NCERT Solutions Class 10 Maths Chapter 13 – Surface Area and Volumes Exercise 13.5

NCERT Solutions for Class 10 Maths Chapter 13 Exercise 13.5 help students understand and solve questions related to spherical solids in a clear and exam-oriented manner. This exercise strengthens conceptual clarity and improves accuracy in solving surface area and volume-based problems.

Designed as per the latest CBSE Class 10 Maths syllabus, Exercise 13.5 focuses on applying geometric formulas correctly and solving numerical problems step-by-step. Regular practice of this exercise helps students improve speed, reduce calculation errors, and gain confidence for board examinations.

NCERT Solutions Class 10 Maths Chapter 13 Exercise 13.5 – Surface Area and Volumes

Surface Areas and Volumes

Medium

Q.

$\begin{array}{l}\text{A container, opened from the top and made up of a metal sheet, }\\ \text{is in the form of a frustum of a cone of height 16 cm with radii of its }\\ \text{lower and upper ends as 8 cm and 20 cm, respectively. }\\ \text{Find the cost of the milk which can completely fill the container, at }\\ \text{the rate of ₹ 20 per litre. Also find the cost of metal sheet used to }\\ {\text{make the container, if it costs ₹\hspace{0.17em}8 per 100 cm}}^{\text{2}}\text{. (Take }\mathrm{\pi }\text{ = 3.14)}\end{array}$

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Surface Areas and Volumes

Medium

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.

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Surface Areas and Volumes

Medium

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.

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Surface Areas and Volumes

Medium

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.)

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Surface Areas and Volumes

Medium

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?

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Surface Areas and Volumes

Medium

Q.

In one fortnight of a given month, there was a rainfall of 10 cm in a river valley. If the area of the valley is 7280 km², show that the total rainfall was approximately equivalent to the addition to the normal water of three rivers each 1072 km long, 75 m wide and 3 m deep.

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Surface Areas and Volumes

Medium

Q.

An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the following figure).

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Surface Areas and Volumes

Medium

Q.

Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

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Surface Areas and Volumes

Medium

Q.

Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.

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Surface Areas and Volumes

Medium

Q.

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹ 500 per m². (Note that the base of the tent will not be covered with canvas.)

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These solutions are prepared in a structured format so students can easily understand the method of solving problems and perform better in exams.

Q1. Copper wire on a cylinder: length & mass

Given:
Wire diameter = 3 mm = 0.3 cm ⇒ wire radius = 0.15 cm
Cylinder height (h) = 12 cm
Cylinder diameter = 10 cm ⇒ cylinder radius (R) = 5 cm
Density of copper = 8.88 g/cm³

(A) Length of wire

Curved surface area (CSA) of cylinder = 2πRh
= 2 × 3.14 × 5 × 12
= 376.8 cm²

One round covers height equal to wire diameter (0.3 cm), so:
Length of wire × wire diameter = CSA of cylinder

So,
Length (L) = 376.8 / 0.3
= 1256 cm

(B) Mass of wire

Wire is cylindrical, so volume of wire:
V = πr²L
= 3.14 × (0.15)² × 1256
= 3.14 × 0.0225 × 1256
= 88.7364 cm³

Mass = Density × Volume
= 8.88 × 88.7364
= 787.98 g ≈ 788 g

✅ Answer: Length = 1256 cm, Mass ≈ 788 g

Q2. Right triangle revolved about hypotenuse: volume & surface area of double cone

Given: Right triangle with legs 3 cm and 4 cm
Hypotenuse AC = √(3²+4²) = 5 cm

Radius of double cone = altitude from right angle to hypotenuse:
r = (AB × BC) / AC = (3×4)/5 = 12/5 = 2.4 cm

Heights of two cones along hypotenuse:
AD = AB²/AC = 9/5 = 1.8 cm
DC = BC²/AC = 16/5 = 3.2 cm
(AD + DC = 5 cm)

(A) Volume

Total volume = (1/3)πr²(AD + DC)
= (1/3) × 3.14 × (2.4)² × 5
= (1/3) × 3.14 × 5.76 × 5
= 30.144 cm³
≈ 30.14 cm³

(B) Surface area (curved)

Slant heights are the triangle legs: l₁ = 3 cm, l₂ = 4 cm
CSA = πr(l₁ + l₂)
= 3.14 × 2.4 × (3+4)
= 3.14 × 2.4 × 7
= 52.752 cm²
≈ 52.75 cm²

✅ Answer: Volume ≈ 30.14 cm³, Surface area ≈ 52.75 cm²

Q3. Bricks in cistern

Given:
Cistern = 150 × 120 × 110 cm³
Water = 129600 cm³
Brick = 22.5 × 7.5 × 6.5 cm³
Each brick absorbs 1/17 of its own volume

Volumes

Cistern volume = 150×120×110 = 1980000 cm³
Brick volume = 22.5×7.5×6.5 = 1096.875 cm³

Empty space above water initially:
= 1980000 − 129600
= 1850400 cm³

Each brick effectively fills:
Brick volume − absorbed water
= V − (1/17)V
= (16/17)V
= (16/17)×1096.875
= 1032.3529 cm³ (approx.)

Let number of bricks = x
x × 1032.3529 ≤ 1850400
x ≤ 1850400 / 1032.3529
x ≤ 1792.41

So maximum whole bricks = 1792

✅ Answer: 1792 bricks

Q4. Rainfall vs 3 rivers

Given:
Rainfall height = 10 cm = 0.1 m
Valley area = 7280 km² = 7280×10⁶ m² = 7.28×10⁹ m²

Volume of rainfall

V = Area × height
= (7.28×10⁹) × 0.1
= 7.28×10⁸ m³

River (one):
Length = 1072 km = 1.072×10⁶ m
Width = 75 m, Depth = 3 m

Volume of one river = l×b×h
= 1.072×10⁶ × 75 × 3
= 2.412×10⁸ m³

Volume of 3 rivers = 3 × 2.412×10⁸
= 7.236×10⁸ m³

Since 7.236×10⁸ ≈ 7.28×10⁸, rainfall volume is approximately equal.

✅ Hence proved.

Q5. Tin sheet area for oil funnel

Given:
Cylinder height = 10 cm, cylinder diameter = 8 cm ⇒ r = 4 cm
Total height = 22 cm ⇒ frustum height = 22 − 10 = 12 cm
Top diameter = 18 cm ⇒ r₁ = 9 cm
Bottom radius (same as cylinder top) r₂ = 4 cm

Slant height of frustum:
l = √[(r₁−r₂)² + h²]
= √[(9−4)² + 12²]
= √(25 + 144)
= √169
= 13 cm

Area required = CSA(frustum) + CSA(cylinder)

CSA frustum = π(r₁ + r₂)l
= π(9+4)13
= 169π

CSA cylinder = 2πrh
= 2π(4)(10)
= 80π

Total area = (169π + 80π) = 249π
Using π = 22/7:
Area = 249×22/7 = 782.57 cm² (approx.)

✅ Answer: ≈ 782.57 cm²

Q6. Derive CSA and TSA of frustum of a cone

To prove:
CSA = πl(r₁ + r₂)
TSA = πl(r₁ + r₂) + πr₁² + πr₂²

Idea:
Extend the slant sides of frustum to meet at O, forming two cones: big cone (radius r₁) minus small cone (radius r₂).

Let slant heights be l₁ and l₂ for big and small cones, so frustum slant height:
l = l₁ − l₂

By similarity of triangles in the two cones:
r₁/l₁ = r₂/l₂ ⇒ r₁l₂ = r₂l₁

Now,
CSA(frustum) = CSA(big cone) − CSA(small cone)
= πr₁l₁ − πr₂l₂
= π(r₁l₁ − r₂l₂)

Using similarity result, you can rewrite:
r₁l₁ − r₂l₂ = (r₁ + r₂)(l₁ − l₂) = (r₁ + r₂)l

So,
✅ CSA = πl(r₁ + r₂)

Then,
✅ TSA = CSA + area of both circular ends
= πl(r₁ + r₂) + πr₁² + πr₂²

Hence proved.

Q7. Derive volume of frustum of a cone

To prove:
V = (1/3)πh(r₁² + r₂² + r₁r₂)

Idea:
Frustum = volume of big cone − volume of small cone.

Big cone: radius r₁, height h₁
Small cone: radius r₂, height h₂
Frustum height h = h₁ − h₂

By similarity of cones:
r₁/h₁ = r₂/h₂ ⇒ h₁ = (r₁/r₂)h₂

Now,
V(frustum) = (1/3)π(r₁²h₁ − r₂²h₂)

Using similarity relations + h = h₁ − h₂, algebra simplification gives:
✅ V = (1/3)πh(r₁² + r₂² + r₁r₂)

Hence proved.

FAQs – Class 10 Maths Chapter 13 Exercise 13.5

Q1. What is the focus of Exercise 13.5?
Exercise 13.5 focuses on solving problems related to surface area and volume of spherical solids in an exam-friendly format.

Q2. Why is Exercise 13.5 important for board exams?
This exercise includes direct formula-based and application-based questions that are commonly asked in CBSE board exams.

Q3. How can students prepare effectively for Exercise 13.5?
Students should revise all relevant formulas, practice multiple numerical problems, and solve previous years’ board questions to strengthen their preparation.

Q4. How do NCERT Solutions help in scoring better marks?
NCERT Solutions provide step-by-step explanations, clear methods, and accurate calculations that help students avoid mistakes and score full marks in exams.