NCERT Solutions for Class 10 Maths Chapter 13, Surface Areas and Volumes, Exercise 13.2, are available here in a downloadable PDF. The solutions in this exercise are prepared by the expert Maths teachers at BYJU’S to make students understand the concepts in a better way.

Topics covered in this article are according to the NCERT syllabus and guidelines. Therefore, Solutions for Maths Class 10 are very helpful for students to prepare for the CBSE Board exams and to score good marks.

### Download the PDF of NCERT Solutions for Class 10 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.2

### Access Other Exercise Solutions of Class 10 Maths Chapter 13 – Surface Areas and Volumes

Exercise 13.1 Solutions 9 Questions (7 long, 2 short)

Exercise 13.3 Solutions 9 Questions (9 long)

Exercise 13.4 Solutions 5 Questions (5 long)

Exercise 13.5 Solutions 7 Questions (7 long)

### Access Answers to NCERT Class 10 Maths Chapter 13 – Surface Areas and Volumes Exercise 13.2

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

**Solution:**

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

The diagram is as follows.

Now, Volume of solid = Volume of conical part + Volume of hemispherical part

We know that the volume of cone = ⅓ πr^{2}h

And,

The volume of the hemisphere = ⅔πr^{3}

So, the volume of the solid will be

= π cm^{3}

**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 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 are nearly the same.)**

**Solution:**

Given,

Height of cylinder = 12–4 = 8 cm

Radius = 1.5 cm

Height of cone = 2 cm

Now, the total volume of the air contained will be = Volume of cylinder+2×(Volume of cone)

∴ The total volume = πr^{2}h+[2×(⅓ πr^{2}h )]

= 18 π+2(1.5 π)

= 66 cm^{3}.

**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 a length of 5 cm and a diameter of 2.8 cm (see figure).**

**Solution:**

It is known that the gulab jamuns are similar to a cylinder with two hemispherical ends.

So, the total height of a gulab jamun = 5 cm.

Diameter = 2.8 cm

So, radius = 1.4 cm

∴ The height of the cylindrical part = 5 cm–(1.4+1.4) cm

= 2.2 cm

Now, the total volume of one gulab jamun = Volume of cylinder + Volume of two hemispheres

= πr^{2}h+(4/3)πr^{3}

= 4.312π+(10.976/3) π

= 25.05 cm^{3}

We know that the volume of sugar syrup = 30% of the total volume

So, the volume of sugar syrup in 45 gulab jamuns = 45×30%(25.05 cm^{3})

= 45×7.515 = 338.184 cm^{3}

**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:**

The volume of the cuboid = length x width x height

We know the cuboid’s dimensions as 15 cm x 10 cm x 3.5 cm

So, the volume of the cuboid = 15 x 10 x 3.5 = 525 cm^{3}

Here, depressions are like cones, and we know

Volume of cone = (⅓)πr^{2}h

Given, radius (r) = 0.5 cm, and depth (h) = 1.4 cm

∴ The volume of 4 cones = 4x(⅓)πr^{2}h

= 1.46 cm^{2}

Now, the volume of wood = Volume of the cuboid – 4 x volume of the cone

= 525-1.46 = 523.54 cm^{2}

**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:**

For the cone,

Radius = 5 cm,

Height = 8 cm

Also,

Radius of sphere = 0.5 cm

The diagram will be like

It is known that,

The volume of the cone = volume of water in the cone

= ⅓πr^{2}h = (200/3)π cm^{3}

Now,

Total volume of water overflow = (¼)×(200/3) π =(50/3)π

The volume of the lead shot

= (4/3)πr^{3 }

= (1/6) π

Now,

The number of lead shots = Total volume of water overflown/Volume of lead shot

= (50/3)π/(⅙)π

= (50/3)×6 = 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 cm ^{3} of iron has approximately 8 g mass.**

**Solution:**

Given that the height of the big cylinder (H) = 220 cm

The radius of the base (R) = 24/2 = 12 cm

So, the volume of the big cylinder = πR^{2}H

= π(12)^{2 }× 220 cm^{3}

= 99565.8 cm^{3}

Now, the height of the smaller cylinder (h) = 60 cm

The radius of the base (r) = 8 cm

So, the volume of the smaller cylinder = πr^{2}h

= π(8)^{2}×60 cm^{3}

= 12068.5 cm^{3}

∴ The volume of iron = Volume of the big cylinder+ Volume of the small cylinder

= 99565.8 + 12068.5

= 111634.5 cm^{3}

We knowthat

Mass = Density x volume

So, the mass of the pole = 8×111634.5

= 893 Kg (approx.)

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

**Solution:**

Here, the volume of water left will be = Volume of the cylinder – Volume of solid

Given,

Radius of cone = 60 cm,

Height of cone = 120 cm

Radius of cylinder = 60 cm

Height of cylinder = 180 cm

Radius of hemisphere = 60 cm

Now,

The total volume of solid = Volume of cone + Volume of hemisphere

Volume of cone = 1/3πr^{2}h = 1/3 × π×60^{2}×120cm^{3} = 144×10^{3}π cm^{3}

Volume of hemisphere = (⅔)×π×60^{3 }cm^{3} = 144×10^{3}π cm^{3}

So, total volume of solid = 144×10^{3}π cm^{3} + 144×10^{3}π cm^{3} = 288 ×10^{3}π cm^{3}

Volume of cylinder = π×60^{2}×180 = 648000 = 648×10^{3} π cm^{3}

Now, the volume of water left will be = Volume of cylinder – Volume of solid

= (648-288) × 10^{3}×π = 1.131 m^{3}

**8. A spherical glass vessel has a cylindrical neck 8 cm long and 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 cm ^{3}. Check whether she is correct, taking the above as the inside measurements and π = 3.14.**

**Solution:**

Given,

For the cylinder part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm

For the spherical part, Radius (r) = (8.5/2) = 4.25 cm

Now, the volume of this vessel = Volume of cylinder + Volume of sphere

= π×(1)^{2}×8+(4/3)π(4.25)^{3}

= 346.51 cm^{3}

Exercise 13.2 of Class 10 Maths has problems based on finding the volumes of the following:

- A solid cone in terms of Pi.
- A structure formed with a cylindrical shape object having a cone on both sides.
- A cylindrical object immersed with some substance occupying the volume of the object.
- A wood of cuboid shape having depressions in it.

Students can visit BYJU’S website at any time at their convenience to get the complete solutions for Chapter 13 of Class 10 Maths. Apart from this, they can also find other learning materials, such as notes, books, question papers, and some tips and hints to do the preparation for the Class 10 CBSE Maths exam.

The questions in Exercise 13.2 are prepared as per the topic, Volume of Combination of Solids, covered in the chapter before this exercise. These Solutions provide easy methods for students to prepare well, which will help them achieve high grades in the Class 10 board exams.