GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Gujarat Board GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.5

Question 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:
In case of cylinder
Diameter = 10 cm
radius r = 5 cm

Diameter of wire which is rounded over cylinder = 3 mm
Length of cylinder = 12 cm
= 12 x 10 mm = 120 mm
Number of round to cover 120 mm
= \(\frac {120}{3}\) = 40
Length of wire in completing one round over the cylinder = 2πr
= 2π x 5 = 10π cm
Length of wire in covering whole surface of the cylinder = Length of wire in completing 40 rounds
= 10 π x 40
= 400 π cm
= 400 x 3.14 = 1256 cm
Radius of copper wire = \(\frac {3}{2}\)mm
r₁ = \(\frac {3}{20}\)cm
Volume of wire = πr²₂h
= π (\(\frac {3}{20}\)) x 400n = 9π² cm³
Mass of wire = 9 π² x 8.88
= 9 x (3.14)² x 8.88 g
= 787.98g ≈ 788g

Question 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 n as found appropriate.)
Solution:
Here
ABC is a right triangle in ∠BAC = 90°

BC² = AB² + AC²
= 3² + 4² = 9 + 16
BC² = 25
BC = 5 cm
Here AD and A’D are radii of two cones which are formed by rotating the right triangle about the hypotenuse BC.
ΔADB – ΔCAB (by AA similarity)
\(\frac {AD}{AC}\) = \(\frac {AB}{BC}\)
⇒ \(\frac {AD}{4}\) = \(\frac {3}{5}\)
⇒ AD = cm (r₁ or r₂ = AD let)
Also \(\frac {BD}{AB}\) = \(\frac {AB}{BC}\)
\(\frac {BD}{3}\) = \(\frac {3}{5}\)
⇒ BD = \(\frac {3}{5}\) cm (h₁ = BD say)
Hence CD = 5 – \(\frac {9}{5}\)
h₂ = \(\frac {16}{5}\)cm (h₂ CD say)
Volume of double cones

And surface area of the double cones
= πr₁l₁ + πr₂l₂ = π (r₁l₁ + r₂l₂)
= 3.14\(\left[\frac{12}{5} \times 3+\frac{12}{5} \times 4\right]\) [l₁ = 3cm , l₂ = 4cm ]
= 3.14 x \(\frac {12}{5}\) [3 + 4]
= \(\frac{3.14 \times 12 \times 7}{5}\) = 52.75 cm²

Question 3.
A cistern, internally measuring 150 cm x 120 cm x 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 x 7.5 cm x 6.5 cm?
Solution:
Volume of cistern = l x b x h
= 150 x 120 x 110
= 1980000 cm³
Volume of water in cistern = 129600 cm³
Volume of cistern to be filled
= Volume of cistern – Volume of water
= 1980000 – 129600
= 1850400 cm³
Let n bricks are needed to fill the cistern up to the brim.
Volume of one brick
= l x b x h
= 22.5 x 7.5 x 6.5
= 1096.875 cm³
Water absorbed by one brick = \(\frac{1096.875}{17}\) cm³
Water absorbed by n brick = \(\frac{n(1096.875)}{17}\) cm³
According to the question,

n = 1792.41
Hence number of bricks cannot be put in the cistern = 1792

Question 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 and each 1072 km long, 75 m wide and 3 m deep.
Solution:
Area of valley = 97280 km²
Rainfall = 10 cm
= \(\frac {10}{100}\) metre = \(\frac{10}{100 \times 1000}\) km
Volume of rainfall = Area of river valley x Rainfall
= 7280 x \(\frac{10}{100 \times 1000}\)
= 0.728 km³
Volume of three river = 3l x b x h
= 3 x 1072 x \(\frac {75}{1000}\) x \(\frac {3}{1000}\)
= 0.7236 km³
Hence, total rainfall and normal water of three rivers are approximately equivalent.

Question 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 figure).

Solution:
Here mease of frustum
r₁ = \(\frac {18}{2}\) = 9 cm
r₁ = \(\frac {8}{2}\) = 4 cm
h = 22 – 10 = 12 cm
Slant height of frustum
l = \(\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}\)
=\(\sqrt{(9-4)^{2}+12^{2}}\) = \(\sqrt{5^{2}+12^{2}}\)
= \(\sqrt{25+144}\) = \(\sqrt{169}\) = 13cm
Area of tin sheet required = Curved surface area of the cylindrical portion + Curved surface area of the frustum portion
= 2πr₂h₂ + π(r₁ + r₂)l
= 2 x π x 4 x 10 + π(4 + 9) x 13
= 80π x 169π = 249π
= 249 x \(\frac{22}{7}\) = \(\frac{5478}{7}\) = 782 \(\frac{4}{7}\) cm²

Question 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 are explained.
Solution:
Let r₁ and r₂ be the radii of circular base, where r₁ > r₂ and h be the height of the frustum of a cone.

Let h₁ be the height of large cone LPQ
Height of small cone = h₁ – h
Let l₁ be the slant height of cone LPQ and l be the frustum of cone. The slant height of Cone LRS
= l₁ – l
In rt. ΔLMQ and rt. ΔLNS
∠1 = ∠1 (common)
∠M = ∠N (each 900)
∴ ΔLMQ – ΔLNS (by AA similarity)

(i) Lateral surface area or curved surface area of a frustum of cone = CSA of large cone – CSA of small cone
= πr₁l₁ – πr₂(l₁ – l)

= πl(r₁ + r₂) = π(r₁ + r₂)l sq. units
(ii) Slant height = \(\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}\)
(iii) Total surface area of frustum of a cone + πr₁r² + πr₂r²
= CSA of frustum of cone + πr²₁ + πr²₂
= π(r₁ + r₂)l + πr²₁ + πr²₂ sq. units

Question 7.
Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
Solution:
Let r₁ and r₂ be radii of two circular bases where r₁ > r₂, h be the height of frustum of a cone.

Let h₁ be the height of large cone LPD and height of small cone LRS = h₁ – h
Let l₁, be the slant height of cone and t be the slant height of frustum of a cone.
Slant height of cone LRS = l₁ – l
In rt. ΔLMQ and rt. ΔLNS (Common)
∠M = ∠N (Each 900)
ΔLMN – ΔLNS (By AA similarity)

Volume of a frustum of a cone = Volume of large cone LPQ – Volume of small cone LRS