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

Haryana State Board HBSE 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.3

Take π = \(\frac{22}{7}\) unless stated otherwise

Question 1.
A metallic sphere of radius 42 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
Solution:
We have,
Radius of sphere (r₁) = 42 cm
Volume of sphere = \(\frac{4}{3}\)πr₁³
= \(\frac{4}{3}\)π × (42)³ cm³
Radius of the cylinder (r₂) = 6 cm
Let height of cylinder be h cm. Then
Volume of cylinder = πr₂²h
= π × 6² × h
Volume of cylinder = Volume of sphere
⇒ π × 6² × h = \(\frac{4}{3}\)π × (4.2)³
⇒ h = \(\frac{\frac{4}{3} \pi \times 4.2 \times 4.2 \times 4.2}{\pi \times 6 \times 6}\)
⇒ h = 4 × 0.7 × 0.7 × 1.4
⇒ h = 2.74 cm
Hence, height of the cylinder = 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:
We have,
Radii of metallic sphere are 6 cm, 8 cm, and 10 cm respectively,.
Volume of 1st sphere (V₁) = \(\frac{4}{3}\)π × (6)³
Volume of IInd sphere (V₂) = \(\frac{4}{3}\)π × (8)³
Volume of IIIrd sphere (V₃) = \(\frac{4}{3}\)π × (10)³
Let the radius of resulting sphere be R cm.
Then
Volume of resulting sphere = V₁ + V₂ + V₃
⇒ \(\frac{4}{3}\)πR³ = \(\frac{4}{3}\)π × (6)³ + \(\frac{4}{3}\)π × (8)³ + \(\frac{4}{3}\)π × (10)³
⇒ \(\frac{4}{3}\)πR³ = \(\frac{4}{3}\)π [6³ + 8³ + 10³]
⇒ R³ = 216 + 512 + 1000
⇒ R³ = 1728.
⇒ R³ = 12 × 12 × 12 = 12³
⇒ R = 12
Hence, radius of resulting sphere = 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.
Solution:
We have,
Radius of well (r) = \(\frac{7}{2}\) = 3.5 m
Depth of the well (h) = 20 m
Volume of earth taken out of the well = πr²h

= \(\frac{22}{7}\) × 3.5 × 3.5 × 20 = 770 m³
Let the height of the platform be h₁ m
Base dimensions of the platform = 22m by 14 m
Volume of earth to form platform = Volume of earth taken put of the well
⇒ 22 × 14 × h₁ = 770
⇒ h₁ = \(\frac{770}{22 \times 14}\) = 2.5 m.
Hence, height of the platform = 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 2011]
Solution:
We have,
Radius of the well (r) = \(\frac{3}{2}\) = 1.5 m
Depth of the well (h) = 14 m

Volume of the earth dug out = πr²h
= π × (1.5)² × 14
Radius of the embankment (R) = 1.5 + 4 = 5.5 m
Let height of the embankment be h’ m
Volume of the earth used for making embankment = Volume of the earth dug out
⇒ πR²h’ – πr²h’ = π × (1.5)² × 14
⇒ πh’ (R² – r²) = π × 1.5 × 1.5 × 14
⇒ πh’ (5.5² – 1.5²) = π × 1.5 × 1.5 × 14
⇒ πh’ [(5.5 + 1.5) (5.5 – 1.5)] = π × 1.5 × 1.5 × 14
⇒ πh’ × 7 × 4 = π × 1.5 × 1.5 × 14
h’ = \(\frac{\pi \times 1.5 \times 1.5 \times 14}{\pi \times 7 \times 4}\)
⇒ h’ = \(\frac{2.25}{2}\)
⇒ h’ = 1.125 m
Hence, height of the embankment = 1.125 m.

Question 5.
A container shaped like a right circular cylinder having diameter 12 cm and 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:
We have
Radius of cylinder (R) = \(\frac{12}{2}\) = 6 cm
Height of cylinder (H) = 15 cm
Volume of ice cream which filled in cylinder = πR²H
= π × 6² × 15 cm³
Radius of conical part (r) = \(\frac{6}{2}\) = 3 cm
Height of conical part (A) = 12 cm
Radius of hemispherical part = 3 cm

Volume of one ice cream cone = Volume of conical part + Volume of hemispherical part
= \(\frac{1}{3}\) πr²h + \(\frac{2}{3}\) πr³
= \(\frac{1}{3}\) πr² (h + 2r)
= \(\frac{1}{3}\) π × 3² (12 + 2 × 3) d
= π × 3 × 18 cm³
Let the number of cones made be n
Volume of ice cream of n cones = Volume of ice cream which filled from cylinder
⇒ π × 3 × 18 × n = π × 6² × 15
n = \(\frac{\pi \times 6 \times 6 \times 15}{3 \times 18 \times \pi}\)
⇒ n = 10.
Hence, number of cones = 10.

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 × 10 cm × 3.5 cm ?
Solution:
We have,
Radius of the coin (r) = \(\frac{1.75}{2}\) cm
Thickness of coin (h) = 2 mm = 0.2 cm
Volume of 1 coin = πr²h
= \(\frac{22}{7} \times\left(\frac{1.75}{2}\right)^2\) × 0.2
Let number of coins made be n.
Volume of n coins = n × \(\frac{22}{7} \times\left(\frac{1.75}{2}\right)^2\) × 0.2 cm³
Volume of the cuboid = 5.5 × 10 × 3.5 cm³
Volume of n coins = Volume of the cuboid

⇒ n = 400.
Hence, required number of coins = 400.

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 heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.
Solution:
We have,
Radius of cylindrical bucket (r₁) = 18 cm
Height of cylindrical bucket (h₁) = 32 cm
Volume of sand in the bucket = πr₁²h₁
= π × 18² × 32 cm³
Height of the conical heap (h₂) = 24
Let radius of conical heap be r₂ cm
Volume of conical heap of sand = \(\frac{1}{3}\) πr₂²h₂
= \(\frac{1}{3}\) × π × r₂² × 24
Volume of conical heap of sand = Volume of sand in the bucket
⇒ \(\frac{1}{3}\) π × r₂² × 24 = π × 18² × 32
r₂² = \(\frac{\pi \times 18 \times 18 \times 32 \times 3}{\pi \times 24}\)
⇒ r₂² = 18 × 18 × 4
⇒ r₂² = 18² × 2²
⇒ r₂ = \(\sqrt{18^2 \times 2^2}\)
⇒ r₂ = 18 × 2 = 36 cm
Slant height of conical heap = \(\sqrt{h_2^2+r_2^2}\)
= \(\sqrt{24^2+36^2}\)
= \(\sqrt{576+1296}\)
= \(\sqrt{1872}\)
= \(\sqrt{12 \times 12 \times 13}\)
= 12√13 cm.

Hence, radius and slant height of conical heap are 36 cm and l2Ji cm respectively.

Question 8.
Water in a canal, 6 m wide and 15 m deep, is flowing with a speed of 10 km/hr.
How much area will it irrigate in 30 minutes, if 8
cm of standing water is needed?
Solution:
We have,
Depth of water of canal = 1.5 m
Width of canal = 6m
Speed of canal = 10 km/h = 10 × 1000 m/h = 10000 m/h
Volume of water flowing through canal in 60 minutes = 10000 × 6 × 1.5 m³
Volume of water flowing through canal in 30 minutes = \(\frac{10000 \times 6 \times 1.5 \times 30}{60}\)
= 5000 × 6 × 15 m³
Height of standing water = 8 cm = \(\frac{8}{100}\) m
Let required area be x m²
Volume of standing water = Volume of water flowing through canal in 30 minutes
\(\frac{x \times 8}{100}\) = 5000 × 6 × 1.5
[∵ Volume of standing water = area × height]
⇒ x = \(\frac{5000 \times 6 \times 1.5 \times 100}{8}\)
⇒ x = 562500 m².
⇒ Area = 562500 m² = \(\frac{562500}{10000}\)
= 56.25 hectares.
Hence, area will irrigate = 562500 m² or 56.25 hectares.

Question 9.
A farmer cormects a pipe of internal diameter 20 cm from i» nal i,t’ r cylindrical tank in her field, which i’ I m in dia rnet and 2 m deep. If water flows through hc pipe at tiw rate of 3 km/h, in how much time will the Lank be filled?
Solution:
We have,
Internal diameter of pipe = 20 cm
Radius of the pipe = \(\frac{20}{2}\) = 10 cm
= \(\frac{10}{100}\) = \(\frac{1}{10}\) m.
Speed of water = 3 km/h = 3000 m/h
Volume of water that flows 1h rough the pipe in 1h = π × (\(\frac{1}{10}\))² × 3000
= 30π m³
Radius of cylindrical tank (r) = \(\frac{10}{2}\) = 5 m
Depth of the tank (h) = 2 m
Volume of the tank = πr²h
= π × 5² × 2
= 50π m³
Time taken to filled the tank = \(\frac{\text { Volume of the tank }}{\text { Volume of water flows in } 1 \mathrm{hr}}\)
= \(\frac{50 \pi}{30 \pi}=\frac{5}{3}\) h
= 1 h 40 minutes.
Hence, required time = 1 h 40 minutes OR 100 minutes.