RBSE Solutions for Class 10 Maths Chapter 16 Surface Area and Volume Ex 16.3

RBSE Solutions for Class 10 Maths Chapter 16 Surface Area and Volume Ex 16.3 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 16 Surface Area and Volume Exercise 16.3.

Board	RBSE
Textbook	SIERT, Rajasthan
Class	Class 10
Subject	Maths
Chapter	Chapter 16
Chapter Name	Surface Area and Volume
Exercise	Exercise 16.3
Number of Questions Solved	14
Category	RBSE Solutions

Rajasthan Board RBSE Class 10 Maths Chapter 16 Surface Area and Volume Ex 16.3

Question 1.
The height and radius of base of a cone are 28 cm and 21 cm respectively. Find its (RBSESolutions.com) curved surface area, total surface area and the volume.
Solution :
Given, height of cone (h) = 28 cm.
Radius of base (r) = 21 cm

Question 2.
The volume of a right circular cone is 1232 cm³ and its (RBSESolutions.com) height is 24 cm. Find the slant height of the cone.
Solution :
Given, height of cone (h) = 24 cm
Let radius of the cone be r.
Volume of the cone = 1232 cm³

∴ radius of the cone = 7 cm
∵ slant height of cone (l)

Hence, the slant height of the cone = 25 cm.

Question 3.
Diameter and slant height of a (RBSESolutions.com) cone are 14 m and 25 m respectively. Find it total surface area.
Solution :
Given, the base diameter of cone = 14 m
∴ Radius (r) = \(\frac { 14 }{ 5 }\) = 7 m
Slant height (d) = 25 m
Total surface area of the cone = πr(l+r)
= \(\frac { 22 }{ 7 }\) × 7 × (25 + 7)
= \(\frac { 22 }{ 7 }\) × 7 × 32
= 22 × 32
= 744 m²
Hence, total surface area of cone 704 m²

Question 4.
The radius of the base of a cone is 14 cm and its slant height is 50 cm. Find the (RBSESolutions.com) curved surface area and total surface area of the cone.
Solution :
Given
The base radius of cone (r) = 14 cm.
Slant height (l) = 50 cm
∴Curved surface area of cone = πrl
= \(\frac { 22 }{ 7 }\) × 14 × 50
= 2200 cm²
And total surface area of cone = πr (l + r)
= \(\frac { 22 }{ 7 }\) × 14 × (50+14)
= 22 × 2 × 64
= 2816 cm²
Hence, curved surface area 2200 cm²
and total surface area = 2816 cm²

Question 5.
The height of a right cone is 8 cm and its radius of the (RBSESolutions.com) base is 6 cm. Find the volume of the cone.
Solution :
Given,
Height of cone (h) = 8 cm
base radius (r) = 6 cm
Volume of the cone (V) = \(\frac { 1 }{ 3 }\)πr²h
= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 6 × 6 × 8
= \(\frac { 2112 }{ 7 }\)
= 301.7 cm³
Hence, volume of the cone = 301.7 cm³

Question 6.
The lateral surface area and the slant height of a (RBSESolutions.com) cone are 1884.4 m² and 12 m respectively. Find its base radius.
Solution :
Given,
Slant height of cone (l) = 12 meter
Lateral surface area = 1884.4 m²
Let the radius of base be r.
∴ πrl = 1884.4
\(\frac { 2112 }{ 7 }\) × 12 × r = 1884.4
r = \(\frac { 1884.4\times 7 }{ 22\times 12 }\)
= 50 m(approx)

Question 7.
Base area of a right cone is 154 cm². If its slant height is 25 cm. (RBSESolutions.com) Find the height of the cone.
Solution :
Given,
Slant height of the cone (l) = 25 cm
Base area = 154 cm²
A cone has a circular base,

Let the height of the cone be h,

Hence, the height of the cone = 24 cm.

Question 8.
Two cones have same diameters of the base. The ratio of (RBSESolutions.com) their slant heights is 5 : 4. If the lateral surface area of the smaller cone is 400 cm². find the lateral surface area of the larger cone.
Solution :
Given,
Two cones have same base diameters. So their radii will also be equal.
Let r₁ = r₂ = r.
Let the slant heights S be l₁ and l₁ then according to the question.
\(\frac { { l }_{ 1 } }{ { l }_{ 2 } }\) = \(\frac { 5 }{ 4 }\)
And lateral surface area of smaller cone (πrl₂) = 400 cm²

lateral surface area of larger cone = \(\frac { { l }_{ 1 } }{ { l }_{ 2 } }\) × 400
= \(\frac { 4 }{ 5 }\) × 100 = 500 cm²

Question 9.
The ratio of slant height and the radius of a cone is 7 : 4. If its lateral (RBSESolutions.com) surface area is 792 cm², then find its radius.
Solution :
Given, slant height (l) : radius (r) = 7 : 4


Hence, radius of the cone = 12 cm.

Question 10.
The circumference of the base of conical tent with a (RBSESolutions.com) height 9 m is 44 m. Find the volume of the air inside it.
Solution :
Given,
Height of the cane (h) = 9 m
Circumference of the base = 44 m
∴ 2πr = 44

Question 11.
A conical vessel with base radius 10 cm. contains some water in it. The water (RBSESolutions.com) level in the vessel is 18 cm high. It is pored into another cylindrical vessel with radius 5 cm. Find the water level in the cylindrical vessel.
Solution :
Base radius of conical vessel (R) = 10 cm
And the height (H) = 18 cm
Volume of conical vessel = \(\frac { 1 }{ 3 }\)πR²H
= \(\frac { 1 }{ 3 }\) × π × (10)² × 18
= π ×100 × 6
= 600 π cm³
Let the water lavel in the cylindrical vessel be h.
Radius (r) = 5 cm.
Now according to the problem
Volume of water in the cylindrical vessel = Volume of water in conical vessel
πr²h = 600 π
or r²h = 600
⇒ (5)²h = 600
⇒ 25 h = 600
h = \(\frac { 600 }{ 25 }\)
= 24 cm
Hence the water level in the cylindrical vessel = 24 cm.

Question 12.
A largest right circular cone ¡s formed by cutting a wooden (RBSESolutions.com) cubical piece of 14 cm side. Find the volume of the cone.
Solution :
Given,
Side of the cube (a) = 14 cm.
The diameter of the largest cube cut off = 14 cm

The radius of ths cone r = \(\frac { 14 }{ 2 }\) = 7 cm
The height of the cone cut off (h) = 14 cm.
∴ Volume of cone = \(\frac { 1 }{ 3 }\)πr²h
\(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × 14
= \(\frac { 2156 }{ 3 }\)
Hence, the volume of cubical piece cut off = 718.67 cm³.

Question 13.
The base radius and height of a cube are 7 cm and 24 cm respectively. Find it (RBSESolutions.com) slant height, lateral surface area, total surface area and volume.
Solution :
Given,
The base radius of cone (e) = 7 cm
Height (h) = 24 cm

Hence, slant height of cone 25 cm, curved surface area = 550 cm², total surface area = 704 cm² and volume = 1232 cm³

Question 14.
Radius and the angle of sector are 12 cm and 120° respectively. By (RBSESolutions.com) making its straight cores a cone is formed. Find its volume.
Solution :
Given,
Radius of the sector OA = OB (r) = 12 cm
Angle of the sector = 120°

We hope the given RBSE Solutions for Class 10 Maths Chapter 16 Surface Area and Volume Ex 16.3 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 16 Surface Area and Volume Exercise 16.3, drop a comment below and we will get back to you at the earliest.