Three students A, B and C of Class 10 are each given an equal amount of Play-Dough to form different shaped solids for a mathematics project. A forms a cone of…

Given,

For the cone formed by A,

Radius (R) = 7 cm

Height (H) = 28 cm

Since, all the three students are given an equal amount of Play-Dough, the volumes of the cone, the sphere and the cylinder are equal.

(a) Volume of sphere = Volume of cone

$$\begin{aligned}\Rightarrow \frac{4}{3}\pi r^3 = \frac{1}{3}\pi R^2 H \\[1em] \Rightarrow \frac{4}{3}\pi r^3 = \frac{1}{3} \times \pi \times 7^2 \times 28 \\[1em] \Rightarrow 4r^3 = 7^2 \times 28 \\[1em] \Rightarrow 4r^3 = 49 \times 28 \\[1em] \Rightarrow r^3 = \frac{1372}{4} \\[1em] \Rightarrow r^3 = 343 \\[1em] \Rightarrow r^3 = 7^3 \\[1em] \Rightarrow r = 7 \text{ cm}.\end{aligned}$$

Hence, radius of the sphere = 7 cm.

(b) Given,

Radius of the cylinder (R') = 14 cm

Let height of cylinder be H'.

Volume of cylinder = Volume of sphere

$$\begin{aligned}\Rightarrow \pi R'^2 H' = \frac{4}{3}\pi r^3 \\[1em] \Rightarrow 14^2 \times H' = \frac{4}{3} \times 7^3 \\[1em] \Rightarrow 196 \times H' = \frac{4 \times 343}{3} \\[1em] \Rightarrow H' = \frac{1372}{3 \times 196} \\[1em] \Rightarrow H' = \frac{7}{3} \\[1em] \Rightarrow H' = 2.33 \text{ cm} \approx 2 \text{ cm}.\end{aligned}$$

Hence, height of the cylinder = 2 cm.

(c) By formula,

Volume of the dough given to each of them = Volume of cone

$$\begin{aligned}= \frac{1}{3}\pi R^2 H \\[1em] = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 28 \\[1em] = \frac{1}{3} \times 22 \times 7 \times 28 \\[1em] = \frac{4312}{3}\\[1em] = 1437.33 \text{ cm}^3 ≈ 1437 \text{ cm}^3.\end{aligned}$$

Hence, volume of the dough given to each of them = 1437 cm³.