NCERT Solutions for Exercise 13.2 Class 10 Maths Chapter 13 - Surface Area and Volumes

NCERT Solutions for Exercise 13.2 Class 10 Maths Chapter 13 - Surface Area and Volumes

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The volume of any three-dimensional object is the space occupied within the object's boundaries. There are some different formulas for calculating the volume of different objects. Similarly, volumes of the combination of two or more objects can be determined. Because sometimes, two or more different or the same types of objects are combined to form a new shape. The volume of the combination of an object is the sum of the volume of the combined object. For example, a toy is made with a right circular cone at the top and a hemisphere at the base, then the sum of the volumes of the right circular cone and the hemisphere is the total volume of the toy.

This Story also Contains

  1. Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2
  2. Assess NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2
  3. Topics Covered in Chapter 12, Surface Areas and Volumes: Exercise 12.2
  4. NCERT Solutions for Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject Wise
NCERT Solutions for Exercise 13.2 Class 10 Maths Chapter 13 - Surface Area and Volumes
exercise 13.2

This exercise has eight questions as per the NCERT Books that require students to visualise the shapes as described in the problem statements, then break up the figure, especially those types of figures whose volume or the formula to find its volume is known to us. So that we can very easily determine their volumes and either add them or subtract them to get the final volume of the resultant figure. 10th class Maths exercise 12.2 NCERT solutions are designed as per the students' demand, covering comprehensive, step-by-step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in-depth understanding of concepts. Students can find all exercises together using the link provided below.

Download Free PDF of NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2

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Assess NCERT Solutions for Class 10 Maths Chapter 12 Exercise 12.2

Q1. 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 $\pi$.

Answer:

The volume of the solid is given by :

The volume of the solid = Volume of the cone + Volume of a hemisphere

The volume of the cone :

$=\ \frac{1}{3} \pi r^2h$

or $=\ \frac{1}{3} \pi \times 1^2\times 1$

or $=\ \frac{\pi}{3}\ cm^3$

And the volume of the hemisphere :

$=\ \frac{2}{3}\pi r^3$

or $=\ \frac{2}{3}\pi \times 1^3$

or $=\ \frac{2\pi}{3}\ cm^3$

Hence, the volume of the solid is :

$=\ \frac{\pi}{3}\ +\ \frac{2\pi}{3}\ =\ \pi\ cm^3$

Q2. 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 aluminium 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.)

Answer:

The volume of air present = Volume of cylinder + 2 (Volume of a cone)

Now, the volume of a cylinder: $=\ \pi r^2h$

or $=\ \pi \left ( \frac{3}{2} \right )^2\times 8$

or $=\ 18\pi \ cm^3$

And the volume of a cone is :

$=\ \frac{1}{3} \pi r^2h$

or $=\ \frac{1}{3} \pi \times \left ( \frac{3}{2} \right )^2\times 2$

or $=\ \frac{3}{2} \pi \ cm^3$

Thus, the volume of air is :

$=\ 18 \pi\ +\ 2\times \frac{3}{2} \pi \ =\ 21\pi$

or $=\ 66\ cm^3$

Q3. 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).

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Answer:

It is clear from the figure that gulab jamun has one cylindrical part and two hemispherical parts.

Thus, the volume of gulab jamun is = the Volume of the cylindrical part + 2 (Volume of the hemisphere )

Now, the volume of the cylinder is $=\ \pi r^2h$

or $=\ \pi\times 1.4^2\times 2.2$

or $=\ 13.55\ cm^3$

And the volume of a hemisphere is :

$\\=\ \frac{2}{3}\pi r^3\\\\=\ \frac{2}{3}\times \pi \times (1.4)^3\\\\=\ 5.75\ cm^3$

Thus, the volume of 1 gulab jamun is $= 13.55 + 2 (5.75) = 25.05 cm^3.$

Hence the volume of 45 gulab jamun $=\ 45(25.05)\ cm^3\ =\ 1127.25\ cm^3$

Further, it is given that one gulab jamun contains sugar syrup up to $30$ $\%$.

So, the total volume of sugar present:

$=\ \frac{30}{100}\times 1127.25\ =\ 338\ cm^3$

Q4. 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.).

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Answer:

The volume of wood is given by = volume of the cuboid - the volume of four cones.

Firstly, the volume of a cuboid: $=\ lbh$

or $=\ 15\times 10\times 3.5$

or $=\ 525\ cm^3$

And, the volume of the cone :

$\\=\ \frac{1}{3} \pi r^2h\\=\ \frac{1}{3} \pi \times (0.5)^2\times 1.4\\\\=\ 0.3665\ cm^3$

Thus the volume of wood is $= 525 + 4 (0.3665) = 523.53 \:cm^3$

Q5. 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.

Answer:

According to the question :

Water spilt from the container = Volume of lead balls.

Let us assume the number of lead balls to be n.

Thus, the equation becomes :

$\frac{1}{4}\times Volume_{cone}\ =\ n\times \frac{4}{3} \pi r^3$

or $\frac{1}{4}\times \frac{1}{3}\pi\times5^2\times 8\ =\ n\times \frac{4}{3} \pi\times 0.5^3$

or $n\ =\ \frac{25\times 8}{16\times \left ( \frac{1}{2} \right )^3}$

or $n\ =\ 100$

Hence, the number of lead shots dropped is 100.

Q6. 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 8g mass. (Use $\pi = 3.14$ )

Answer:

The pole can be divided into one large cylinder and one small cylinder.

Thus, the volume of the pole = volume of the large cylinder + volume of the small cylinder

$=\ \pi r_l^2h_l\ +\ \pi r_s^2h_s$

or $=\ \pi \times 12^2\times 220\ +\ \pi \times 8^2\times 60$

or $=\ \pi \times \left ( 144\times 220\ +\ 64\times 60 \right )$

or $=\ 3.14\times 35520$

or $=\ 111532.5\ cm^3$

Now, according to the question mass of the pole is :

$=\ 8\times 111532.5$

or $=\ 892262.4\ g\ =\ 892.262\ Kg$

Q7. 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 of the cylinder is 60 cm and its height is 180 cm.

Answer:

It is clear from the question that the required volume is :

The volume of water (left) =Volume of a cylinder - Volume of solid

Now the volume of the cylinder is $=\ \pi r^2h$

or $=\ \pi\times (60)^2\times 180\ cm^3$

And the volume of the solid is :

$\\=\ \frac{1}{3} \pi r^2h\ +\ \frac{2}{3}\pi r^3\\\\=\ \frac{1}{3} \times \pi \times (60)^2\times 120\ +\ \frac{2}{3}\times \pi \times (60)^3\\\\=\ \pi (60)^2\times 80\ cm^3$

Thus, the volume of water left :

$\\=\ \pi (60)^2\times 180\ -\ \pi (60)^2\times 80\\\\=\ \pi (60)^2\times 100\\\\=\ 1131428.57\ cm^3\\\\=\ 1.131\ m^3$

Q8. 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 $\pi = 3.14$.

Answer:

The volume of the vessel is given by :

The volume of the vessel = Volume of sphere + Volume of the cylindrical part

Now, the volume of the sphere is :

$\\=\ \frac{4}{3}\pi r^3\\\\=\ \frac{4}{3}\pi \left ( \frac{8.5}{2} \right )^3\\\\=\ 321.55\ cm^3$

And the volume of the cylinder is:-

$\\=\ \pi r^2h\\\\=\ \pi \times (1)^2\times 8\\\\=\ 25.13\ cm^3$

Thus the volume of the vessel is $= 321.55 + 25.13 = 346.68\: cm^3$




Also Read:

Topics Covered in Chapter 12, Surface Areas and Volumes: Exercise 12.2

  • Volume: The amount of space occupied by any three-dimensional shape is called volume.
  • Volume of a Combination of Solids: The volume of the combined object is determined by applying the formula for volumes for different three-dimensional objects.
  • Conversion of Solids from One Shape to Another: When one shape is melted to make another shape. For example, convert a cone to a hemisphere after melting it.
  • Frustum of a Cone: When the top portion of the cone is cut off, it becomes a frustum, and in this exercise, a question related to this topic is also included

Also See:

NCERT Exemplar Solutions of Class 10 Subject Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

Q: What is the core concept of NCERT solutions for Class 10 Maths exercise 13.2?
A:

The core concept behind this exercise is to find the volume combination of solids. We have to remember the formula for finding the volume of cuboid, sphere, cylinder etc

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