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

1. 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.

Solution:

Given that, Diameter of cylinder = 10 cm
So, radius of the cylinder (r) = 10/2 cm = 5 cm
∴ Length of wire in completely one round = 2πr = 3.14×5 cm = 31.4 cm
It is given that diameter of wire = 3 mm = 3/10 cm
∴ The thickness of cylinder covered in one round = 3/10 m
Hence, the number of turns (rounds) of the wire to cover 12 cm will be-

Now, the length of wire required to cover the whole surface = length of wire required to complete 40 rounds
40 x 31.4 cm = 1256 cm
Radius of the wire = 0.3/2 = 0.15 cm
Volume of wire = Area of cross-section of wire × Length of wire
= π(0.15)²×1257.14
= 88.898 cm³
We know,
Mass = Volume × Density
= 88.898×8.88
= 789.41 gm

2. 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)

Solution: Draw the diagram as follows:

Let us consider the ABA
Here, AS = 3 cm, AC = 4 cm
So, Hypotenuse BC = 5 cm
We have got 2 cones on the same base AA’ where the radius = DA or DA’
Now, AD/CA = AB/CB
By putting the value of CA, AB and CB we get,
AD = 2/5 cm
We also know,
DB/AB = AB/CB
So, DB = 9/5 cm
As, CD = BC-DB,
CD = 16/5 cm
Now, volume of double cone will be

Solving this we get,
V = 30.14 cm³
The surface area of the double cone will be

= 52.75 cm²

3. A cistern, internally measuring 150 cm × 120 cm × 100 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 being 22.5 cm × 7.5 cm × 6.5 cm?

Solution: Given that the dimension of the cistern = 150 × 120 × 110
So, volume = 1980000 cm³
Volume to be filled in cistern = 1980000 – 129600
= 1850400 cm3
Now, let the number of bricks placed be “n”
So, volume of n bricks will be = n×22.5×7.5×6.5
Now as each brick absorbs one-seventeenth of its volume, the volume will be
= n/(17)×(22.5×7.5×6.5)
For the condition given in the question,
The volume of n bricks has to be equal to volume absorbed by n bricks + Volume to be filled in cistern
Or, n×22.5×7.5×6.5 = 1850400+n/(17)×(22.5×7.5×6.5)
Solving this we get,
n = 1792.41

4. 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.

Solution: From the question, it is clear that
Total volume of 3 rivers = 3×[(Surface area of a river)×Depth]
Given, Surface area of a river = [1072×(75/1000)] km
And, Depth = (3/1000) km
Now, volume of 3 rivers = 3×[1072×(75/1000)]×(3/1000)
= 0.7236 km³
Now, volume of rainfall = total surface area × total height of rain

= 0.7280 km³
For the total rainfall was approximately equivalent to the addition to the normal water of three rivers, the volume of rainfall has to be equal to volume of 3 rivers.
But, 0.7280 km³ = 0.7236 km³
So, the question statement is true.

5. 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 Fig.).

Solution: Given,
Diameter of upper circular end of frustum part = 18 cm
So, radius (r₁) = 9 cm
Now, the radius of the lower circular end of frustum (r₂) will be equal to the radius of the circular end of the cylinder
So, r₂ = 8/2 = 4 cm
Now, height (h₁) of the frustum section = 22 – 10 = 12 cm
And,
Height (h₂) of cylindrical section = 10 cm (given)
Now, the slant height will be-

Or, l = 13 cm
Area of tin sheet required = CSA of frustum part + CSA of cylindrical part
= π(r₁+r₂)l+2πr₂h₂
Solving this we get,
Area of tin sheet required = 782 4/7 cm²

6. 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.

Solution: Consider the diagram

The frustum of a cone can be viewed as a difference of two right circular cones OAB and OCD.
Let h₁ and l₁ be the height and slant height of cone OAB and h₂ and l₂ be the height and slant height of cone OCD respectively.
In ΔAPO and ΔDQO
∠APO = ∠DQO = 90° (Since both cones are right circular cones)
∠AOP = ∠DOQ (Common)
Therefore, ΔAPO ∼ ΔDQO (AA criterion of similarity)
AP/DQ = AO/DO = OP/OQ (Corresponding sides of similar triangles are proportional)
⇒ r₁/r₂ = l₁/l₂ = h₁/h₂
⇒ r₁/r₂ = l₁/l₂ or ⇒ r₂/r₁ = l₂/l₁
Subtracting 1 from both sides we get
r₁ / r₂ – 1 = l₁ / l₂ – 1
(r₁ – r₂)/r₂ = (l₁ – l₂) / l₂
(r₁ – r₂)/r₂ = l/l₂ [From diagram, l₁ – l₂ = l]
l₂ = lr₂/(r₁ – r₂)…. (i)
or r₂ / r₁ = l₂ / l₁
Subtracting 1 from both sides we get
r₂ / r₁ – 1 = l₂ / l₁ – 1
(r₂ – r₁) / r₁ = (l₂ – l₁) / l₁
(r₁ – r₂) / r₁ = (l₁ – l₂) / l₁
(r₁ – r₂) / r₁ = l / l₁
l₁ = lr₁/(r₁ – r₂) …. (ii)

(i) CSA of frustum of cone = CSA of cone OAB – CSA of cone OCD
= πr₁l₁ – πr₂l₂
= π (r₁l₁ – r₂l₂)
= π [(r₁ × lr₁/(r₁ – r₂) – r₂ × lr₂/(r₁ – r₂)] [Using (i) and (ii)]
= π [(lr₁² – lr₂²)/(r₁ – r₂)]
= π [l(r₁² – r₂²)/(r₁ – r₂)]
= π [l(r₁ – r₂)(r₁ + r₂)/(r₁ – r₂)] [Since, a² – b² = (a – b)(a + b)]
= π l(r₁ + r₂)
TSA of frustum of cone = CSA of frustum + Area of lower circular end + Area of top circular end
= πl (r₁ + r₂) + πr₁² + πr₁²
Therefore, CSA of the frustum of the cone = πl (r₁ + r₂)
TSA of the frustum of the cone = πl (r₁+ r₂) + πr₁² + πr₂²

7. Derive the formula for the volume of the frustum of a cone.

Solution: Consider the same diagram as the previous question.

The frustum of a cone can be viewed as a difference of two right circular cones OAB and OCD.
Let h₁ and l₁ be the height and slant height of cone OAB and h₂ and l₂ be the height and slant height of cone OCD respectively.
In ΔAPO and ΔDQO
∠APO = ∠DQO = 90° (Since both cones are right circular cones)
∠AOP = ∠DOQ (Common)
Therefore, ΔAPO ∼ ΔDQO ( A.A criterion of similarity)
AP/DQ = AO/DO = OP/OQ (Corresponding sides of similar triangles are proportional)
⇒ r₁/r₂ = l₁/l₂ = h₁/h₂
⇒ r₁/r₂ = h₁/h₂ or ⇒ r₂/r₁ = h₂/h₁
Subtracting 1 from both sides
r₁/r₂ – 1 = h₁/h₂ – 1
(r₁ – r₂)/r₂ = (h₁ – h₂)/h₂
(r₁ – r₂)/r₂ = h/h₂ ….(i) [From figure h₁ – h₂ = h]
h₂ = hr₂/(r₁ – r₂) …… (i)
Now, considering r₂/r₁ = h₂/h₁
Subtracting 1 from both sides we get
r₂/r₁ – 1 = h₂/h₁ – 1
(r₂ – r₁)/r₁ = (h₂ – h₁)/h₁
(r₁ – r₂)/r₁ = (h₁ – h₂)/h₁
(r₁ – r₂)/r₁ = h/h₁
h₁ = hr₁/(r₁ – r₂) ….(ii)
Volume of frustum of cone = Volume of cone OAB – Volume of cone OCD
= 1/3 πr₁²h₁ – 1/3 πr₂²h₂
= 1/3π [r₁²h₁ – r₂²h₂]
= 1/3π [r₁² × hr₁/(r₁ – r₂) – r₂² × hr₂(r₁ – r₂)] [Using (i) and (ii)]
= 1/3π [hr₁³/(r₁ – r₂) – hr₂³/(r₁ – r₂)]
= 1/3πh [(r₁³ – r₂³)/(r₁ – r₂)]
= 1/3πh [(r₁ – r₂)(r₁² + r₁r₂ + r₂²)/(r₁ – r₂)] [Since, (a³ – b³) = (a – b)(a² + ab + b²)]
= 1/3πh (r₁² + r₂² + r₁r₂)
Hence proved, volume of the frustum = 1/3πh (r₁² + r₂² + r₁r₂)