The NCERT Solutions for Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.3 focus on understanding the curved surface area and total surface area of cylinders. This exercise helps students apply formulas to calculate areas of real-life objects like pipes, tins, and pillars. By solving these problems, learners enhance their ability to visualize three-dimensional shapes and develop practical skills essential for geometry and mensuration topics.
The step-by-step solutions provided here follow the latest NCERT guidelines and are designed to clear doubts quickly. These solutions not only help students secure good marks in exams but also strengthen their problem-solving skills and boost confidence in handling complex questions related to surface area and volume.
Obtain NCERT Solutions Class 9 Maths Chapter 11 Surface Area and Volume Exercise 11.3 in PDF format for download. These solutions have been developed by subject experts and follow the latest guidelines of the CBSE.
Here is a quick overview and access to solutions for all exercises from Chapter 11, Surface Area and Volume.
(Assume π=22/7 unless stated otherwise)
1. Find the volume of the right circular cone with
(i) radius 6 cm, height 7 cm
(ii) radius 3.5 cm, height 12 cm
Sol.
The volume of a cone is given by V = (1/3)πr²h.
(i) Radius (r) = 6 cm, Height (h) = 7 cm.
Volume = (1/3) × (22/7) × (6)² × 7 cm³
= (1/3) × 22 × 36 cm³
= 22 × 12 cm³
= 264 cm³.
(ii) Radius (r) = 3.5 cm = 7/2 cm, Height (h) = 12 cm.
Volume = (1/3) × (22/7) × (7/2)² × 12 cm³
= (1/3) × (22/7) × (49/4) × 12 cm³
= (1/3) × 22 × 7 × 3 cm³
= 22 × 7 cm³
= 154 cm³.
2. Find the capacity in litres of a conical vessel with
(i) radius 7 cm, slant height 25 cm.
(ii) height 12 cm, slant height 13 cm.
Sol.
(i) Radius (r) = 7 cm, Slant height (ℓ) = 25 cm.
First, find the height (h):
r² + h² = ℓ²
(7)² + h² = (25)²
49 + h² = 625
h² = 625 − 49
h² = 576
h = √576 = 24 cm.
Volume of cone = (1/3)πr²h
= (1/3) × (22/7) × (7)² × 24 cm³
= (1/3) × 22 × 7 × 24 cm³
= 22 × 7 × 8 cm³
= 1232 cm³.
Capacity in litres: 1 L = 1000 cm³.
Capacity = 1232/1000 L = 1.232 L.
(ii) Height (h) = 12 cm, Slant height (ℓ) = 13 cm.
First, find the radius (r):
r² + h² = ℓ²
r² + (12)² = (13)²
r² + 144 = 169
r² = 169 − 144
r² = 25
r = √25 = 5 cm.
Volume of cone = (1/3)πr²h
= (1/3) × (22/7) × (5)² × 12 cm³
= (1/3) × (22/7) × 25 × 12 cm³
= (22 × 25 × 4)/7 cm³
= 2200/7 cm³.
Capacity in litres:
Capacity = (2200/7) cm³ = (2200/7)/1000 L = 2200/7000 L = 22/70 L = 11/35 L.
3. The height of a cone is 15 cm. If its volume is 1570 cm³, find the radius of the base. (Use π=3.14)
Sol.
Height (h) = 15 cm.
Volume = 1570 cm³.
We know Volume = (1/3)πr²h.
1570 = (1/3) × 3.14 × r² × 15
1570 = 3.14 × r² × 5
1570 = 15.70 × r²
r² = 1570 / 15.70
r² = 100
r = √100 = 10 cm.
4. If the volume of a right circular cone of height 9 cm is 48π cm³, find the diameter of its base.
Sol.
Height (h) = 9 cm.
Volume = 48π cm³.
We know Volume = (1/3)πr²h.
48π = (1/3) × π × r² × 9
48 = (1/3) × r² × 9 (Cancel π from both sides)
48 = 3r²
r² = 48/3
r² = 16
r = √16 = 4 cm.
Diameter (d) = 2r = 2 × 4 = 8 cm.
5. A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
Sol.
For conical pit:
Diameter = 3.5 m.
Radius (r) = 3.5/2 m = 1.75 m.
Depth (h) = 12 m.
Capacity of the conical pit = (1/3)πr²h
= (1/3) × (22/7) × (1.75)² × 12 m³
= (1/3) × (22/7) × 3.0625 × 12 m³
= 22 × 3.0625 × (12/21) m³ (Incorrect calculation in original, 12/3 = 4, then divide by 7)
Let's recalculate:
= (1/3) × (22/7) × (1.75 × 1.75) × 12 m³
= (1/3) × (22/7) × (7/4 × 7/4) × 12 m³
= (1/3) × (22/7) × (49/16) × 12 m³
= 22 × (7/16) × 4 m³
= 22 × (7/4) m³ = 11 × 3.5 m³ = 38.5 m³.
Capacity in kilolitres: 1 m³ = 1 kilolitre (kl).
Capacity = 38.5 kl.
6. The volume of a right circular cone is 9856 cm³. If the diameter of the base is 28 cm, find
(i) height of the cone
(ii) slant height of the cone
(iii) curved surface area of the cone
Sol.
Given Volume = 9856 cm³.
Diameter of base = 28 cm.
Radius (r) = 28/2 = 14 cm.
(i) To find the height (h):
Volume = (1/3)πr²h
9856 = (1/3) × (22/7) × (14)² × h
9856 = (1/3) × (22/7) × 196 × h
9856 = (22 × 28) × h / 3
9856 = 616h / 3
h = (9856 × 3) / 616
h = 48 cm.
(ii) To find the slant height (ℓ):
ℓ² = h² + r²
ℓ² = (48)² + (14)²
ℓ² = 2304 + 196
ℓ² = 2500
ℓ = √2500 = 50 cm.
(iii) To find the curved surface area (CSA):
CSA = πrℓ
= (22/7) × 14 × 50 cm²
= 22 × 2 × 50 cm²
= 22 × 100 cm²
= 2200 cm².
7. A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
Sol.
When a right triangle is revolved about one of its perpendicular sides, the solid formed is a cone.
The side about which it revolves becomes the height (h).
The other perpendicular side becomes the radius (r) of the base.
The hypotenuse becomes the slant height (ℓ).
Here, the triangle is revolved about the side 12 cm.
So, Height (h) = 12 cm.
Radius (r) = 5 cm.
Slant height (ℓ) = 13 cm. (This is consistent with 5-12-13 being a Pythagorean triplet).
Volume of the cone = (1/3)πr²h
= (1/3) × π × (5)² × 12 cm³
= (1/3) × π × 25 × 12 cm³
= π × 25 × 4 cm³
= 100π cm³.
8. If the triangle ABC in the question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Question 7 and 8.
Sol.
When the triangle ABC (sides 5 cm, 12 cm, 13 cm) is revolved about the side 5 cm:
Height (h) = 5 cm.
Radius (r) = 12 cm.
Slant height (ℓ) = 13 cm.
Volume of the new cone = (1/3)πr²h
= (1/3) × π × (12)² × 5 cm³
= (1/3) × π × 144 × 5 cm³
= π × 48 × 5 cm³
= 240π cm³.
Ratio of the volumes of the two solids (Volume from Q7 : Volume from Q8):
Volume (Q7) = 100π cm³
Volume (Q8) = 240π cm³
Ratio = (100π) / (240π) = 100/240 = 10/24 = 5/12.
The ratio is 5:12.
9. A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
Sol.
For the conical heap of wheat:
Diameter = 10.5 m.
Base Radius (r) = 10.5/2 m = 5.25 m.
Height (h) = 3 m.
First, find its volume:
Volume = (1/3)πr²h
= (1/3) × (22/7) × (5.25)² × 3 m³
= (1/3) × (22/7) × 27.5625 × 3 m³
= 22 × (27.5625/7) m³
= 22 × 3.9375 m³
= 86.625 m³.
Next, find the area of the canvas required. This is the curved surface area (CSA) of the cone.
We need the slant height (ℓ):
ℓ = √(r² + h²) = √((5.25)² + (3)²) m
ℓ = √(27.5625 + 9) m
ℓ = √36.5625 m
ℓ ≈ 6.0467 m (approximately 6.05 m as in original solution)
Area of the canvas required (CSA) = πrℓ
= (22/7) × 5.25 × 6.05 m²
= 22 × 0.75 × 6.05 m²
= 16.5 × 6.05 m²
= 99.825 m².
Thus, the required area of the canvas is 99.825 m².
(Session 2025 - 26)