The DAV Maths Book Class 8 Solutions Pdf and DAV Class 8 Maths Chapter 14 Worksheet 6 Solutions of Mensuration offer comprehensive answers to textbook questions.
DAV Class 8 Maths Ch 14 WS 6 Solutions
Question 1.
Find the volume of a right circular cylinder if
(i) radius of its base = 7 cm, height = 1.5 dm
(ii) area of base = 154 cm2, height = 1.5 cm
(iii) circumference of base = 1.32 m, height = 25 cm
Solution:
(i) Here r = 7 cm, h = 1.5 dm = 150 cm
∴ The volume of the cylinder = πr2h
= \(\frac{1}{2}\) × 7 × 7 × 150
= 23100 cm3
(ii) Here πr2 = 154 cm2 and h = 1.5 cm
∴ The volume of the cylinder = πr2h
= 154 × 1.5
= 231 cm3
(iii) Here 2πr = 1.32 m and h = 25 cm
2πr = 132 cm
⇒ 2 × \(\frac{22}{7}\) × r = 132
⇒ r = \(\frac{132 \times 7}{2 \times 22}\)
⇒ r = 21 cm
∴ The volume of the cylinder = πr2h
= \(\frac{22}{7}\) × 21 × 21 × 25
= 346.50 cm3
Question 2.
How many cubic metres of the earth must be dug out to sink a healthy 16 m in depth and 7 m in diameter?
Solution:
Here h = 16 m and r = \(\frac{7}{2}\) m
∴ The volume of the earth to be dug out = πr2h
= \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 16\)
= 616 m3
Hence the required volume = 616 m3
Question 3.
A water tank is cylindrical in shape and the diameter of its base is 28 m. If it is 7 meters deep, how many kilolitres of water can it hold?
Solution:
Here r = \(\frac{28}{2}\) = 14 m and h = 7 m
∴ The volume of the tank = πr2h
= \(\frac{22}{7}\) × 14 × 14 × 7
= 4312 m3
= 4312 kl [∵ 1 m3 = 1 kl]
Question 4.
The radius of the base of a cylindrical oil can is 4 m. Find its height if it can contain 1408 kilolitres of oil.
Solution:
1 m3 = 1 kl
∴ Volume of the oil = 1408 kl = 1408 m3
∴ πr2h = 1408
⇒ \(\frac{22}{7}\) × 4 × 4 × h = 1408
⇒ h = \(\frac{1408 \times 7}{22 \times 4 \times 4}\)
⇒ h = 28 m
Hence, the required height = 28 m
Question 5.
The radius and height of a cylinder are in the ratio 5 : 7 and its volume is 550 cm3. Find its radius.
Solution:
Let the radius be 5x cm and the height be 7x cm.
∴ Volume = πr2h
⇒ 550 = \(\frac{22}{7}\) × 5x × 5x × 7x
⇒ 550 = 550x3
⇒ x3 = 1
⇒ x = 1
Hence, the required radius = 5 × 1 = 5 cm
Question 6.
The thickness of a hollow metallic cylinder is 2 cm. It is 35 cm long and its inner radius is 12 cm. Find the metal volume required to make the cylinder, assuming it is open, at either end.
Solution:
The inner radius, r2 = 12 cm
thickness = 2 cm
∴ outer radius, r2 = 12 + 2 = 14 cm
height h = 35 cm
∴ Volume of the metal required = \(\pi r_2^2 h-\pi r_1^2 h\)
= \(\pi\left(r_2^2-r_1^2\right) h\)
= \(\frac{22}{7}\) × (142 – 122) × 35
= \(\frac{22}{7}\) × (196 – 144) × 35
= \(\frac{22}{7}\) × 52 × 35
= 5720 cm3
Hence, the required metal = 5720 cm3
Question 7.
A rectangular sheet of paper is rolled along its length to make a cylinder. The sheet is 33 cm long and 32 cm wide. A circular sheet of paper is attached to the bottom of the cylinder formed. Find the capacity of the cylinder so developed.
Solution:
When the sheet is rolled along its length
∴ 2πr = 33
⇒ 2 × \(\frac{22}{7}\) × r = 33
⇒ r = \(\frac{33 \times 7}{2 \times 22}=\frac{21}{4}\) cm and h = 32 cm
∴ The volume of the cylinder so formed = πr2h
= \(\frac{22}{7} \times \frac{21}{4} \times \frac{21}{4} \times 32\)
= 2772 cm3
Hence, the required volume = 2772 cm3
Question 8.
Two solids of right cylindrical shape are 49 cm and 35 cm high and their base diameters are 16 cm and 14 cm respectively. Both are melted and moulded into a single cylinder, 56 cm high. Find its base diameter.
Solution:
The volume of the first cylinder V1 = πr2h
= \(\frac{22}{7}\) × 8 × 8 × 49
= 9856 cm3
The volume of the second cylinder V2 = πr2h
= \(\frac{22}{7}\) × 7 × 7 × 35
= 5390 cm3
Volume of the moulded cylinder = V1 + V2
= 9856 + 5390
= 15246 cm3
∴ πr2h = 15246
⇒ \(\frac{22}{7}\) × r2 × 56 = 15246
⇒ r2 = \(\frac{15246 \times 7}{22 \times 56}=\frac{693}{8}\)
⇒ r2 = 86.6
⇒ r = 9.3 cm
Hence, the required diameter = 2 × 9.3 = 18.6 cm
Question 9.
The volume of a metallic cylindrical pipe is 748 cm3. Its length is 14 cm and its external radius is 9 cm. Find its thickness.
Solution:
Let its internal radius be r1 cm and external radius be r2 cm.
∴ Volume of the metallic pipe = \(\pi\left[r_2^2-r_1^2\right] \times h\)
⇒ 748 = \(\frac{22}{7}\left[9^2-r_1^2\right] \times 14\)
⇒ \(\frac{748 \times 7}{22 \times 14}=9^2-r_1^2\)
⇒ 17 = 81 – \(r_1^2\)
⇒ \(r_1^2\) = 81 – 17 = 64
⇒ r1 = 8 cm
Hence, the required thickness = r2 – r1 = 9 – 8 = 1 cm
Question 10.
The volume of a cylinder is 150π cu. cm and its height is 6 cm. Find the areas of its total surface and lateral (curved) surface.
Solution:
The volume of the Cylinder = πr2h
⇒ 150π = π × r2 × 6
⇒ r2 = \(\frac{150 \pi}{\pi \times 6}\) = 25
⇒ r = 5 cm
∴ T.S.A. = 2πr[h + r]
= 2 × \(\frac{22}{7}\) × 5 [6 + 5]
= 2 × \(\frac{22}{7}\) × 5 × 11
= \(\frac{2420}{7}\) cm2
C.S.A. = 2πrh
= 2 × \(\frac{22}{7}\) × 5 × 6
= \(\frac{1320}{7}\) cm2
Hence, the required T.S.A. = \(\frac{2420}{7}\) cm2 and C.S.A. = \(\frac{1320}{7}\) cm2
Question 11.
The curved surface area of a cylindrical pillar is 264 m2 and its volume is 924 m3. Find the diameter and the height of the pillar.
Solution:
Curved surface, 2πrh = 264 …..(i)
Volume, πr2h = 924 …..(ii)
Dividing (ii) by (i), we get
\(\frac{\pi r^2 h}{2 \pi r h}=\frac{924}{264}\)
⇒ \(\frac{r}{2}=\frac{7}{2}\)
⇒ r = 7 cm
Putting the value of r in eq. (i)
2 × \(\frac{22}{7}\) × 7 × h = 264
⇒ h = \(\frac{264 \times 7}{2 \times 22 \times 7}\) = 6 cm
Hence, the required diameter = 2 × 7 = 14 cm and the height = 6 cm