The given figure shows the cross-section of a cone, a cylinder and a hemisphere all with the same diameter 10 cm, and the other dimensions are as shown.

Calculate :

(i) the total surface area,

(ii) the total volume of the solid and

(iii) the density of the material if its total weight is 1.7 kg.

(i) Given,

Diameter of cone, cylinder and hemisphere = 10 cm

Radius of cone, cylinder and hemisphere (r) = $\dfrac{10}{2}$ = 5 cm.

From figure,

Height of cone (h) = 12 cm

By formula,

⇒ l² = r² + h²

⇒ l² = 5² + 12²

⇒ l² = 25 + 144

⇒ l² = 169

⇒ l = $\sqrt{169}$ = 13 cm.

Total surface area = Surface area of cone + Surface area of cylinder + Surface area of hemisphere

= πrl + 2πrh + 2πr²

= πr(l + 2h + 2r)

= $\dfrac{22}{7} \times 5 \times (13 + 2 \times 12 + 2 \times 5)$

= $\dfrac{110}{7} \times (13 + 24 + 10)$

= $\dfrac{110 \times 47}{7}$

= 738.57 cm².

Hence, surface area of figure = 738.57 cm².

(ii) From figure,

Volume of figure = Volume of cone + Volume of cylinder + Volume of hemisphere

$= \dfrac{1}{3}πr^2h + πr^2h + \dfrac{2}{3}πr^3 \\[1em] = πr^2\Big(\dfrac{1}{3}h + h + \dfrac{2}{3}r\Big) \\[1em] = \dfrac{22}{7} \times 5^2 \times \Big(\dfrac{1}{3} \times 12 + 12 + \dfrac{2}{3} \times 5\Big) \\[1em] = \dfrac{550}{7} \times \Big(4 + 12 + \dfrac{10}{3}\Big) \\[1em] = \dfrac{550}{7} \times \dfrac{58}{3} \\[1em] = \dfrac{31900}{21} \\[1em] = 1519.05 \text{ cm}^3.$

Hence, volume of figure = 1519.05 cm³.

(iii) By formula,

Density = $\dfrac{\text{Mass}}{\text{Volume}}$

Given,

Mass = 1.7 kg = 1700 gm

Substituting values we get,

Density = $\dfrac{1700}{1519.05}$ = 1.12 g/cm³.

Hence, density of material = 1.12 g/cm³.