NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 13 |

Chapter Name |
Surface Areas and Volumes |

Exercise |
Ex 13.2 |

Number of Questions Solved |
8 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2

**Unless stated otherwise, take π = \(\frac { 22 }{ 7 }\)**

**Question 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 n.

**Solution:**

Radius of the hemisphere = 1 cm

Volume of the hemisphere = \(\frac { 2 }{ 3 }\)πr³ = \(\frac { 2 }{ 3 }\)π(1)³ = \(\frac { 2 }{ 3 }\)πcm³

Radius of base of the cone = 1 cm

Height of the cone = 1 cm

Volume of the cone = \(\frac { 1 }{ 3 }\)πr²h = \(\frac { 1 }{ 3 }\)πr² x 1 = \(\frac { 1 }{ 3 }\)π cm³

Total volume of the solid = Volume of the hemisphere + Volume of the cone

= \(\frac { 2 }{ 3 }\)π cm³ + \(\frac { 1 }{ 3 }\)π cm³ = π cm³

**Question 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 to be nearly the same.)

**Solution:**

Volume of air contained in the model = Total volume of the solid

Diameter of base of each cone = 3 cm

∴ Radius of base of each cone = \(\frac { 3 }{ 2 }\)

Height of each cone = 2 cm

**Question 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 figure).

**Solution:**

Volume of one piece of gulab jamun

= Volume of the cylindrical portion + Volume of the two hemispherical ends 1 2 8

Radius of each hemispherical portion = \(\frac { 2.8 }{ 2 }\) = 1.4 cm

Volume of both hemispherical ends = 2 x 5.74 cm³ = 11.48 cm³

Height of the cylindrical portion = (total height) – (radius of both hemispherical ends)

= 5 cm – 2(1.4) cm = 5 cm – 2.8 cm = 2.2 cm

Radius of the cylindrical portion = 1.4 cm

Volume of the cylindrical portion of gulab jamun = πr²h

= \(\frac { 22 }{ 7 }\) x (1.4)² x 2.2cm³

= \(\frac{22 \times 1.4 \times 1.4 \times 2.2}{7}\)cm³ = 13.55 cm³

Total volume of one gulab jamun = Volume of the two hemispherical ends + Volume of the cylindrical portion

= 11.48 cm³ + 13.55 cm³ = 25.03 cm³

Volume of sugar syrup = 30% of volume of gulab jamun

= \(\frac { 30 }{ 100 }\) x 25.03 cm³ = 7.50 cm³

∴ Volume of sugar syrup in 45 gulab jamuns

= 45 (volume of sugar syrup in one gulab jamun)

= 45 x 7.50 cm³ = 337.5 cm³ = 338 cm³ approx.

**Question 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 figure).

**Solution:**

Radius of one conical depression = 0.5 cm

Depth of one conical depression = 1.4 cm

Volume of cuboidal box = l x b x h

= 15 x 10 x 3.5 cm³

= 525 cm³

Remaining volume of box = Volume of cubical box – Volume of four conical depressions

= 525 cm³ – 1.464 cm³ = 523.5 cm³

**Question 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:**

**Question 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)

**Solution:**

Given: radius of 1^{st} cylinder = 12 cm

and height of 1^{st} cylinder = 220 cm

∴ Volume of 1^{st} cylinder = πr²h

= π(12)² (220) cm³

= 144 x 220π cm³

= 144 x 220 x 3.14 cm³

= 99475.2 cm³ … (i)

Given: radius of 2^{nd} cylinder = 8 cm

and height of 2^{nd} cylinder = 60 cm

∴ Volume of 2^{nd} cylinder = πr²h

= π(8)² (60) cm³ = 64 x 60π cm³

= 64 x 60 x 3.14 cm³

= 12057.6 cm³ … (ii)

Total volume of solid = Volume of 1^{st} cylinder + Volume of 2^{nd} cylinder

= 99475.2 cm³ + 12057.6 cm³ = 111532.8 cm³

Given: mass of 1 cm³ of iron = 8 g

∴ Mass of 111532.8 cm³ of iron = 111532.8 x 8 g

= 892262.4 g = 892.262 kg

**Question 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:**

Radius of hemisphere = 60 cm

When solid (hemisphere + conical) is kept in cylindrical solid, then volume of water left in cylinder

= Volume of cylinder – (Volume of hemisphere + Volume of cone)

= [π(60)² x 180 – π(60)² x 80] cm³

= π(60)² [180 – 80]cm³ = π x 3600 x 100 cm³ = 1130400 cm³ = 1.130 m³

**Question 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:**

Volume of water the glass vessel can hold = 345 cm³ (Measured by the child)

Radius of the cylindrical part = \(\frac { 2 }{ 2 }\) = 1 cm

Height of the cylindrical part = 8 cm

∴ Volume of the cylindrical part = πr²h

= 3.14 x (1)² x 8 cm³

Diameter of the spherical part Radius = 8.5 cm

∴ Radius = \(\frac { 8.5 }{ 2 }\) cm = \(\frac { 85 }{ 20 }\)

Total volume of the glass vessel = Volume of the cylindrical part + Volume of the spherical part

= 25.12 cm³ + 321.39 cm³ = 346.51 cm³

Volume measured by child is 345 cm³, which is not correct. Correct volume is 346.51 cm³.

We hope the NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.2, drop a comment below and we will get back to you at the earliest.