GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Gujarat Board GSEB Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 10 Maths Chapter 13 Surface Areas and Volumes Ex 13.3

Question 1.
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
In case of sphere, r = 4.2 cm
Volume of sphere = \(\frac {4}{3}\)πr³ = \(\frac {4}{3}\)π(4.2)³
In case of cylinder, r = 6 cm and h = ?
Volume of cylinder = πr³h = π x 6³ x h
Volume of cylinder = Volume of sphere
⇒ π x 6² x h = \(\frac {4}{3}\)π (4.2)³
h = \(\frac{4 \times 4.2 \times 4.2 \times 4.2}{3 \times 6 \times 6}\)
h = 4 x 1.4 x 0.7 x 0.7
h = 2.74 cm

Question 2.
Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Solution:
Radii of solid sphere are
r₁ = 6 cm
r₂ = 8 cm
and r₃ = 10 cm
Let radius of resulting sphere be r.
Volume of three spheres = Volume of resulting sphere
\(\frac {4}{3}\)πr³₁ + \(\frac {4}{3}\)πr³₂ + \(\frac {4}{3}\)πr³₃ = \(\frac {4}{3}\)πr³
\(\frac {4}{3}\)π [r³₁ + r³₂ + r³₃] = \(\frac {4}{3}\)πr³
⇒ r³₁ + r³₂ + r³₃ = r³
6³ + 8³ + 10³ = r³
r³ = 216 + 512 + 1000
r³ = 1728
r = \(\sqrt[3]{1728}\)
r = 12 cm

Question 3.
A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform. [NCERT 20121]
Solution:
In case of cylindrical well
Diameter = 7 m
r = \(\frac {7}{2}\)m
Depth of well h = 20 m
Volume of well or earth dug out = πr²h
= \(\frac {22}{7}\) x \(\frac {7}{2}\) x \(\frac {7}{2}\) x 20 = 11 x 7 x 10 = 770 m³
In case of cuboidal platform
l = 22 m
b = 14 m
Let height of platform be h m.
Then volume of earth spread out in the form of cuboidal plate form = Volume of earth dug out from the well
l x b x h = 770
22 x 14 x h = 770
h = \(\frac{770}{22 \times 14}\) = \(\frac{70}{2 \times 14}\)
h = 2.5 m

Question 4.
A well of diameter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment. [CBSE 20121
Solution:
In case of cylindrical well
Diameter = 3 m
Radius r₁ = \(\frac {3}{2}\)m
Depth h = 14m
Volume of earth dug out from well
= πr²₁h
= π (\(\frac {3}{2}\))² x 14
In case of circular embankment
Radius of embankment be R
= r + 4
= \(\frac {3}{2}\) + 4 = \(\frac {11}{2}\) m
Let height of embankment be H.
Volume of embankment = Volume of earth dug out
πR²H – πr²H = πr²₁h
π(R² – r²)H = π(\(\frac {3}{2}\))² x 14

H = \(\frac {9}{8}\)
H = 1.125 m
Hence, the height of the embankment is 1.125 m.

Question 5.
A container shaped like a right circular cylinder having a diameter 12 cm and a height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
Solution:
In case of right circular cylinder, diameter = 12 cm
r₁ = 6 cm
h₁ = 15 cm
Volume of ice cream in right circular cylinder
= πr₁²h₁
= π x 6² x 15
In case of cone
Diameter = 6 cm
Radius r₂ = 3 cm
Height h₂ = 12 cm
Volume of cone = \(\frac {1}{3}\) πr₂²h₂
= \(\frac {1}{3}\) π x 3₂ x 12 = 36π cm³
Volume of hemisphere = \(\frac {2}{3}\) πr³
= \(\frac {2}{3}\)π(3)³ = 187πcm³
Number of cones filled with ice cream

Question 6.
How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm x 10 cm x 3.5 cm?
Solution:
Silver coins
Diameter = 1.75 cm
Radius r₁ = \(\frac {1.75}{2}\) cm = \(\frac {1.75}{200}\) = \(\frac {7}{8}\)cm
Thickness of coin h₁ = 2 mm = \(\frac {2}{10}\)cm = \(\frac {1}{5}\)cm
Volume of a coin = πr²₁h₁
= π x (\(\frac {7}{8}\))² x \(\frac {1}{5}\)
= \(\frac {22}{7}\) x \(\frac {49}{64}\) x \(\frac {1}{5}\)
= \(\frac{11 \times 7}{32 \times 5}\) cm²
In case of cuboid
Length (l) = 5.5 cm
Breadth (b) = 10 cm
Height (h) = 3.5 cm
Volume of cuboid = l x b x h
= 5.5 x 10 x 3.5 cm³
= 192.5 cm³

Hence, 400 silver coins must be method.

Question 7.
A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical help of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
In case of cylindrical bucket
Radius (r) of bucket = 18 cm
Height h = 32 cm
Volume of sand filled in cylindrical bucket
= πr²h
= π x 18² x 3²
= 10368 π cm³
In case of conical heap
Height (H) = 24 cm
Volume of conical heap = Volume of cylindrical bucket
\(\frac {1}{3}\)πR²H = 10368π
\(\frac {1}{3}\) x R² x 24 = 10368
R² = \(\frac{10368 \times 3}{24}\) = 1296
R = \(\sqrt{1296}\)
R = 36 cm
Hence radius of conical heap of sand = 36 cm
I = \(\sqrt{\mathrm{R}^{2}+\mathrm{H}^{2}}\) = \(\sqrt{36^{2}+24^{2}}\)
= \(\sqrt{1296+576}\) = \(\sqrt{1872}\) = 12 \(\sqrt{13}\)
Hence, slant height of sand = 12\(\sqrt{13}\) cm.

Question 8.
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?
Solution:
In case of canal
Width of canal (b) = 6 m
Depth of canal (h) = 1.5 m
Speed of flow of water = 10 km/h
= \(\frac{10 \times 1000 \mathrm{m}}{60 \mathrm{min}}\) = \(\frac{500}{3}\) m / minute
speed of water in 30 minute
= \(\frac{500}{3}\) x 30 = 5000 m
Volume of water flowed in 30 minutes
= l x b x h
= 6 x \(\frac{3}{2}\) x 5000 = 45000 m²
Volume of water stand in a field to irrigate = Volume of water flowed in 30 minutes
⇒ Area of field x \(\frac{8}{100}\) = 45000
( ∴ height = 8 cm = \(\frac{8}{100}\) m)
Area of field = \(\frac{45000 x 100}{8}\) = 562500
1 hectare = 10000 m²
Area of field irrigate = \(\frac{562500}{10000}\) = 56.25 hectares

Question 9.
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?
Solution:
Diameter in case of cylindrical pipe = 20 cm
Radius (r) = 10 cm = \(\frac{10}{100}\) m = 0.1 m
Speed of flow of water in pipe = 3 km / h
= \(\frac{3 \times 1000}{60 \times 60}\) = \(\frac{5}{6}\) m / s
Volume of water flowed per second
= πr²h = π(0.1)² x \(\frac{5}{6}\)
In case of tank
Diameter = 10 m
R = 5m
Depth H = 2m
Volume of tank = πr²h
= π x 5² x 2 = 50π
Time taken to fill the tank

= \(\frac{50 \pi}{\pi \times 0.1 \times 0.1 \times \frac{5}{6}}\)
= \(\frac{5000 \times 6}{1 \times 1 \times 5}\)
= \(\frac{6000}{60}\) minutes = 100 minutes
= \(\frac{100}{60}\) hour = 1 hour 40 minutes