Rbse Solutions for Class 9 Maths Chapter 11 Exercise 11.4 | Volume of Sphare

Rbse Solutions for Class 9 Maths Chapter 11 Exercise 11.4 | Volume of Sphare

1. Find the surface area of a sphere of radius:

(i) $10.5 \text{ cm}$ (ii) $5.6 \text{ cm}$ (iii) $14 \text{ cm}$

Solution

The formula for the surface area of a sphere is $\mathbf{A = 4\pi r^2}$.

(i) Radius $r = 10.5 \text{ cm}$

$$A = 4 \times \frac{22}{7} \times (10.5)^2$$

$$A = 4 \times \frac{22}{7} \times 110.25$$

$$A = 4 \times 22 \times 15.75 = 88 \times 15.75$$

$$\mathbf{\text{Answer: } 1386 \text{ cm}^2}$$

(ii) Radius $r = 5.6 \text{ cm}$

$$A = 4 \times \frac{22}{7} \times (5.6)^2$$

$$A = 4 \times \frac{22}{7} \times 31.36$$

$$A = 4 \times 22 \times 4.48 = 88 \times 4.48$$

$$\mathbf{\text{Answer: } 394.24 \text{ cm}^2}$$

(iii) Radius $r = 14 \text{ cm}$

$$A = 4 \times \frac{22}{7} \times (14)^2$$

$$A = 4 \times 22 \times 2 \times 14 = 88 \times 28$$

$$\mathbf{\text{Answer: } 2464 \text{ cm}^2}$$

2. Find the surface area of a sphere of diameter:

(i) $14 \text{ cm}$ (ii) $21 \text{ cm}$ (iii) $3.5 \text{ m}$

Solution

The radius $r$ is half the diameter $d$. $r = d/2$.

(i) Diameter $d = 14 \text{ cm}$

1. Radius: $r = 7 \text{ cm}$
2. Area: $A = 4 \times \frac{22}{7} \times 7^2 = 4 \times 22 \times 7$$$\mathbf{\text{Answer: } 616 \text{ cm}^2}$$

(ii) Diameter $d = 21 \text{ cm}$

1. Radius: $r = 10.5 \text{ cm}$
2. Area: $A = 4 \times \frac{22}{7} \times (10.5)^2$$$\mathbf{\text{Answer: } 1386 \text{ cm}^2}$$ (Same calculation as 1(i))

(iii) Diameter $d = 3.5 \text{ m}$

1. Radius: $r = 1.75 \text{ m}$
2. Area: $A = 4 \times \frac{22}{7} \times (1.75)^2$$$A = 4 \times \frac{22}{7} \times 3.0625 = 4 \times 22 \times 0.4375$$$$\mathbf{\text{Answer: } 38.5 \text{ m}^2}$$

3. Find the total surface area of a hemisphere of radius $10 \text{ cm}$. (Use $\pi = 3.14$)

Solution

The formula for the total surface area of a hemisphere (TSA) is $\mathbf{TSA = 3\pi r^2}$.

1. Radius: $r = 10 \text{ cm}$
2. TSA:$$TSA = 3 \times 3.14 \times (10)^2$$$$TSA = 3 \times 3.14 \times 100$$$$TSA = 3 \times 314$$$$\mathbf{\text{Answer: } 942 \text{ cm}^2}$$

4. The radius of a spherical balloon increases from $7 \text{ cm}$ to $14 \text{ cm}$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

Solution

Let $r_1$ be the initial radius and $r_2$ be the final radius.

• $r_1 = 7 \text{ cm}$
• $r_2 = 14 \text{ cm}$

$$\text{Ratio} = \frac{\text{Initial Surface Area} (A_1)}{\text{Final Surface Area} (A_2)} = \frac{4\pi r_1^2}{4\pi r_2^2}$$

$$\text{Ratio} = \frac{r_1^2}{r_2^2} = \frac{7^2}{14^2} = \left(\frac{7}{14}\right)^2 = \left(\frac{1}{2}\right)^2$$

$$\mathbf{\text{Answer: } \frac{1}{4} \text{ or } 1:4}$$

5. A hemispherical bowl made of brass has an inner diameter of $10.5 \text{ cm}$. Find the cost of tin-plating it on the inside at the rate of $₹16 \text{ per } 100 \text{ cm}^2$.

Solution

Tin-plating on the inside means we need to find the Curved Surface Area (CSA) of the hemisphere.

1. Inner Radius ($r$): $r = \frac{10.5}{2} = 5.25 \text{ cm}$
2. Inner CSA: $\text{CSA} = 2\pi r^2$$$\text{CSA} = 2 \times \frac{22}{7} \times (5.25)^2$$$$\text{CSA} = 44 \times 0.75 \times 5.25 = 173.25 \text{ cm}^2$$
3. Total Cost:$$\text{Cost} = \text{Area} \times \text{Rate} = 173.25 \text{ cm}^2 \times \frac{₹16}{100 \text{ cm}^2}$$$$\text{Cost} = 1.7325 \times 16$$$$\mathbf{\text{Answer: } ₹27.72}$$

6. Find the radius of a sphere whose surface area is $154 \text{ cm}^2$.

Solution

The formula for the surface area of a sphere is $A = 4\pi r^2$.

1. Solve for $r$:$$154 = 4 \times \frac{22}{7} \times r^2$$$$154 = \frac{88}{7} r^2$$$$r^2 = \frac{154 \times 7}{88} = \frac{7 \times 7}{4} = \frac{49}{4}$$$$r = \sqrt{\frac{49}{4}} = \frac{7}{2}$$$$\mathbf{\text{Answer: } 3.5 \text{ cm}}$$

7. The diameter of the moon is approximately one-fourth of the diameter of the earth. Find the ratio of their surface areas.

Solution

The ratio of surface areas of two spheres is the square of the ratio of their diameters (or radii).

$$\frac{\text{Area of Moon}}{\text{Area of Earth}} = \left(\frac{\text{Diameter of Moon}}{\text{Diameter of Earth}}\right)^2$$

$$\frac{\text{Area of Moon}}{\text{Area of Earth}} = \left(\frac{1}{4}\right)^2 = \frac{1}{16}$$

$$\mathbf{\text{Answer: } \frac{1}{16} \text{ or } 1:16}$$

8. A hemispherical bowl is made of steel $0.25 \text{ cm}$ thick. The inner radius of the bowl is $5 \text{ cm}$. Find the outer curved surface area of the bowl.

Solution

We need the Curved Surface Area (CSA) of the outer surface.

1. Inner Radius ($r_i$): $5 \text{ cm}$
2. Thickness ($t$): $0.25 \text{ cm}$
3. Outer Radius ($r_o$): $r_o = r_i + t = 5 + 0.25 = 5.25 \text{ cm}$
4. Outer CSA: $\text{CSA} = 2\pi r_o^2$$$\text{CSA} = 2 \times \frac{22}{7} \times (5.25)^2$$$$\text{CSA} = 44 \times 0.75 \times 5.25 = 173.25 \text{ cm}^2$$

$$\mathbf{\text{Answer: } 173.25 \text{ cm}^2}$$

9. A right circular cylinder just encloses a sphere of radius $r$. Find:

(i) surface area of the sphere,

(ii) curved surface area of the cylinder,

(iii) ratio of the areas obtained in (i) and (ii).

Solution

When a cylinder just encloses a sphere of radius $r$:

• Cylinder Radius ($R$): $R = r$
• Cylinder Height ($H$): $H = 2r$

(i) Surface Area of the Sphere:

$$\mathbf{\text{Answer: } 4\pi r^2}$$

(ii) Curved Surface Area of the Cylinder:

$$\text{CSA}_{\text{cylinder}} = 2\pi R H$$

$$\text{CSA}_{\text{cylinder}} = 2\pi (r) (2r)$$

$$\mathbf{\text{Answer: } 4\pi r^2}$$

(iii) Ratio of the Areas obtained in (i) and (ii):

$$\text{Ratio} = \frac{\text{Area of Sphere}}{\text{CSA of Cylinder}} = \frac{4\pi r^2}{4\pi r^2}$$

$$\mathbf{\text{Answer: } 1:1}$$