NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2

NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2 helps in extending our knowledge beyond the calculation of volumes of single solids. It helps in understanding the combinations of two or more basic three-dimensional geometric shapes. The exercise is focused on the calculation of the total volume of these shapes by different arithmetic operations. The concept not only helps in solving exercise 12.2 but also in solving the practical problems related to these shapes. So, let’s understand this key concept of Chapter 12: Surface area and Volumes of solids. 

1.0Download NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2 : Free PDF

2.0Introductions to Volumes of Combination of Solids

Before going into the formulas and procedures to calculate the volume of combined solids, it is imperative that students first grasp the fundamental concept behind the idea. A combination of solids is a three-dimensional object that is made by combining two or more fundamental geometric solids like cuboids, cylinders, cones, hemispheres, etc.

Compared to surface area computation, which does not involve overlapping areas, volume computation is simpler. The total volume of a composite solid is merely the volume of the constituent solids that make up the figure. There is no subtraction here since every segment of the volume adds to the total space of the object.

3.0NCERT Class 10 Maths Exercise 12.2 Overview: Key Concepts

The exercise involves calculating volumes of various composite solids formed with the combination of different basic 3D figures such as hemispheres, cones, cuboids, etc. Let’s break down the key concepts used for calculating volumes in each question of the exercise:

1. Volume of a Cone and Hemisphere Combination:

In the exercise, the most used composite solid is the combination of a cone and hemisphere. If a solid is a cone on top of a hemisphere, and they both have the same radius, the total volume is found by adding the volume of the cone and the volume of the hemisphere. The volume formulas for them are:

$\text{The total Volume of this Solid}=\text{Volume of the Cone +Volume of the Hemisphere}$

$\text{The total Volume of this Solid}=\frac{1}{3}{r}^{2}h+\frac{2}{3}{r}^3$

Here, r is the radius of the cone and hemisphere, which is the same for both figures. 

2. Volume of a Cylindrical Model with Conical Ends:

When a 3D solid consists of a cylinder with two cones attached at both ends of the cylinder, we calculate the volume of the individual solid and then sum both of them. For a question like this, the volume of the solid on the end needs to be doubled, as there must be two ends of the first solid. It is calculated as: 

$\text{The total Volume of this Solid}=\text{Volume of Cylinder}+2\times VolumeofConicalEnds$

$\text{The total Volume of this Solid}=\pi r_1^{2}h+2\times \frac{1}{3}\pi r_2^2{h}_2$

Here, 

• $r_1^2$ is the radius of the Cylinder, $r_2^2$ is the radius of conical ends, and these can be equal if given. 
• $h_1$ and $h_2$ are the heights of the cylinder and cone, respectively. These, too, can be equal if given in the question. 

3. Volume of a Cuboid with Conical Depressions:

Every question in the exercise doesn’t necessarily involve the addition of volume by other solids. In some cases, volume can be decreased by making a depression in the first solid. For example, when a conical depression is made in the cuboid by cutting out the cone, it will significantly decrease the volume of the solid. It can be calculated as: 

$\text{The total Volume of this Solid}=\text{Volume of Cuboid-Volume of Conical Depression}$

$\text{The total Volume of this Solid}=l\times b\times h-\frac{1}{3}\pi r^{2}h$

Practical Application of Combination of Solids: 

The exercise also helps calculate some real-life objects in the form of composite solids. These types of questions require combining different topics of mathematics. The exercise includes an example of calculating the volume of syrup in Gulab Jamuns, which is shaped as the composite figure of a cylinder with ends as hemispheres with the same radius as the cylinder. The Volume of the syrup, as per the example, is 30% of the total volume of all Gulab jamuns, which can be calculated as: 

$\text{The total Volume of this Solid}=\text{Volume of Cylinder}+2\times \text{Volume of Hemispherical Ends}$

$\text{The total Volume of this Solid}=\pi r^{2}h+2\times \frac{2}{3}\pi r^3$

To increase your knowledge and practice the principles, begin practising problems from NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volume Exercise 12.2 today, and gain a solid grip on finding the volumes of composite solids.

4.0NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.2: Detailed Solutions

1. A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Sol.

Here, r = 1 cm and h = 1 cm.

Volume of the conical part = (1/3)πr²h and volume of the hemispherical part = (2/3)πr³

Therefore, Volume of the solid shape = (Volume of cone + Volume of hemisphere)

= (1/3)πr²h + (2/3)πr³ = (1/3)πr²(h + 2r)

= (1/3)π(1)²[1 + 2(1)] cm³

= (1/3)π × 1 × 3 cm³

= π cm³

2. Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminum sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same).

Sol. Volume of the cylindrical part = π × (1.5)² × 8 cm³ = 18π cm³

Volume of each conical part = (1/3) x π x (1.5)² x 2 cm³ = (3/2)π cm³

Therefore, the volume of the air = The volume of cylindrical part + The volumes of two conical parts

= 18π + 2 x (3/2)π cm³ = 21π cm³

= 21 x (22/7) cm³ = 66 cm³

3. A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm. (see fig.)

Sol. Since, a gulab jamun is like a cylinder with hemispherical ends.

Total height of the gulab jamun =5 cm.

Diameter =2.8 cm

⇒ Radius =1.4 cm

Therefore, Length (height) of the cylindrical part = 5 cm - (1.4 + 1.4) cm = 5 cm - 2.8 cm = 2.2 cm.

Now, volume of the cylindrical part = πr²h and volume of both the hemispherical ends = 2(2/3 πr³) = 4/3 πr³.

Therefore, Volume of a gulab jamun = πr²h + 4/3 πr³ = πr²(h + 4/3 r)

= (22/7) × (1.4)² [2.2 + (4/3)(1.4)] cm³

= (22/7) × (14/10) × (14/10) [(22/10) + (56/30)] cm³

= (22/7) × (14/10) × (14/10) [(22/10) + (56/30)] cm²

= (44 × 14 / 100) × (122 / 30) cm³

Volume of 45 gulab jamuns = 45 × [(44 × 14 / 100) × (122 / 30)] cm³

= (15 × 44 × 14 × 122 / 1000) cm³

Since, the quantity of syrup in gulab jamuns = 30% of [volume]

= 30% of [(15 × 44 × 14 × 122 / 1000)] cm³

= (30 / 100) × (15 × 44 × 14 × 122 / 1000) cm³

= 338.184 cm³

= 338 cm³ (approx)

4. A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see fig.).

Solution:

Radius of conical cavity = 0.5 cm and depth (i.e., vertical height) = 1.4 cm

Volume of wood taken out to make one cavity = (1/3)πr²h = (1/3) × (22/7) × (0.5)² × (1.4) cm³

= (1/3) × (22/7) × (1/4) × (14/10) cm³ = 11/30 cm³

Volume of wood taken out to make four cavities = 4 × (11/30) cm³ = 44/30 cm³

Volume of the wood in the pen stand = Volume of cuboid - Volume of four cavities

= (15 × 10 × 3.5) - (44/30) cm³

= (525 - 1.47) cm³ (approx.)

= 523.53 cm³ (approx.)

5. A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one fourth of the water flows out. Find the number of lead shots dropped in the vessel.

Solution:

Height of the conical vessel (h) = 8 cm

Base radius (r) = 5 cm

Volume of water in conical vessel = (1/3) x π x r² x h

= (1/3) x (22/7) x (5)² x 8 cm³

= 4400/21 cm³

Now, Total volume of lead shots = (1/4) x (4400/21) cm³ = 1100/21 cm³

Since, radius of spherical lead shot (r) = 0.5 cm

Therefore, Volume of 1 lead shot = (4/3) x π x r³

= (4/3) x (22/7) x (5/10) x (5/10) x (5/10) cm³

Therefore, Number of lead shots = (Total volume of lead shots) / (Volume of 1 lead shot)

= (1100/21) / ((4 x 22 x 5 x 5 x 5) / (3 x 7 x 1000))

= 100

Thus, the required number of lead shots = 100.

6. A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8 g mass. (Use π = 3.14)

Sol.

The first cylindrical part has a height of 220 cm and a radius of 12 cm.

Its volume = π × (12)² × 220 cm³.

The second cylindrical part has height 60 cm and radius 8 cm.

Its volume = π × (8)² × 60 cm³.

Total volume = {144 × 220 + 64 × 60} π cm³

= 35520 π cm³ = 35520 × 3.14 cm³

= 111532.8 cm³

Total weight (at the rate of 8 gm per 1 cm³)

= (111532.8 × 8) / 1000 kg = 111.5328 × 8 kg

= 892.2624 kg

7. A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius is 60 cm and its height is 180 cm.

Solution:

Height of the conical part = 120 cm

Base radius of the conical part = 60 cm

Therefore, Volume of the conical part = (1/3)πr²h

= (1/3) × (22/7) × 60² × 120 cm³

Radius of the hemispherical part = 60 cm.

Therefore, Volume of the hemispherical part = (2/3)πr³

= (2/3) × (22/7) × (60)³ cm³

Therefore, Volume of the solid = [Volume of conical part] + [Volume of hemispherical part]

= [(1/3) × (22/7) × 60² × 120] + [(2/3) × (22/7) × 60³] cm³

= (2/3) × (22/7) × 60² [60 + 60] cm³

= (2/3) × (22/7) × 60 × 60 × 120 cm³

= 6336000/7 cm³

Volume of water in the cylinder = πr²h

= (22/7) × 60 × 60 × 180 = 14256000/7 cm³

Therefore, Volume of water left in the cylinder

= Volume of cylinder - Volume of solid

= (14256000/7 - 6336000/7) cm³

= 7920000/7 cm³

= 1131428.57142 cm³

= 1131428.57142 / 1000000 m³

= 1.13142857142 m³

= 1.131 m³ (approx.)

8. A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

Solution:

The cylinder neck has length = 8 cm and radius = 1 cm.

Volume of the cylinder part = π(1)² × 8 cm³ = 8π cm³

The radius of the spherical part = 8.5/2 cm = 4.25 cm

Volume of the spherical part = (4/3)π × (4.25)³

Total volume of water = Volume of cylindrical part + Volume of spherical part.

= 8π + (4/3) × (4.25)³π cm³

= 8 × 3.14 + (4/3) × 3.14 × (4.25)³ cm³

= 25.12 + 321.38 (approx.)

= 346.5 cm³ (approx.)

So, 345 cm³ is not correct.

5.0Benefits of Class 10 Maths Chapter 12 Surface Areas And Volumes Exercise 12.2

• Helps in calculating areas and volumes of objects like cylinders, cones, spheres, and cuboids.
• Enhances logical thinking and mathematical reasoning.
• The skills learned here are useful in higher classes and competitive exams like JEE and Olympiads.