Chapter 13 of Maths Class 10 is Surface Area and Volume. Exercise 13.2 Class 10 Solutions are provided here . These are as per the latest CBSE Pattern.

## Class 10 Maths Chapter 13

Chapter 13 of Maths Class 10 is Surface Area and Volume. Here we have provided the descriptive solutions of each question of exercise 13.2 class 10 maths . These ncert solutions are useful for students in their examination.

## Exercise 13.2 Class10 Maths Solution

Subject | Mathematics |

Class | 10 |

Chapter Name | Chapter 13 ( Surface Area and Volume ) |

Exercise | Exercise 13.2 (Volume of a Combination of Solids) |

Total Questions | 8 Questions |

Text Book | NCERT Text Book (Updated Text Book ) |

**Exercise 13.2 Class 10 Maths Question 1. 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 π.**

**Solution** :

According to question, the figure is – cone standing on a hemisphere.

The radius of cone and radius of Hemisphere is given as 1 cm.

Height of Cone = Radius = 1 cm

Volume of the figure = ?

Now, We have divided the total figure in two parts I and II ,

Volume of Part I (Part I is a Hemisphere with radius = 1 cm)

Volume of Hemisphere = \( \frac{2}{3}\ \pi\ r^3 \)

= \( \frac{2}{3}\ \pi\ 1^3 \)

= \( \frac{2}{3}\ \pi\ \) cm^{3}

Volume of Part II ( Part II is a Cone of radius and height1 cm )

Volume of Cone = \( \frac{1}{3}\ \pi\ r^2 h \)

= \( \frac{1}{3}\ \pi\ 1^2 \ * 1 \)

= \( \frac{1}{3}\ \pi \)

Total Volume of the figure = Volume of Part I + Volume of Part II

= \( \frac{2}{3}\ \pi\ +\ \frac{1}{3}\ \pi\ =\ \pi\ {\rm cm}^3 \)

**Exercise 13.2 Class 10 Maths Question 2 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.)**

**Solution**

The shape of the figure is given below ( as mentioned in question ).

Diameter of the figure 3 cm

Total Height of the figure =12 cm

We have divided the figure in three parts .

Part I – Upper Cone

Radius = 1.5 cm

Height (h1) = 2 cm

Volume of Upper Cone = 1/3 pi r sq h

= 1/3 pi (1.5)sq (2) = 1.5 pi cm^{3}

Part II – Lower Cone

Radius = 1.5 cm

Height (h1) = 2 cm

Volume of Upper Cone = 1/3 pi r sq h

= 1/3 pi (1.5)sq (2) = 1.5 pi cm^{3}

Part III – Middle Cylinder

Radius = 1.5 cm

Height of the cylinder = 12 – ( 2 + 2 ) = 8 cm

Volume of Cylinder = pi rsq h2

= pi (1.5)sq (8)

= 18 pi

Total Volume of the Figure :

=Volume of Part I + Volume of part II + Volume of Part III

=1.5 pi + 1.5 pi + 18 pi = 21 pi = 66 cm^{3}.

**Exercise 13.2 Class 10 Maths Question 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. 13.15).**

**Solution**

The Gulab jamun is of the figure as shown below ( cylinder with two hemispherical ends).

Total Volume of Gulab jamun = Volume of Cylindrical part + Volume of Upper Hemispherical part + Volume of Lower Cylindrical Part

Part 1 ( Cylindrical Part)

Radius = 1.4 cm

Height = 5-(1.4+ 1.4) = 2.2 cm

Volume of Cylindrical part

= pi rsq h

= pi (1.4)sq ( 2.2)

=

Part 2 ( Upper Hemispherical part of Gulab jamun )

Radius = 1.4 cm

Volume of Hemisphere = 2/3 pi r^{3}

= 2/3 pi (1.4) ^{3}

=

Part 3 ( Lower Hemispherical part of Gulab jamun )

Radius = 1.4 cm

Volume of Hemisphere = 2/3 pi r^{3}

= 2/3 pi (1.4) ^{3}

=

Total Volume of 1 Gulab jamun = Volume of Cylindrical part + Volume of Upper Hemispherical part + Volume of Lower Cylindrical Part

= pi ( 1.96 ) ( 12.2) / 3 cm3

Volume of 45 Gulab Jamun = 45 * pi ( 1.96 ) ( 12.2) / 3 cm3

= 112.7.28 cm3

Volume of Syrup in 45 Gulab Jamun

= 30 /100 * 1127.28 = 338 cm3

**Exercise 13.2 Class 10 Maths Question 4 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. 13.16).**

**Solution**

Depth of each cone = 1.4 cm

Radius of each cone = 0.5 cm

Volume of Wood = Volume of Cuboid – 4* ( volume of cone)

= lbh – 4 * 1/3 * pi r2 h

=15 * 10 * 3.5 – 4 * 1/3 * 22/7 * 1/4 * 1.4

= 523.533

**Exercise 13.2 Class 10 Maths Question 5. 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.**

**Solution**

According to Question :

Height of Cone = 8 cm

Radius = 5 cm

Volume of Cone = 1/3 pi r2h

=1/3pi(5)^2 (8)

= 200/3 pi cm3

Radius of sphere lead shots = 0.5 cm

Volume of spherical lead shots =4/3 pi R3

=4/3 pi (0.5)^3

= pi/6 cm3

The volume of water that flows out = 1/4 volume of cone

= 1/4 ( 200pi)/3 cm3 = 50pi/3 cm3

Let the number of lead shots dropped in the vessel = x

Number of lead shots = x = ( 50pi/3) / ( pi/6)

x = 100

**Exercise 13.2 Class 10 Maths Question 6 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 cm ^{3} of iron has approximately 8g mass. (Use π = 3.14)**

**Solution**

According to Question

Height of Upper Cylinder = 220 cm

Radius of Upper Cylinder = = 8 cm

Height of lower Cylinder = H = 60 cm

Radius of Lower Cylinder = R = 12 cm

Volume of Solid Iron Pole = Volume of Upper Cylinder + Volume of Lower Cylinder

= pi r2 h + piR2H =

3.14 * 12^2 * 220+3.14 +8*8*60 = 111532.8 cm3

Mass of the pole = 111532.8 * 8g = 892.2624 kg.

**Exercise 13.2 Class 10 Maths Question 7. 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**.

**Solution**

According to Question

Height of Cone = 120 cm

Radius of Hemisphere = 60 cm

Height of Cylinder = 120 + 60 = 180 cm

Radius of Cylinder = 60cm

Thus,

Volume of Cone = 1/3 pi r2h

=1/3*3.14 *60*60*120

=452160 cm3

Volume of Hemisphere = =2/3pir3 = 2/3 *3.144 * 60^3 = 452160

Volume of Cylinder = pi r2 h

= 3.14 * 60*60*180 = 2034720 cm3

Water left in the cylinder = Volume of Cylinder – Volume of Cone – Volume of Hemisphere

= 2034720 – 452160 -452160 = 1130400cm3

**Exercise 13.2 Class 10 Maths Question 8 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 π = 3.14.**

**Solution**

According to Question :

Height of Cylindrical neck = 8 cm

Radius of Cylindrical Neck = 1 cm

Diameter of Sphere = 8.5 cm

Volume of Water the vessel can hold = Volume of Sphere + Volume of Cylindrical Neck

= 4/3 pir3 + pi r2h

346.51 cm3

Hence the calculations done students is correct.