NCERT Solutions Class 10 Maths Chapter 13 – Surface Area and Volumes Exercise 13.3

NCERT Solutions for Class 10 Maths Chapter 13 – Surface Area and Volumes Exercise 13.3 are designed to help students apply the concepts of surface area and volume to solve problems involving composite solids that involve conical, cylindrical, and spherical shapes. This exercise helps students solve more complex problems involving 3D objects formed by the combination of different geometric solids.

Exercise 13.3 covers:

NCERT Solutions Class 10 Maths Chapter 13 – Surface Area and Volumes Exercise 13.3

Surface Areas and Volumes

Medium

Q.

A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

View Solution

Surface Areas and Volumes

Medium

Q.

Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

View Solution

Surface Areas and Volumes

Medium

Q.

A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

View Solution

Surface Areas and Volumes

Medium

Q.

A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.

View Solution

Surface Areas and Volumes

Medium

Q.

A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.

View Solution

Surface Areas and Volumes

Medium

Q.

How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

View Solution

Surface Areas and Volumes

Medium

Q.

A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

View Solution

Surface Areas and Volumes

Medium

Q.

Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

View Solution

Surface Areas and Volumes

Medium

Q.

A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?

View Solution

Surface Areas and Volumes

Medium

Q.

A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.

View Solution

Surface Areas and Volumes

Medium

Q.

The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.

View Solution

• Finding the surface area and volume of composite shapes, which includes combining different solids like cones, spheres, and cylinders.

• Applying the formulas for volume and surface area to complex geometric figures in real-life applications.

• Problem-solving involving the combination of solids and calculations of missing dimensions.

The solutions are step-by-step, helping students to grasp the concept of composite solid geometry and apply their knowledge effectively in Class 10 exams.

Q1. A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Difficulty Level: Medium

Known/Given:

• Radius of the metallic sphere = 4.2 cm

• Radius of the cylinder = 6 cm

Unknown:

• The height of the cylinder.

Reasoning:
Since a metallic sphere is melted and recast into the shape of a cylinder, their volumes must be the same.

• Volume of the sphere = Volume of the cylinder

Using the formulae:

• Volume of the sphere:

  $\frac{4}{3} \pi r^3$34​πr3

• Volume of the cylinder:

  $\pi r^2 h$πr2h

Solution:

• Radius of the sphere:

  $r = 4.2$r=4.2 cm

• Radius of the cylinder:

  $r = 6$r=6 cm

• Let the height of the cylinder be

  $h$h.

Equating the volumes:

$$\frac{4}{3} \pi (4.2)^3 = \pi (6)^2 h$$

34​π(4.2)3=π(6)2h

Simplifying:

$$\frac{4}{3} \pi (74.088) = \pi (36) h$$

34​π(74.088)=π(36)h

$$\frac{296.352}{3} = 36h$$

3296.352​=36h

$$98.784 = 36h$$

98.784=36h

$$h = \frac{98.784}{36} = 2.74 \, \text{cm}$$

h=3698.784​=2.74cm

Q2. Metallic spheres of radii 6 cm, 8 cm, and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

Difficulty Level: Medium

Known/Given:

• Radii of three metallic spheres are 6 cm, 8 cm, and 10 cm.

Unknown:

• The radius of the resulting sphere.

Reasoning:
Since three metallic spheres are melted and recast into a single solid sphere, the volume of the resulting sphere is the sum of the volumes of the three spheres.

Using the formula for the volume of a sphere:

• Volume of the sphere:

  $\frac{4}{3} \pi r^3$34​πr3

Solution:

• Volume of the resulting sphere = Sum of the volumes of the three spheres.

$$\text{Volume of the resulting sphere} = \frac{4}{3} \pi (6)^3 + \frac{4}{3} \pi (8)^3 + \frac{4}{3} \pi (10)^3$$

Volume of the resulting sphere=34​π(6)3+34​π(8)3+34​π(10)3

$$= \frac{4}{3} \pi (216 + 512 + 1000)$$

=34​π(216+512+1000)

$$= \frac{4}{3} \pi (1728)$$

=34​π(1728)

$$= 2304 \pi$$

=2304π

Now, let the radius of the resulting sphere be

$r$

r.

$$\frac{4}{3} \pi r^3 = 2304 \pi$$

34​πr3=2304π

Simplifying:

$$\frac{4}{3} r^3 = 2304$$

34​r3=2304

$$r^3 = \frac{2304 \times 3}{4} = 1728$$

r3=42304×3​=1728

$$r = \sqrt[3]{1728} = 12 \, \text{cm}$$

r=31728​=12cm

Q3. A 20 m deep well with diameter 7 m is dug, and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

Difficulty Level: Medium

Known/Given:

• Depth of the well = 20 m

• Diameter of the well = 7 m

• Length of the platform = 22 m

• Breadth of the platform = 14 m

Unknown:

• The height of the platform.

Reasoning:
The volume of earth dug from the well is equal to the volume of earth used to form the platform.

Volume of the well = Volume of the platform

Using the formulae for volumes:

• Volume of the well (cylinder):

  $\pi r^2 h$πr2h

• Volume of the platform (cuboid):

  $l \times b \times h$l×b×h

Solution:

• Radius of the well:

  $r = \frac{7}{2} = 3.5 \, \text{m}$r=27​=3.5m

• Height of the well:

  $h = 20 \, \text{m}$h=20m

Volume of the well:

$$V_{\text{well}} = \pi (3.5)^2 (20) = \pi (12.25) (20) = 245 \pi \, \text{m}^3$$

Vwell​=π(3.5)2(20)=π(12.25)(20)=245πm3

Let the height of the platform be

$h$

h.
Volume of the platform:

$$V_{\text{platform}} = 22 \times 14 \times h = 308 h \, \text{m}^3$$

Vplatform​=22×14×h=308hm3

Equating the volumes:

$$245 \pi = 308 h$$

245π=308h

$$h = \frac{245 \pi}{308}$$

h=308245π​

$$h \approx \frac{245 \times 3.14}{308} = 2.5 \, \text{m}$$

h≈308245×3.14​=2.5m

Q4. A well of diameter 3m is dug 14m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4m to form an embankment. Find the height of the embankment.

Difficulty Level: Medium

Known/Given:

• Depth of the well = 14 m

• Diameter of the well = 3 m

• Width of the circular ring of the embankment = 4 m

Unknown:

• The height of the embankment.

Reasoning:
The volume of earth dug from the well is equal to the volume of the earth used to form the embankment.

The shape of the well is cylindrical, and the embankment is a hollow cylinder (circular ring) with inner and outer radii.

Volume of the earth dug from the well = Volume of the hollow cylindrical embankment.

Solution:

• Radius of the well:

  $r = \frac{3}{2} = 1.5 \, \text{m}$r=23​=1.5m

• Depth of the well:

  $h = 14 \, \text{m}$h=14m

• Inner radius of the embankment:

  $r = 1.5 \, \text{m}$r=1.5m

• Outer radius of the embankment:

  $R = 1.5 + 4 = 5.5 \, \text{m}$R=1.5+4=5.5m

Volume of the well:

$$V_{\text{well}} = \pi (1.5)^2 (14) = 31.5 \pi \, \text{m}^3$$

Vwell​=π(1.5)2(14)=31.5πm3

Volume of the hollow cylindrical embankment:

$$V_{\text{embankment}} = \pi \left[ (5.5)^2 - (1.5)^2 \right] h = \pi (30.25 - 2.25) h = \pi (28) h$$

Vembankment​=π[(5.5)2−(1.5)2]h=π(30.25−2.25)h=π(28)h

Equating the volumes:

$$31.5 \pi = \pi (28) h$$

31.5π=π(28)h

$$h = \frac{31.5}{28} \approx 1.125 \, \text{m}$$

h=2831.5​≈1.125m

FAQs: Class 10 Maths Chapter 13 – Surface Area and Volumes Exercise 13.3

Q1. What is the main focus of Exercise 13.3?
Answer:
Exercise 13.3 focuses on calculating the surface area and volume of composite solids, which are solids formed by combining cylinders, cones, spheres, and other basic solids. This exercise helps students apply multiple geometric formulas to find the total area and volume.

Q2. What are the formulas to calculate the surface area and volume of composite solids?
Answer:
The formulas to calculate surface area and volume of basic solids in composite shapes are:

• Surface area of a cylinder:

  $A = 2\pi r(h + r)$A=2πr(h+r)

• Volume of a cylinder:

  $V = \pi r^2 h$V=πr2h

• Surface area of a cone:

  $A = \pi r(l + r)$A=πr(l+r)

• Volume of a cone:

  $V = \frac{1}{3} \pi r^2 h$V=31​πr2h

• Surface area of a sphere:

  $A = 4\pi r^2$A=4πr2

• Volume of a sphere:

  $V = \frac{4}{3} \pi r^3$V=34​πr3

• Surface area of a hemisphere:

  $A = 3\pi r^2$A=3πr2

• Volume of a hemisphere:

  $V = \frac{2}{3} \pi r^3$V=32​πr3

Q3. How do I calculate the surface area of a composite solid?
Answer:
To calculate the surface area of a composite solid:

1. Break the composite shape into simpler components (e.g., a cone and a cylinder).

2. Find the surface area of each part using the appropriate formulas.

3. Add or subtract the surface areas based on whether parts are external or internal.

Q4. How is the volume of a composite solid calculated?
Answer:
To calculate the volume of a composite solid:

1. Divide the solid into simpler shapes.

2. Find the volume of each shape using the respective formulas.

3. Add or subtract the volumes based on the problem's requirements.

Q5. How do NCERT Solutions help with exam preparation?
Answer:
NCERT Solutions provide clear, detailed explanations and step-by-step methods for solving complex problems involving composite solids. By practicing these solutions, students can improve their understanding of geometry and apply formulas confidently, which is essential for success in Class 10 exams.