, is designed.
The height and diameter of the cylindrical part are $14 \text{ cm}$ and $6 \text{ cm}$ respectively and the diameter of the spherical ball on the top is $7 \text{ cm}$.
Based on the above information, answer the following questions :
(i) Find the total height of the trophy excluding the wooden part.
(ii) Find the difference between the radius of sphere and that of cylinder.
(iii) (a) If the cylindrical part and spherical part are separated and gold plated overall, find the total surface area to be gold plated.
OR
(iii) (b) Find the volume of the metal used in making the trophy, assuming that the metal is completely filled in it.
The height and diameter of the cylindrical part are $14 \text{ cm}$ and $6 \text{ cm}$ respectively and the diameter of the spherical ball on the top is $7 \text{ cm}$.
Based on the above information, answer the following questions :
(i) Find the total height of the trophy excluding the wooden part.
(ii) Find the difference between the radius of sphere and that of cylinder.
(iii) (a) If the cylindrical part and spherical part are separated and gold plated overall, find the total surface area to be gold plated.
OR
(iii) (b) Find the volume of the metal used in making the trophy, assuming that the metal is completely filled in it.
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Solution:
(i) Total height $= 14 + 7 = 21 \text{ cm}$ (1 Mark)
(ii) Required difference $= 3.5 - 3 = 0.5 \text{ cm}$ (1 Mark)
(iii) (a) Total surface area to be gold plated $= 2\pi rh + 2\pi r^2 + 4\pi R^2$ (1 Mark)
$= 2\pi(3 \times 14 + 3^2 + 2 \times 3.5^2)$ (1 Mark)
$= \frac{3322}{7} \text{ cm}^2$ or $474.57 \text{ cm}^2$
OR
(iii) (b) Volume of the metal used $= \pi r^2 h + \frac{4}{3} \pi R^3$ (1 Mark)
$= \pi (3^2 \times 14 + \frac{4}{3} \times 3.5^3)$ (1 Mark)
$= \frac{1727}{3} \text{ cm}^3$ or $575.67 \text{ cm}^3$
(i) Total height $= 14 + 7 = 21 \text{ cm}$ (1 Mark)
(ii) Required difference $= 3.5 - 3 = 0.5 \text{ cm}$ (1 Mark)
(iii) (a) Total surface area to be gold plated $= 2\pi rh + 2\pi r^2 + 4\pi R^2$ (1 Mark)
$= 2\pi(3 \times 14 + 3^2 + 2 \times 3.5^2)$ (1 Mark)
$= \frac{3322}{7} \text{ cm}^2$ or $474.57 \text{ cm}^2$
OR
(iii) (b) Volume of the metal used $= \pi r^2 h + \frac{4}{3} \pi R^3$ (1 Mark)
$= \pi (3^2 \times 14 + \frac{4}{3} \times 3.5^3)$ (1 Mark)
$= \frac{1727}{3} \text{ cm}^3$ or $575.67 \text{ cm}^3$