ML Aggarwal Mensuration Chapter Test Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of Chapter Test Questions for Mensuration as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.
ML Aggarwal Mensuration Chapter Test Class 10 ICSE Maths Solutions
| Board | ICSE |
| Subject | Maths |
| Class | 10th |
| Chapter-17 | Mensuration |
| Writer / Book | Understanding |
| Topics | Solutions of Chapter Test |
| Academic Session | 2024-2025 |
Mensuration Chapter Test
ML Aggarwal Class 10 ICSE Maths Solutions
Question 1. A cylindrical container is to be made of tin sheet. The height of the container is 1 m and its diameter is 70 cm. If the container is open at the top and the tin sheet costs Rs 300 per m2, find the cost of the tin for making the container.
Answer :
Height of container opened at the top (h) = 1 m = 100 cm
and diameter = 70 cm
∴Radius (r) = 70/2 = 35 cm
∴Total surface area = 2πrh + πr2
= πr(2h + r)
= 22/7 × 35(2×100 + 35) cm2
= 110(200 + 35)
= 110 × 235 cm2
= (110 × 235)/(100 × 100)m2
= 517/200 m2
∴ Area of sheet required = 517/200 m2
Cost of 1 m2 sheet = ₹ 300
∴ Total cost = 517/200 × 300
= ₹ 1551/2
= ₹ 775.50
Question 2. A cylinder of maximum volume is cut out from a wooden cuboid of length 30 cm and cross-section of square of side 14 cm. Find the volume of the cylinder and the volume of wood wasted.
Answer :
Dimensions of the wooden cuboid = 30 cm × 14 cm × 14 cm
Volume = 30 × 14 × 14 = 5880 cm3
Largest size of cylinder cut out of the wooden cuboid will be of diameter = 14 cm
And height = 30 cm
∴ Radius of cylinder = 14/2= 7 cm
Volume of cylinder = πr2h
= 22/7 × 7 × 7 × 30 cm3
= 4620 cm3
∴ Volume of wooden wasted = 5880 – 4620 = 1260 cm3
Question 3. Find the volume and the total surface area of a cone having slant height 17 cm and base diameter 30 cm. Take π = 3.14.
Answer :
Slant height of a cone (l) = 17 cm
Diameter of base = 30 cm
Radius (r) = 30/2 = 15 cm
Question 4. Find the volume of a cone given that its height is 8 cm and the area of base 156 cm2.
Answer :
Height of a cone = 8 cm
Area of base = 156 cm
.’. Volume = 1/3 × area of base × height
= 1/3 × 156 × 8
= 1248/3 cm3
= 416 cm3
Question 5. The circumference of the edge of a hemispherical bowl is 132 cm. Find the capacity of the bowl.
Answer :
Circumference of the edge of bowl = 132 cm
Radius of a hemispherical bowl
= 132/2π = (132 × 7)/(2 × 22)
= 21 cm
Now volume of the bowl = 2/3πr3
= 2/3 × 22/7 × (21)3 cm3
= 2/3 × 2/7 × 9261 cm3
= 19404 cm3
Question 6. The volume of a hemisphere is 2425.1/2 cm2. Find the curved surface area.
Answer :
Volume of a hemisphere = 2425.1/2 cm3
= 4851/2 cm3
Let radius = r, then
2/3πr3 = 4851/2
⇒ 2/3 × 22/7 × r3 = 4851/2
⇒ r3 = (4851 × 3 × 7)/(2 × 2 × 22)
= 9261/8
= (21/2)3
∴ r = 21/2 cm
∴ Curved surface area = 2πr2
= 2 × 22/7 × 21/2 × 21/2 cm2
= 693 cm2
Question 7. A solid wooden toy is in the shape of a right circular cone mounted on a hemisphere. If the radius of the hemisphere is 4.2 cm and the total height of the toy is 10.2 cm, find the volume of the toy
Answer :
A wooden solid toy is of a shape of a right circular cone
mounted on a hemisphere.
Radius of hemisphere (r) = 4.2 cm
Total height = 10.2 cm
Mensuration Chapter Test
ML Aggarwal Class 10 ICSE Maths Solutions
Page 439
Question 8. A solid is in the form of a cylinder with hemispherical ends. The total height of the solid is 19 cm and the diameter of the cylinder is 7 cm. Find the volume and the total surface area of the solid.
Answer :
Radius of cylinder = 7/2cm
and height of cylinder = 19 – 2 × 7/2 cm
= 19 – 7 = 12 cm
and radius of hemisphere = 7/2 cm
= 4/3 × 22/7 × (7/2)3 + 22/7 × (7/2)2 × 12 cm3
= 4/3 × 22/7 × (7×7×7)/(2×2×2) + 22/7 × 7/2 × 7/2 × 12 cm3
= 539/3 + 462
= (539 + 1386)/3
= 1925/2 cm3
= 641.2/3 cm3
And total surface area of the solid = 2×2πr2 + 2πrh
= 4πr2 + 2πrh
= 2πr(2r + h)
= 2× 22/7 × 7/2(2×7/2 + 12)
= 22(7 + 12)
= 22 × 19 cm2
= 418 cm2
Question 9. The radius and height of a right circular cone are in the ratio 5 : 12. If its volume is 2512 cm , find its slant height. (Take π = 3.14).
Answer :
Let radius of cone (r) = 5x
then height (h) = 12x
Volume = 1/3 πr2h
= 1/3(3.14)×(5x)2 × 12x
We know, volume = 2512 cm3
⇒ 1/3(3.14)× 25x2 × 12x = 2512
⇒ 1/3 × 3.14 × 300x3 = 2512
x3 = (2513×3)/(3.14×300)
= (2512×3×100)/(314×300)
= 8
= (2)3
∴ x = 2
∴ Radius of cone (r) = 5 × 2 = 10 cm
And height (h) = 12 × 2 = 24 cm
= 26 cm
Question 10. A cone and a cylinder are of the same height. If diameters of their bases are in the ratio 3 : 2, find the ratio of their volumes.
Answer :
Let height of cone and cylinder = h
Diameter of the base of cone = 3x
Diameter of base of cylinder = 2x
Volume of cone = 1/3π(r1)2 h
= 1/3 π×(3x/2)2 × h
= 1/3 π× 9/4x2 ×h
= ¾ πx2h
And volume of cylinder = πr2h
= π(2x/2)2 h
= πx2h
= ¾ : 1
⇒ 3 : 4
Question 11. A solid cone of base radius 9 cm and height 10 cm is lowered into a cylindrical jar of radius 10 cm, which contains water sufficient to submerge the cone completely. Find the rise in water level in the jar.
Answer :
Radius of the cone (r) = 9 cm
Height of the cone (h) = 10 cm
Volume of water filled in cone
= 1/3 πr2h
= 1/3π(9)2 × 10 cm3
= 810/3 π
= 270 π cm3
Now radius of the cylindrical jar = 10 cm
Let h be the height of water in the jar
∴ πr2h = 270π
⇒ π(10)2h = 270π
⇒ 100 πh = 270π
⇒ h = 270π/100π
= 2.7 cm
Question 12. An iron pillar has some part in the form of a right circular cylinder and the remaining in the form of a right circular cone. The radius of the base of each of cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if one cu. cm of iron weighs 7.8 grams.
Answer :
Radius of the base of cone = 8 cm
And radius of cylinder = 8 cm
Height of cylindrical part (h1) = 240 cm
And height of conical part (h2) = 36 cm
Volume of the iron pillar = 1/3 πr2h2 + πr2h1
= πr2(1/3h2 + h1)
= 22/7 × 8 × 8 [1/3 ×36 + 240] cm3
= 1408/7 [36/3 + 240] cm3
= 1408/7 [252] cm3
= 1408/7 × 252
= 1408 × 36
= 50688 cm3
Weight of 1 cm3 = 7.8 gm
∴ Total weight of the pillar = 50688 × 7.8 gm
= 395366.4 gm
= 395.3664 kg
Question 13. A circus tent is made of canvas and is in the form of right circular cylinder and a right circular cone above it. The diameter and height of the cylindrical part of the tent are 126 m and 5 m respectively. The total height of the tent is 21 m. Find the total cost of the tent if the canvas used costs Rs 36 per square metre.
Answer :
Diameter of the cylindrical part = 126 m
Radius (r) = 126/2 = 63m
Height of cylindrical part = 5 m
Total height of the tent = 21 m
∴ Height of conical portion = 21 – 5
= 16
Surface area of the tent = 2πrh + πrl
= πr (2h + l)
= 22/7 × 63(2×5 + 65)
= 198 × (10 + 65)
= 198 × 75 m2
= 14850 m2
Cost of one 1 sq. m cloth = ₹ 36
∴ Total cost = Rs 14850 × 36
= ₹ 534600
Question 14. The entire surface of a solid cone of base radius 3 cm and height 4 cm is equal to the entire surface of a solid right circular cylinder of diameter 4 cm. Find the ratio of their
(i) curved surfaces
(ii) volumes.
Answer :
Radius of the base of a cone (r) = 3 cm
Height (h) = 4 cm
= 5
= πr(l + r)
= 22/7 × 3(5 + 3) cm2
= 66/7 × 8
= 528/7 cm2
Diameter of cylinder = 4 cm
∴ Radius (r1) = 4/2 = 2 cm
Total surface area = 528/7 cm2
Let h be the height, then
∴ 2πr1h1 + 2πr2 = 528/7
⇒ 2πr(h1 + r) = 528/7
⇒ 2 × 22/7 × 2(h1 + 2) = 528/7
⇒ h1 + 2 = 528/7 × 7/(2×22×2)
⇒ h1 + 2 = 6
⇒ h1 = 6 – 2 = 4 cm
(i) Ratio between curved surface of cone and cylinder = πrl : 2πr1h1
= πrl : 2πr1h1
= π×3×5 : 2×π×2×4
= 15 : 16
(ii) Ratio between their volumes
= 1/3 πr2h : πr12h1
= 1/3 π×3×3×4 : π×2×4
= 3 : 4
Question 15. A cone is 8.4 cm high and the radius of its base is 2.1 cm. It is melted and recast into a sphere. Find the radius of the sphere.
Answer :
Radius of base of a cone (r) = 2. 1 cm
and height (h) = 8.4 cm
Volume = 1/3πr2h
= 1/3π × (2.1)2 × (8.4) cm3
= π × 4.41 × 2.8 cm3
= 12.348 π cm3
∴ Volume of sphere = 12.348 π cm3
Radius = Volume/(4/3.π)1/3
= [12.348π × 3]/[4 × π]1/3
= (9.261)1/3
= (2.1 × 2.1 × 2.1)1/3
= 2.1 cm
Question 16. Find the least number of coins of diameter 2.5 cm and height 3 mm which are to be melted to form a solid cylinder of radius 3 cm and height 5 cm.
Answer :
Radius of a cylinder (r) = 3 cm
Height (h) = 5 cm
Volume = πr2h = π×3×3×5 = 45π cm2
Diameter of a coins = 2.5 cm
∴ Radius (r1) = 2.5/2 = 1.25 cm
And height (h1) = 3 mm = 3/10 cm
∴ Volume of a coin = πr12h1
= π×1.25×1.25×3/10 cm3
= 0.46875π cm3
∴ Number of coins required = 45π/0.4687π
= 45/0.46875
= 96 coins
Question 17. A hemisphere of lead of radius 8 cm is cast into a right circular cone of base radius 6 cm. Determine the height of the cone correct to 2 places of decimal.
Answer :
Radius of hemisphere = 8 cm
Volume = 2/3πr3 cm3
= 2/3 π ×(8)3 cm3
= 2/3 π ×512 cm3
= 1024/3 π cm3
∴ Volume of right circular cone = 1024/3 π cm3
Radius = 6 cm
Let h be the height of the cone
∴ 1/3 πr2 h = 1024/3 π
⇒ 1/3 π×(6)2h = 1024/3 π
⇒ 12 πh = 1024/3 π
⇒ h = 1024π/(3×12π)
= 256/9
= 28.44 cm
Question 18. A vessel in the form of a hemispherical bowl is full of water. The contents are emptied into a cylinder. The internal radii of the bowl and cylinder are respectively 6 cm and 4 cm. Find the height of the water in the cylinder.
Answer :
Radius of hemispherical bowl = 6 cm
.’. Volume of the water in the bowl
= 2/3 πr3
= 2/3 π×(6)3 cm3
= 144 π cm3
∴ Volume of water in the cylinder = 144π cm3
Radius of the cylinder = 4 cm
Let h be the height of water
∴ πr2h = 144π
⇒ (4)2h = 144
⇒ 16h = 144
∴ h = 144/16 = 9
Hence,
height of water in the cylinder = 9 cm
Question 19. A sphere of diameter 6 cm is dropped into a right circular cylindrical vessel partly filled with water. The diameter of the cylindrical vessel is 12 cm. If the sphere is completely submerged in water, by how much will the level of water rise in the cylindrical vessel?
Answer :
Radius of sphere = 6/2 = 3 cm
Let height of water raised = h cm
Now volume of sphere = 4/3 πr3
= 4/3 π(3)3 cm3
= 36 π cm3
And volume of water in the cylinder = 36 π cm3
∴ πr2h = 36 π
⇒ (6)2h = 36
⇒ 36 h = 36
⇒ h = 1
∴ Height of raised water = 1 cm
Question 20. A solid sphere of radius 6 cm is melted into a hollow cylinder of uniform thickness. If the external radius of the base of the cylinder is 5 cm and its height is 32 cm, find the uniform thickness of the cylinder.
Answer :
Radius of solid sphere = 6 cm
Volume of solid sphere
= 4/3 πr3
= 4/3 × π × (6)3 cm3
= 288π cm3
∴ Volume of hollow cylinder = 288π cm3
External radius of cylinder (R) = 5 cm
And height (h) = 32 cm
Let r be the inner radius
∴ Volume = π(R2 – r2)h
∴ π(R2 – r2)h = 288 π
⇒ [(5)2 – r2] × 32 = 288
⇒ 25 – r2 = 288/32
⇒ 25 – r2 = 9
⇒ r2 = 25 – 9 = 16
⇒ r2 = (4)2
∴ r = 4
∴ Thickness of hollow cylinder = R – r
= 5 – 4
= 1 cm
–: End of ML Aggarwal Mensuration Chapter Test Class 10 ICSE Maths Solutions :–
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