Directions:

The surface area of a solid metallic sphere is 900 π cm².

Based on this information, answer the following questions:

69. If the given sphere is melted and recast into 3 smaller spheres of equal volumes, then the radius of each smaller sphere is :

(a) 5 cm
(b) 5 $\sqrt{3}$ cm
(c) 5 $\sqrt[3]{3}$ cm
(d) 5 $\sqrt[3]{9}$ cm

70. If the given sphere is cut into two hemispheres, then how much does the total surface area get increased? (Take π = 3.14) :

(a) no change
(b) 706.5 cm²
(c) 1015 cm²
(d) 1413 cm²

71. If the given sphere is melted and recast into solid right cones, each of radius 2.5 cm and height 8 cm, how many cones are formed?

(a) 135
(b) 270
(c) 405
(d) 540

72. If the given sphere is melted and recast into small spheres each of radius 0.5 cm, then the number of spheres formed is :

(a) 1350
(b) 2700
(c) 13500
(d) 27000

69. Given,

Let radius of sphere be R cm.

Surface area of a solid metallic sphere = 900 π cm²

∴ 4πR² = 900 π

⇒ 4R² = 900

⇒ R² = $\dfrac{900}{4}$

⇒ R² = 225

⇒ R = $\sqrt{225}$

⇒ R = 15 cm

Let radius of small spheres be r cm.

Since, the given sphere is melted and recast into 3 smaller spheres of equal volumes.

Volume of sphere = 3 × Volume of small spheres

$\therefore \dfrac{4}{3}π\text{R}^3 = 3 \times \dfrac{4}{3}π\text{r}^3 \\[1em] \Rightarrow \text{R}^3 = 3 \times \text{r}^3 \\[1em] \Rightarrow 15^3 = 3 \times \text{r}^3 \\[1em] \Rightarrow 3375 = 3 \times \text{r}^3 \\[1em] \Rightarrow \text{r}^3 = \dfrac{3375}{3} \\[1em] \Rightarrow \text{r}^3 = 1125 \\[1em] \Rightarrow \text{r} = \sqrt[3]{1125} \\[1em] \Rightarrow \text{r} = \sqrt[3]{125 \times 9} \\[1em] \Rightarrow \text{r} = 5\sqrt[3]{9} \text{ cm.}$

Hence, Option (d) is the correct option.

70. When a sphere is cut into two hemispheres, two new circular faces are exposed.

∴ Radius of hemisphere = R = 15 cm.

The increase in total surface area = Area of two circular faces

= 2 × πR²

= 2 × 3.14 × 15²

= 2 × 3.14 × 225

= 1413 cm²

Hence, Option (d) is the correct option.

71. Let, radius of cone be a = 2.5 cm

Height of cone, h = 8 cm

Let solid right cones formed be n.

Since, the given sphere is melted and recast into solid right cones.

∴ Volume of sphere = n × Volume of cone

$\therefore \dfrac{4}{3}π \text{R}^3 = \text{n} \times \dfrac{1}{3}π\text{a}^2 \text{h} \\[1em] \Rightarrow 4 \text{R}^3 = \text{n} \times \text{a}^2 \text{h} \\[1em] \Rightarrow \text{n} = \dfrac{4 \text{R}^3}{\text{a}^2 \text{h}} \\[1em] \Rightarrow \text{n} = \dfrac{4 \times 15^3}{2.5^2 \times 8} \\[1em] \Rightarrow \text{n} = \dfrac{4 \times 3375}{6.25 \times 8} \\[1em] \Rightarrow \text{n} = \dfrac{13500}{50} \\[1em] \Rightarrow \text{n} = 270.$

Hence, Option (b) is the correct option.

72. Let radius of small spheres be b = 0.5 cm.

Let number of small spheres formed be n.

Since, the given sphere is melted and recast into small spheres.

∴ Volume of sphere = n × Volume of small spheres

$\therefore \dfrac{4}{3}π \text{R}^3 = \text{n} \times \dfrac{4}{3}π \text{b}^3 \\[1em] \Rightarrow \text{R}^3 = \text{n} \times \text{b}^3 \\[1em] \Rightarrow \text{n} = \dfrac{\text{R}^3}{\text{b}^3} \\[1em] \Rightarrow \text{n} = \dfrac{15^3}{0.5^3} \\[1em] \Rightarrow \text{n} = \dfrac{3375}{0.125} \\[1em] \Rightarrow \text{n} = 27000.$

Hence, Option (d) is the correct option.