ML Aggarwal Mensuration Exe-17.5 Class 10 ICSE Maths Solutions

ML Aggarwal Mensuration Exe-17.5 Class 10 ICSE Maths Solutions. We Provide Step by Step Answer of  Exe-17.5 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 Exe-17.5 Class 10 ICSE Maths Solutions

Board	ICSE
Subject	Maths
Class	10th
Chapter-17	Mensuration
Writer / Book	Understanding
Topics	Solutions of Exe-17.5
Academic Session	2024-2025

Mensuration Exe-17.5

ML Aggarwal Class 10 ICSE Maths Solutions

Page 434

Question 1. The diameter of a metallic sphere is 6 cm. The sphere is melted and drawn into a wire of uniform cross-section. If the length of the wire is 36 m, find its radius.

Answer :

Diameter of metallic sphere = 6 cm

Radius (r) = 6/2 = 3 cm

Volume = 4/3 πr³

= 4/3 × π × (3)³ cm³

= 4/3 × π × 3 × 3 × 3 cm³ = 36π cm³

∴ Volume of wire = 36π cm³

Length of wire (h) = 36 m

Let r be the radius of the wire

∴ πr²h = 36 π

⇒ r² × 36 × 100 = 36

⇒ r² = 1/100 = (1/10)²

⇒ r = 1/10 cm = 1 mm

Question 2. A solid metallic sphere of radius 6 cm is melted and made into a solid cylinder of height 32 cm. Find the : 

(i) radius of the cylinder
(ii) curved surface area of the cylinder

Answer :

Radius of metallic sphere (R) = 6 cm
Height of cylinder (h) = 32 cm
Volume of cylinder = Volume of metallic sphere

Curved Surface area of the year = 2πrh
= 2 × 3.1 × 3 × 32 = 595.2 cm²

Question 3. A solid metallic hemisphere of radius 8 cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.

Answer :

Radius of a solid hemisphere (r) = 8 cm

Volume of the hemisphere, V = (2/3)πr³

= (2/3) × 8³

= (1024/3)π cm³

Radius of cone, R = 6 cm

Since hemisphere is melted and recasted into a cone, the volume remains the same.

Volume of the cone, (1/3)πR²h = (1024/3)π

⇒ (1/3) × π × 6² × h = (1024/3)π

⇒ 36h = 1024

⇒ h = 1024/36 = 28.44 cm

Hence the height of the cone is 28.44 cm.

Question 4. A rectangular water tank of base 11 m x 6 m contains water upto a height of 5 m. if the water in the tank is transferred to a cylindrical tank of radius 3.5 m, find the height of the water level in the tank.

Answer :

Base of a water tank = 11 m × 6 m
Height of water level in it (h) = 5 m
Volume of water =11 × 6 × 5 = 330 m³

Volume of cylindrical tank = πr²h

πr²h = 330

⇒ (22/7) × 3.5² × h = 330

⇒ (22/7) × 12.25 × h = 330

⇒ h = 330 × 7/22 × 12.25 = 8.57 m

Hence

the height of the water level in the tank is 8.57 m.

Question 5. The volume of a cone is the same as that of the cylinder whose height is 9 cm and diameter 40 cm. Find the radius of the base of the cone if its height is 108 cm.

Answer :

Diameter of a cylinder = 40 cm
Radius (r) = 40/2 = 20 cm
Height(h) = 9 cm

∴ Volume = πr²h

= π × 20 × 20 × 9 cm³

= 3600π cm³

Now volume of cone = 3600π cm³

Height of the cone = 108 cm

Hence

the radius of the cone is 10 cm.

Question 6. Eight metallic spheres, each of radius 2 cm, are melted and cast into a single sphere. Calculate the radius of the new (single) sphere.

Answer :

Radius of each metallic sphere (r) = 2 cm

Volume of a sphere = (4/3)πr³

= (4/3)×π × 2³

= (4/3)×π × 8

= (32/3)π cm³

Volume of 8 spheres = 8 × (32/3)π

= (256/3)π cm³

Let R be radius of new sphere.

Volume of the new sphere = (4/3)πR³

Since 8 spheres are melted and casted into a single sphere, volume remains same.

(4/3)πR³ = (256/3)π

⇒4R³ = 256

⇒R³ = 256/4 = 64

Taking cube root

⇒ R = 4 cm

Hence the radius of the new sphere is 4 cm.

Question 7. A hemispherical bowl of diameter 7.2 cm is filled completely with chocolate sauce. This sauce is poured into an inverted cone of radius 4.8 cm. Find the height of the cone. 

Answer :

Question 8. Two spheres of the same metal weigh 1 kg and 7 kg. The radius of the smaller sphere is 3 cm. The two spheres are melted to form a single big sphere. Find the diameter of the big sphere.

Answer :

Radius of smaller sphere = 3 cm
Let r be the radius of a larger sphere

Mass of the smaller sphere, m₁ = 1 kg

Mass of the bigger sphere, m₂ = 7 kg

The spheres are melted to form a new sphere.

So the mass of new sphere, m₃ = 1 + 7 = 8 kg

Density of smaller sphere = density of new sphere

Let V₁ be volume of smaller sphere and V₃ be volume of bigger sphere.

m₁/V₁ = m₃/V₃

⇒ 1/V₁ = 8/V₃

⇒ V₁/V₃ = 1/8 …(i)

Given radius of the smaller sphere, r = 3 cm

V₁ = (4/3)πr³

⇒ V₁ = (4/3) π× 3³

⇒ V₁ = 36π

Let R be radius of new sphere.

V₃ = (4/3)πR³

⇒ V₁/ V₃ = 36 ÷ (4/3)πR³

⇒ V₁/ V₃ = 27/R³ …(ii)

From (i) and (ii)

1/8 = 27/R³

⇒ R³ = 27 × 8 = 216

Taking cube root on both sides,

R = 3 x 2 = 6 cm
Diameter of big sphere = 2 x 6 = 12 cm

Question 9. A hollow copper pipe of inner diameter 6 cm and outer diameter 10 cm is melted and changed into a solid circular cylinder of the same height as that of the pipe. Find the diameter of the solid cylinder.

Answer :

Inner diameter of a hollow pipe = 6 cm
and outer diameter = 10 cm
Inner radius (r) = 6/2 = 3cm

Outer radius, R = 10/2 = 5 cm

Let h be the height of the pipe.

Volume of pipe = π(R² – r²)h

= π(5² – 3²) × h

= πh(25 – 9)

= πR²h

16 = R²

⇒ R = 4

∴ Diameter of solid cylinder = 4 × 2 = 8 cm

Question 10. A hollow sphere of internal and external diameter 4 cm and 8 cm respectively, is melted into a cone of base diameter 8 cm. Find the highest of the cone.

Answer :

Internal diameter of hollow sphere = 4 cm

So, the internal radius of hollow sphere = 2 cm

External diameter of hollow sphere = 8 cm

So, the external radius of hollow sphere = 4 cm

We know that, Volume of the hollow sphere 4/3 π × (4³ – 2³)          … (i) 

Also given,

Diameter of the cone = 8 cm

So, the radius of the cone = 4 cm

Let the height of the cone be x cm

Volume of the cone 1/3 π × 4² × h          ….. (ii)

As the volume of the hollow sphere and cone are equal. We can equate equations (i) and (ii) 

So, we get

4/3 π × (4³ – 2³) = 1/3 π × 4² × h

4 x (64 – 8) = 16 x h h = 14

Therefore, the height of the cone so obtained will have a height of 14 cm.

Question 11. A well with inner diameter 6 m is dug 22 m deep. Soil taken out of it has been spread evenly all round it to a width of 5 m to form an embankment. Find the height of the embankment.

Answer :

Inner diameter of a well = 6 m

Radius of the well, r = 6/2 = 3 m

Depth (h) = 22 m

Volume of the soil dug out of well = πr²H

= 3² × 22×π

= 198π m³

Width of the embankment = 5 m

Inner radius of embankment, r = 3 m

Outer radius of embankment, R = 3 + 5 = 8 m

Let h be height of the soil embankment.

Volume of the soil embankment = π(R² – r²)h

= (8² – 3²)π×h

= (64 – 9)π×h

= 55h

Volume of the soil dug out = volume of the soil embankment

198 = 55h

⇒ h = 198/55

⇒ h = 3.6 m

Hence, the height of the soil embankment is 3.6 m.

Question 12. A cylindrical can of internal diameter 21 cm contains water. A solid sphere whose diameter is 10.5 cm is lowered into the cylindrical can. The sphere is completely immersed in water. Calculate the rise in water level, assuming that no water overflows.

Answer :

Internal diameter of cylindrical can = 21 cm
Radius (R) = 21/2 cm

Diameter of sphere = 10.5 cm

Radius of the sphere, r = 10.5/2 = 21/4 cm

Let the rise in water level be h.

Rise in volume of water = Volume of sphere immersed

πR²h = (4/3)πr³

⇒ (21/2)²πh = (4/3) ×π (21/4)³

⇒ (21/2)×(21/2)×h = (4/3)×(21/4)×(21/4)×(21/4)

⇒ h = 21/12

⇒ h = 7/4

⇒ h = 1.75 cm

Hence,

the rise in water level is 1.75 cm.

Question 13. There is water to a height of 14 cm in a cylindrical glass jar of radius 8 cm. Inside the water there is a sphere of diameter 12 cm completely immersed. By what height will the water go down when the sphere is removed?

Answer :

Given radius of the glass jar, R = 8 cm

Diameter of the sphere = 12 cm

Radius of the sphere, r = 12/2 = 6 cm

When the sphere is removed from the jar, volume of water decreases.

Let h be the height by which water level will decrease.

Volume of water decreased = Volume of the sphere

πR²h = (4/3)πr³

⇒ 8²πh = (4/3)π6³

⇒ h = (4/3) × 6 × 6 × 6/(8×8)

= 18/4 = 9/2 = 4.5 cm

Hence,

the height by which water level decreased is 4.5 cm.

---

Mensuration Exe-17.5

ML Aggarwal Class 10 ICSE Maths Solutions

Page 435

Question 14. A vessel in the form of an inverted cone is filled with water to the brim. Its height is 20 cm and diameter is 16.8 cm. Two equal solid cones are dropped in it so that they are fully submerged. As a result, one-third of the water in the original cone overflows. What is the volume of each of the solid cone submerged ? (2006)

Answer :

Height of conical vessel (h) = 20 cm
and diameter = 16.8 cm

height of the cone, h = 20 cm

Diameter of the cone = 16.8 cm

Radius of the cone, r = 16.8/2 = 8.4 cm

Volume of water in the vessel = (1/3)πr²h

= (1/3) ×π× 8.4² × 20

= (1/3) × (22/7) × 8.4 × 8.4 × 20

= 1478.4 cm³

One third of volume of water in the vessel = (1/3)× 1478.4

= 492.8 cm³

One third of volume of water in the vessel = Volume of water over flown

Volume of water over flown = volume of two equal solid cones dropped into the vessel.

Volume of two equal solid cones dropped into the vessel = 492.8 cm³

Volume of one solid cone dropped into the vessel = 492.8/2 = 246.4 cm³

Hence,

the volume of each of the solid cone submerged is 246.4 cm³.

Question 15. A solid metallic circular cylinder of radius 14 cm and height 12 cm is melted and recast into small cubes of edge 2 cm. How many such cubes can be made from the solid cylinder?

Answer :

Radius of a solid metallic cylindrical (r) = 14 cm
and height (h) = 12 cm

Volume of the cylinder = πr²h

= 14² × 12 ×π

= 196 × 12 ×π

= 2352× π

= 2352 × 22/7

= 7392 cm³

Edge of the cube, a = 2 cm

Volume of cube = a³

= 2³ = 8 cm³

Number of cubes made from solid cylinder = 7392/8 = 924.

Question. How many shots each having diameter 3 cm can be made from a cuboidal lead solid of dimensions 9 cm x 11 cm x 12 cm?

Answer :

Diameter of a shot = 3 cm

dimensions of the cuboidal solid = 9 cm × 11 cm × 12 cm

Volume of the cuboidal solid = 9 × 11 × 12 = 1188 cm³

Radius (r) = 3/2 cm
Volume of one shot = (4/3)πr³

= (4/3) × π×1.5³

= (4/3) × (22/7) × 1.5³

= 297/21 cm³

Number of shots made from cuboidal lead of solid = 1188 ÷ (297/21)

= 1188 × 21/297 = 84

Therefore,

the number of shots made from cuboidal lead of solid is 84.

Question 17. A solid metal cylinder of radius 14 cm and height 21 cm is melted down and recast into spheres of radius 3.5 cm. Calculate the number of spheres that can be made.

Answer :

Radius of a solid metallic cylinder (r) = 14 cm
and height (h) = 21 cm
Volume of cylinder = πr²h

R = 3.5 cm

Volume of the metal cylinder = πr²h

= (22/7) × 14² × 21

= 22 × 2 × 14 × 21

= 12936 cm³

Volume of sphere = (4/3)πR³

= (4/3) × (22/7) × 3.5³

= 11 × 49/3

= 539/3 cm³

Number of spheres that can be made = Volume of the metal cylinder/Volume of sphere

= 12936 ÷ 539/3

= 12936 × 3/539

= 72

Question 18. A metallic sphere of radius 10.5 cm is melted and then recast into small cenes, each of radius 3.5 cm and height 3 cm. Find the number of cones thus obtained. (2005)

Answer :

Radius of a metallic sphere (r) = 10.5 cm
Volume = (4/3)πR³

= (4/3)× π× 10.5³ 

= 1543.5π cm³

Radius of cone, r = 3.5 cm

Height of the cone, h = 3 cm

Volume of the cone = (1/3) πr²h

= (1/3)× π × 3.5² × 3

= 12.25π cm³

Number of cones made from sphere = Volume of sphere / volume of cone

= 1543.5π/12.25π

= 126

Question 19. A certain number of metallic cones each of radius 2 cm and height 3 cm are melted and recast in a solid sphere of radius 6 cm. Find the number of cones. (2016)

Answer :

Radius of each cone (r) = 2 cm
and height (h) = 3 cm

Volume of cone = (1/3)πr²h

= (1/3) ×π× 2² × 3

= 4π cm³

Radius of the solid sphere, R = 6 cm

Volume of the solid sphere = (4/3)πR³

= (4/3)×π×6³

= 4/3 ×π × 6 × 6 × 6

= 288π cm³

Number of cones made from sphere = Volume of solid sphere / volume of the cone

= 288π/4π

= 72

Question 20. A vessel is in the form of an inverted cone. Its height is 11 cm and the radius of its top, which is open, is 2.5 cm. It is filled with water up to the rim. When some lead shots, each of which is a sphere of radius 0.25 cm, are dropped into the vessel, 2/5 of the water flows out. Find the number of lead shots dropped into the vessel. (2003)

Answer :

Radius of the top of the inverted conical vessel (R) = 2.5 cm
and height (h)= 11 cm

Volume of the cone = (1/3)πr²h

= (1/3) ×π ×2.5² × 11

= (11/3) ×π × 6.25 cm³

When lead shots are dropped into vessel, (2/5) of water flows out.

Volume of water flown out = (2/5) × (11/3) × 6.25× π

= (22/15) × 6.25 × π

= (137.5/15)π cm³

Radius of sphere, R = 0.25 cm = ¼ cm

Volume of sphere = (4/3)πR³

= (4/3) ×π× (1/4)³

= 48π cm³

Number of lead shots dropped = Volume of water flown out/Volume of sphere

= (137.5/15)π ÷ 48π

= (137.5/15)π × π/48

= 137.5 × 48/15

= 440

Question 21. The surface area of a solid metallic sphere is 616 cm². It is melted and recast into smaller spheres of diameter 3.5 cm. How many such spheres can be obtained? (2007)

Answer :

Surface area of a metallic sphere = 616 cm²

4πR² = 616

⇒ 4 × (22/7)R² = 616

⇒ R² = 616 × 7/4 × 22

⇒ R² = 49

⇒ R = 7

Volume of the solid metallic sphere = (4/3)πR³

= (4/3) ×π× 7³

= (1372/3)π cm³

Diameter of smaller sphere = 3.5 cm

So radius, r = 3.5/2 = 7/4 cm

Volume of the smaller sphere = (4/3)πr³

= (4/3) ×π× (7/4)³

= (343/48)π cm³

Number of spheres made = Volume of the solid metallic sphere/ Volume of the smaller sphere

= (1372/3)π ÷ (343/48)π

= (1372 × 48)/(3 × 343)

= 64

number of spheres made is 64.

Question 22. The surface area of a solid metallic sphere is 1256 cm². It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate.

(i) the radius of the solid sphere.
(ii) the number of cones recast. (Use π = 3.14).

Answer :

(i) Surface area of a solid metallic sphere = 1256 cm²

4πR² = 1256

⇒ 4 × 3.14 × R² = 1256

⇒ R² = 1256/4 × 3.14

⇒ R² = 100

⇒ R = 10

Hence, the radius of solid sphere is 10 cm.

(ii) Volume of the solid sphere = (4/3)R³

= (4/3) ×π× 10³

= (4000/3)×π cm³

= 12560/3 cm³

Radius of the cone, r = 2.5 cm

Height of the cone, h = 8 cm

Volume of the cone = (1/3)πr²h

= (1/3) × 3.14 × 2.5² × 8

= 157/3 cm³

Number of cones made = Volume of the solid sphere/ Volume of the cone

= (12560/3) ÷ (157/3)

= (12560/3) × (3/157)

= 12560/157

= 80

number of cones is 80.

Question 23. A cylindrical can whose base is horizontal and of radius 3.5 cm contains sufficient water so that when a sphere is placed in the can, the water just covers the sphere. Given that the sphere just fits into the can, calculate :

(i) the total surface area of the can in contact with water when the sphere is in it.
(ii) the depth of the water in the can before the sphere was put into the can. Given your answer as proper fractions.

Answer :

(i) Radius of a cylindrical can = 3.5 cm

Radius of the sphere = 3.5 cm
and height of water level in the can = 3.5 × 2 = 7 cm

Height of cylinder, h = 7 cm

Total surface area of can in contact with water = curved surface area of cylinder + base area of cylinder.

= 2πrh + πr²

= πr(2h + r)

= 22/7 × 3.5 × (2×7 + 3.5)

= (22/7) × 3.5 × (14 + 3.5)

= 11 × 17.5

= 192.5 cm²

Hence the surface area of can in contact with water is 192.5 cm².

(ii) Let the depth of the water before the sphere was put be d.

Volume of cylindrical can = volume of sphere + volume of water

πr²h = (4/3)πr³ + πr²d

⇒ πr²h = πr²{(4/3)r + d)}

⇒ h = (4/3)r + d

⇒ d = h – (4/3)r

⇒ d = 7 – (4/3) × 3.5

⇒ d = (21 – 14)/3

⇒ d = 7/3

–: End of ML Aggarwal Mensuration Exe-17.5 Class 10 ICSE Maths Solutions :–

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