Rbse Solutions for Class 9 Maths Chapter 11 Exercise 11.3 | Volume of Cone

Rbse Solutions for Class 9 Maths Chapter 11 Exercise 11.3 | Volume of Cone

1. Volume of the Right Circular Cone

(i) Radius $r = 6 \text{ cm}$, Height $h = 7 \text{ cm}$

$$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (6)^2 \times 7$$

$$V = \frac{1}{3} \times 22 \times 36 = 22 \times 12$$

$$\mathbf{V = 264 \text{ cm}^3}$$

(ii) Radius $r = 3.5 \text{ cm}$, Height $h = 12 \text{ cm}$

$$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (3.5)^2 \times 12$$

$$V = \frac{1}{3} \times \frac{22}{7} \times 12.25 \times 12$$

$$V = 22 \times 1.75 \times 4$$

$$\mathbf{V = 154 \text{ cm}^3}$$

---

2. Capacity in Litres of a Conical Vessel

(i) Radius $r = 7 \text{ cm}$, Slant height $l = 25 \text{ cm}$

1. Find Height ($h$):$$h = \sqrt{l^2 – r^2} = \sqrt{25^2 – 7^2} = \sqrt{625 – 49} = \sqrt{576}$$$$h = 24 \text{ cm}$$
2. Volume ($V$):$$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 7^2 \times 24$$$$V = 22 \times 7 \times 8$$$$V = 1232 \text{ cm}^3$$
3. Capacity in Litres: ($1 \text{ L} = 1000 \text{ cm}^3$)$$\text{Capacity} = \frac{1232}{1000}$$$$\mathbf{\text{Capacity} = 1.232 \text{ Litres}}$$

(ii) Height $h = 12 \text{ cm}$, Slant height $l = 13 \text{ cm}$

1. Find Radius ($r$):$$r = \sqrt{l^2 – h^2} = \sqrt{13^2 – 12^2} = \sqrt{169 – 144} = \sqrt{25}$$$$r = 5 \text{ cm}$$
2. Volume ($V$):$$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 5^2 \times 12$$$$V = \frac{22 \times 25 \times 4}{7} = \frac{2200}{7} \text{ cm}^3$$
3. Capacity in Litres:$$\text{Capacity} = \frac{2200}{7} \times \frac{1}{1000} = \frac{2200}{7000} = \frac{22}{70}$$$$\mathbf{\text{Capacity} = \frac{11}{35} \text{ Litres}}$$

---

3. Find the Diameter

Given: $h = 15 \text{ cm}$, $V = 1570 \text{ cm}^3$. (Use $\pi = 3.14$)

1. Use Volume formula to find $r$:$$V = \frac{1}{3} \pi r^2 h$$$$1570 = \frac{1}{3} \times 3.14 \times r^2 \times 15$$$$1570 = 3.14 \times r^2 \times 5$$
2. Solve for $r^2$:$$r^2 = \frac{1570}{3.14 \times 5} = \frac{1570}{15.7} = 100$$$$r = 10 \text{ cm}$$
3. Find Diameter ($d$):$$d = 2r = 2 \times 10$$$$\mathbf{d = 20 \text{ cm}}$$

---

4. Find the Diameter

Given: $h = 9 \text{ cm}$, $V = 48\pi \text{ cm}^3$.

1. Use Volume formula to find $r$:$$V = \frac{1}{3} \pi r^2 h$$$$48\pi = \frac{1}{3} \pi r^2 (9)$$$$48\pi = 3\pi r^2$$
2. Solve for $r$:$$r^2 = \frac{48\pi}{3\pi} = 16$$$$r = 4 \text{ cm}$$
3. Find Diameter ($d$):$$d = 2r = 2 \times 4$$$$\mathbf{d = 8 \text{ cm}}$$

---

5. Capacity in Kilolitres

Given: Diameter $d = 3.5 \text{ m}$, Height (depth) $h = 12 \text{ m}$.

1. Radius ($r$):$$r = \frac{3.5}{2} = 1.75 \text{ m}$$
2. Volume ($V$):$$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (1.75)^2 \times 12$$$$V = \frac{1}{3} \times \frac{22}{7} \times 3.0625 \times 12$$$$V = 22 \times 0.4375 \times 4$$$$\mathbf{V = 38.5 \text{ m}^3}$$
3. Capacity in Kilolitres (kL): ($1 \text{ m}^3 = 1 \text{ kL}$)$$\mathbf{\text{Capacity} = 38.5 \text{ kilolitres}}$$

---

6. Height, Slant Height, and CSA

Given: $V = 9856 \text{ cm}^3$, Diameter $d = 28 \text{ cm}$.

(i) Height of the cone ($h$):

1. Radius ($r$): $r = \frac{28}{2} = 14 \text{ cm}$
2. Use Volume formula:$$V = \frac{1}{3} \pi r^2 h$$$$9856 = \frac{1}{3} \times \frac{22}{7} \times 14^2 \times h$$$$9856 = \frac{1}{3} \times 22 \times 2 \times 14 \times h = \frac{616}{3} h$$$$h = \frac{9856 \times 3}{616}$$$$\mathbf{h = 48 \text{ cm}}$$

(ii) Slant height of the cone ($l$):

$$l = \sqrt{r^2 + h^2} = \sqrt{14^2 + 48^2}$$

$$l = \sqrt{196 + 2304} = \sqrt{2500}$$

$$\mathbf{l = 50 \text{ cm}}$$

(iii) Curved Surface Area of the cone (CSA):

$$\text{CSA} = \pi r l = \frac{22}{7} \times 14 \times 50$$

$$\text{CSA} = 22 \times 2 \times 50$$

$$\mathbf{\text{CSA} = 2200 \text{ cm}^2}$$

---

7. Volume of Solid (Revolved about $12 \text{ cm}$ side)

Given: Right $\triangle ABC$ sides $5 \text{ cm}, 12 \text{ cm}, 13 \text{ cm}$.

When revolved about the side $12 \text{ cm}$, the cone formed has:

• Height ($h$): $12 \text{ cm}$
• Radius ($r$): $5 \text{ cm}$

1. Volume ($V$):$$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times 5^2 \times 12$$$$V = \pi \times 25 \times 4$$$$\mathbf{V = 100\pi \text{ cm}^3}$$

---

8. Volume and Ratio (Revolved about $5 \text{ cm}$ side)

When revolved about the side $5 \text{ cm}$, the cone formed has:

• Height ($h$): $5 \text{ cm}$
• Radius ($r$): $12 \text{ cm}$

1. Volume ($V’$):$$V’ = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \pi \times 12^2 \times 5$$$$V’ = \frac{1}{3} \times \pi \times 144 \times 5 = \pi \times 48 \times 5$$$$\mathbf{V’ = 240\pi \text{ cm}^3}$$
2. Ratio of Volumes ($\text{Q7} : \text{Q8}$):$$\text{Ratio} = \frac{V}{V’} = \frac{100\pi}{240\pi} = \frac{10}{24}$$$$\mathbf{\text{Ratio} = \frac{5}{12} \text{ or } 5:12}$$

---

9. Volume and Canvas Area for Wheat Heap

Given: Diameter $d = 10.5 \text{ m}$, Height $h = 3 \text{ m}$.

1. Radius ($r$): $r = \frac{10.5}{2} = 5.25 \text{ m}$

(i) Volume of heap ($V$):

$$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times (5.25)^2 \times 3$$

$$V = \frac{22}{7} \times 27.5625 = 22 \times 3.9375$$

$$\mathbf{V = 86.625 \text{ m}^3}$$

(ii) Area of canvas ($\text{CSA}$):

1. Slant height ($l$):$$l = \sqrt{r^2 + h^2} = \sqrt{(5.25)^2 + 3^2}$$$$l = \sqrt{27.5625 + 9} = \sqrt{36.5625}$$$$\mathbf{l = 6.0467 \text{ m}}$$ (Your provided solution used $6.05 \text{ m}$ which is a slight approximation.)
2. Curved Surface Area ($\text{CSA}$):$$\text{CSA} = \pi r l = \frac{22}{7} \times 5.25 \times 6.0467$$$$\text{CSA} \approx 22 \times 0.75 \times 6.0467 = 16.5 \times 6.0467$$$$\mathbf{\text{CSA} \approx 99.77 \text{ m}^2}$$

(Note: If you use $l = 6.05 \text{ m}$ as intended in your provided solution, the result is $\frac{22}{7} \times 5.25 \times 6.05 = 99.825 \text{ m}^2$. Both values are acceptable depending on the rounding of the slant height.)