NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.4 focus on important problems related to volume of solids and the concept of conversion of one solid into another. This exercise is highly important for CBSE Class 10 board exams, as it tests your understanding of volume formulas and logical application of mathematical concepts.
In Exercise 13.4, students learn how to apply formulas of sphere, cylinder, cone, and hemisphere to solve questions where solids are melted and reshaped. Since the shape changes but the volume remains constant, students must carefully apply the correct formulas and solve equations to find missing dimensions.
NCERT Solutions Class 10 Maths Chapter 13 – Surface Area and Volumes Exercise 13.4
Q.
Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?
Q.
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
Q.
A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Q.
The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Q.
A fez, the cap used by the Turks, is shaped like the frustum of a cone (see the following figure). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
Q.
Q.
A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
Q.
A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.
Q.
A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)
Q.
A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
NCERT Solutions Class 10 Maths Chapter 13 – Surface Area and Volumes Exercise 13.4
These solutions are designed as per the latest NCERT Class 10 Maths syllabus, helping students strengthen conceptual clarity and improve accuracy in exam-oriented questions. Regular practice of Chapter 13 Exercise 13.4 ensures better problem-solving speed and confidence in board examinations.
Q1. Capacity of the Drinking Glass
Given:
Height (h) = 14 cm
Diameter of larger base = 4 cm → Radius (r₁) = 2 cm
Diameter of smaller base = 2 cm → Radius (r₂) = 1 cm
Formula (Volume of Frustum):
V = (1/3) πh (r₁² + r₂² + r₁r₂)
Calculation:
V = (1/3) × (22/7) × 14 × (2² + 1² + 2×1)
= (1/3) × (22/7) × 14 × (4 + 1 + 2)
= (1/3) × (22/7) × 14 × 7
= (1/3) × 22 × 14
= 308/3
Capacity = 102.67 cm³ (approx.)
Q2. Curved Surface Area of Frustum
Given:
Slant height (l) = 4 cm
Circumference of larger base = 18 cm
Circumference of smaller base = 6 cm
Using C = 2πr
r₁ = 18 / (2π) = 9/π
r₂ = 6 / (2π) = 3/π
Formula (CSA of Frustum):
CSA = π (r₁ + r₂) l
CSA = π (9/π + 3/π) × 4
= π (12/π) × 4
= 48 cm²
Answer: 48 cm²
Q3. Area of Material Used for Fez
Given:
r₁ = 10 cm
r₂ = 4 cm
l = 15 cm
Fez is open at bottom.
Area required = CSA of frustum + Area of top circle
CSA = π (r₁ + r₂) l
= (22/7)(10 + 4)(15)
= (22/7)(14)(15)
= 660 cm²
Area of top circle = πr₂²
= (22/7)(4²)
= (22/7)(16)
= 352/7
Total Area = 660 + 352/7
= 4972/7
= 710.29 cm² (approx.)
Q4. Cost of Milk & Metal Sheet
Given:
h = 16 cm
r₁ = 20 cm
r₂ = 8 cm
π = 3.14
Step 1: Volume of Frustum (Milk Capacity)
V = (1/3) πh (r₁² + r₂² + r₁r₂)
= (1/3)(3.14)(16)(400 + 64 + 160)
= (1/3)(3.14)(16)(624)
= 10449.92 cm³
Convert to litres:
1 litre = 1000 cm³
Milk = 10.45 litres
Cost of milk = 10.45 × 20
= ₹ 209
Step 2: Cost of Metal Sheet
First find slant height:
l = √[(r₁ − r₂)² + h²]
= √[(20 − 8)² + 16²]
= √(144 + 256)
= √400
= 20 cm
CSA = π (r₁ + r₂) l
= 3.14 (28)(20)
= 1758.4 cm²
Area of bottom = πr₂²
= 3.14(64)
= 200.96 cm²
Total metal area = 1959.36 cm²
Cost per 100 cm² = ₹ 8
Cost = (1959.36 / 100) × 8
= ₹ 156.75
Final Answer:
Milk cost = ₹ 209
Metal sheet cost = ₹ 156.75
Q5. Length of Wire
Given:
Cone height = 20 cm
Vertical angle = 60°
Cut at mid height → frustum height = 10 cm
Wire diameter = 1/16 cm
Step 1: Find Radii
tan 30° = 1/√3
Lower radius r₁ = 20/√3
Upper radius r₂ = 10/√3
Frustum height h = 10 cm
Step 2: Volume of Frustum
V = (1/3) πh (r₁² + r₂² + r₁r₂)
After simplification,
Volume = (7000/9) π cm³
Step 3: Volume of Wire (Cylinder)
Radius of wire = 1/32 cm
Volume of cylinder = πr²H
Equating volumes:
π(1/32)² H = (7000/9) π
H = 7964.4 m
Final Answer:
Length of wire = 7964.4 m
FAQs: Class 10 Maths Chapter 13 Exercise 13.4
Q1. What is the key concept used in Exercise 13.4?
The key concept is conservation of volume. When a solid is converted into another shape, the total volume remains the same.
Q2. Which formulas are important for Exercise 13.4?
Students must remember the volume formulas of:
-
Cylinder: V = πr²h
-
Cone: V = (1/3)πr²h
-
Sphere: V = (4/3)πr³
-
Hemisphere: V = (2/3)πr³
Q3. Why is Exercise 13.4 important for exams?
Questions based on conversion of solids are frequently asked in board exams and carry good marks. Strong conceptual understanding helps in scoring full marks.
Q4. How can I prepare effectively for Chapter 13 Exercise 13.4?
Revise all volume formulas thoroughly, practice multiple conversion-based problems, and solve previous years’ questions for better exam preparation.