# Mensuration

Chapters

- ap-scert-10th-class-maths-lesson-1-Real-Numbers
- ap-scert-10th-class-Maths-Lesson-2-Sets
- ap-scert-10th-class-Maths-Lesson-3-Polynomials
- ap-scert-10th-class-Maths-Lesson-4-Pair-of-Linear-Equations-in-Two-Variables
- ap-scert-10th-class-Maths-Lesson-5-Quadratic-Equations
- ap-scert-10th-class-Maths-Lesson-6-Progressions
- ap-scert-10th-class-Maths-Lesson-7-Coordinate-Geometry
- ap-scert-10th-class-Maths-Lesson-8-Similar-Triangles
- ap-scert-10th-class-Maths-Lesson-9-Tangents-and-Secants-to-a-Circle
- ap-scert-10th-class-Maths-Lesson-10-Mensuration
- ap-scert-10th-class-Maths-Lesson-11-Trigonometry
- ap-scert-10th-class-Maths-Lesson-12-Applications-of-Trigonometry
- ap-scert-10th-class-Maths-Lesson-13-Probability
- ap-scert-10th-class-Maths-Lesson-14-Statistics

- ap-scert-10th-class-maths-lesson-1-Real-Numbers
- ap-scert-10th-class-Maths-Lesson-2-Sets
- ap-scert-10th-class-Maths-Lesson-3-Polynomials
- ap-scert-10th-class-Maths-Lesson-4-Pair-of-Linear-Equations-in-Two-Variables
- ap-scert-10th-class-Maths-Lesson-5-Quadratic-Equations
- ap-scert-10th-class-Maths-Lesson-6-Progressions
- ap-scert-10th-class-Maths-Lesson-7-Coordinate-Geometry
- ap-scert-10th-class-Maths-Lesson-8-Similar-Triangles
- ap-scert-10th-class-Maths-Lesson-9-Tangents-and-Secants-to-a-Circle
- ap-scert-10th-class-Maths-Lesson-10-Mensuration
- ap-scert-10th-class-Maths-Lesson-11-Trigonometry
- ap-scert-10th-class-Maths-Lesson-12-Applications-of-Trigonometry
- ap-scert-10th-class-Maths-Lesson-13-Probability
- ap-scert-10th-class-Maths-Lesson-14-Statistics

- Solutions
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- Question Papers
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Solutions

A joker’s cap is in the form of right circular cone whose base radius is 7 cm and height is 24 cm. Find the area of the sheet required to make 10 such caps.

Answer:

Radius of the cap (r) = 7 cm

Height of the cap (h) = 24 cm

= 25

∴ l = 25 cm.

Lateral surface area of the cap = Cone = πrl

L.S.A. = 22/7 × 7 × 25 = 550 cm^{2}.

∴ Area of the sheet required for 10 caps = 10 x 550 = 5500 cm^{2}.

A sports company was ordered to prepare 100 paper cylinders without caps for shuttle cocks. The required dimensions of the cylinder are 35 cm length / height and its radius is 7 cm. Find the required area of thin paper sheet needed to make 100 cylinders.

Answer:

Radius of the cylinder, r = 7 cm

Height of the cylinder, h = 35 cm

T.S.A. of the cylinder with lids at both ends = 2πr(r+h)

= 2 × 22/7 × 7 × (7 + 35)

= 2 × 22/7 × 7 × 42 = 1848 cm^{2}.

Area of thin paper required for 100 cylinders = 100 × 1848

= 184800 cm^{2}

=184800/100×100 m^{2}

= 18.48 m^{2}.

Find the volume of right circular cone with radius 6 cm. and height 7 cm.

Answer:

Base radius of the cone (r) = 6 cm.

Height of the cone (h) = 7 cm

Volume of the cone = 1/3 πr^{2}h

=1/3 × 22/7 × 6 × 6 × 7

= 264 c.c. (Cubic centimeters)

∴ Volume of the right circular cone = 264 c.c.

The lateral surface area of a cylinder is equal to the curved surface area of a cone. If their base be the same, find the ratio of the height of the cylinder to slant height of the cone.

Answer:

Base of cylinder and cone be the same.

CSA / LSA of cylinder = 2πrh

CSA of cone = πrl

The lateral surface area of a cylinder is equal to the curved surface area of cone.

∴ 2πrh = πrl

⇒ h/l = πr/2πr

⇒ h/l = 1/2

∴ h : l = 1 : 2

A self help group wants to manufacture joker’s caps (conical caps) of 3 cm radius and 4 cm height. If the available colour paper sheet is 1000 cm

Answer:

Radius of the cap (conical cap) (r) = 3 cm

Height of the cap (h) = 4 cm

= 5 cm

C.S.A. of the cap = πrl

= 22/7 × 3 × 5

≃ 47.14 cm^{2}

Number of caps that can be made out of 1000 cm^{2}=1000/47.14 ≃ 21.27

∴ Number of caps = 21.

A cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are in the ratio of 3 : 1.

Answer:

Given dimensions are:

Cone:

Radius = r

Height = h

Volume (V) = 1/3 πr^{2}h

Cylinder:

Radius = r

Height = h

Volume (V) = πr^{2}h

Ratio of volumes of cylinder and cone = πr^{2}h : 1/3πr^{2}h

= 1 :1/3

= 3 : 1

Hence, their volumes are in the ratio = 3 : 1.

A solid iron rod has cylindrical shape. Its height is 11 cm. and base diameter is 7 cm. Then find the total volume of 50 rods?

Answer:

Diameter of the cylinder (d) = 7 cm

Radius of the base (r) = 7/2 = 3.5 cm

Height of the cylinder (h) = 11 cm

Volume of the cylinder V = πr^{2}h

= 22/7 × 3.5 × 3.5 × 11 = 423.5 cm^{3}

∴ Total volume of 50 rods = 50 × 423.5 cm^{3}= 21175 cm^{3}.

A heap of rice is in the form of a cone of diameter 12 m. and height 8 m. Find its volume? How much canvas cloth is required to cover the heap? (Use π = 3.14)

Answer:

Diameter of the heap (conical) (d) = 12 cm

∴ Radius = d/2 = 12/2 = 6 cm

Height of the cone (h) = 8 m

Volume of the cone, V =1/3 πr^{2}h

=1/3 × 22/7 × 6 × 6 × 8

= 301.71 m^{3}.

The curved surface area of a cone is 4070 cm

Answer:

C.S.A. of a cone = πrl = 4070 cm^{2}

Diameter of the cone (d) = 70 cm

Radius of the cone = r =d/2 = 70/2 = 35 cm

Let its slant height be ‘l’.

By problem,

πrl = 4070 cm^{2}

22/7 × 35 × l = 4070

110 l = 4070

l = 4070/110 = 37 cm

∴ Its slant height = 37 cm.

A toy is in the form of a cone mounted on a hemisphere. The diameter of the base and the height of the cone are 6 cm and 4 cm respectively. Determine the surface area of the toy. (Use π = 3.14)

Answer:

Diameter of the base of the cone d = 6 cm.

∴ Radius of the base of the cone

r = d/2 = 6/2 = 3 cm

Height of the cone = h = 4 cm

= 5 cm

∴ C.S.A of the cone = πrl

= 22/7 × 3 × 5

=330/7 cm^{2}

Radius of the hemisphere = d/2 = 6/2 = 3 cm

C.S.A. of the hemisphere = 2πr^{2}

= 2 × 22/7 × 3 × 3

=396/7

Hence the surface area of the toy = C.S.A. of cone + C.S.A. of hemisphere

= 330/7 + 396/7

= 726/7 ≃ 103.71 cm^{2}.

A solid is in the form of a right circular cylinder with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cm and the heights of the cylindrical and conical portions are 10 cm and 6 cm respectively. Find the total surface area of the solid. [Use π = 3.14]

Answer:

Total surface area = C.S.A. of the cone + C.S.A. of cylinder + C.S.A of the hemisphere.

Cone:

Radius (r) = 8 cm

Height (h) = 6 cm

= 10 cm

C.S.A. = πrl

= 22/7 × 8 × 10

=1760/7 cm^{2}

Cylinder:

Radius (r) = 8 cm;

Height (h) = 10 cm

C.S.A. = 2πrh

= 2 × 22/7 × 8 × 10

= 3520/7 cm^{2}

Hemisphere:

Radius (r) = 8 cm

C.S.A. = 2πr^{2}

= 2 × 22/7 × 8 × 8

= 2816/7 cm^{2}

∴ Total surface area of the given solid

= 1760/7 + 3520/7 + 2816/7

T.S.A. = 8096/7 = 1156.57 cm^{2}.

A medicine capsule is ih the shape of a cylinder with two hemispheres stuck to each of its ends. The length of the capsule is 14 mm. and the width is 5 mm. Find its surface area.

Answer:

Surface area of the capsule = C.S.A. of 2 hemispheres + C.S.A. of the cylinder

i) Now for Hemisphere:

Radius (r) = d/2 = 5/2 = 2.5 mm

C.S.A of each hemisphere = 2πr^{2}

C.S.A of two hemispheres

= 2 × 2πr^{2}= 4πr^{2}

= 2 × 22/7 × 5/2 × 5/2

= 550/7

= 78.57 mm^{2}.

ii) Now for Cylinder:

Length of capsule = AB =14 mm

Then height (length) cylinder part = 14 - 2(2.5)

h = 14 - 5 = 9 mm

Radius of cylinder part (r) = 5/2

Now C.S.A of cylinder part = 2πrh

= 2 × 22/7 × 52 × 9

= 900/7

= 141.428 mm^{2}

Now total surface area of capsule

= 78.57 + 141.43 = 220 mm^{2}

Two cubes each of volume 64 cm

Answer:

Given, volume of the cube.

V = a^{3}= 64 cm^{3}

∴ a^{3}= 4 × 4 × 4 = 4^{3}, Hence a = 4 cm

When two cubes are added, the length of cuboid = 2a = 2 × 4 = 8 cm,

breadth = a = 4 cm.

height = a = 4 cm is formed.

∴ T.S.A. of the cuboid

= 2 (lb + bh + lh)

= 2(8 × 4 + 4 × 4 + 8 × 4)

= 2(32 + 16 + 32)

= 2 × 80

= 160 cm^{2}

∴ The surface area of resulting cuboid is 160 cm^{2}.

A storage tank consists of a circular cylinder with a hemisphere stuck on either end. If the external diameter of the cylinder be 1.4 m. and its length be 8 m. Find the cost of painting it on the outside at rate of Rs. 20 per m

Answer:

Total surface area of the tank = 2 × C.S.A. of hemisphere + C.S.A. of cylinder.

Hemisphere:

Radius (r) = d/2 = 1.4/2 = 0.7 m

C.S.A. of hemisphere = 2πr^{2}

= 2 × 22/7 × 0.7 × 0.7

= 3.08 m^{2}.

2 × C.S.A. = 2 × 3.08 m^{2}= 6.16 m^{2}

Cylinder:

Radius (r) = d/2 = 1.4/2 = 0.7 m

Height (h) = 8 m

C.S.A. of the cylinder = 2πrh

= 2 × 22/7 × 0.7 × 8

= 35.2 m^{2}

∴ Total surface area of the storage tank = 35.2 + 6.16 = 41.36 m^{2}

Cost of painting its surface area @ Rs. 20 per sq.m, is

= 41.36 × 20 = Rs. 827.2.

A hemisphere is cut out from one face of a cubical wooden block such that the diameter of the hemisphere is equal to the length of the cube. Determine the surface area of the remaining solid.

Answer:

Let the length of the edge of the cube = a units

T.S.A. of the given solid = 5 × Area of each surface + Area of hemisphere

Square surface:

Side = a units

Area = a^{2} sq. units

5 × square surface = 5a^{2}sq. units

Hemisphere:

Diameter = a units;

Radius = a/2

C.S.A. = 2πr^{2}

= 2π(a/2)2

= 2πa^{2}/4= πa^{2}/2 sq. units

Total surface area = 5a^{2}+πa^{2}/2= a^{2}(5+π2) sq. units.

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm and its base radius is of 3.5 cm, find the total surface area of the article.

Answer:

Surface area of the given solid = C.S.A. of the cylinder + 2 × C.S.A. of hemisphere.

If we take base = radius

Cylinder:

Radius (r) = 3.5 cm

Height (h) = 10 cm

C.S.A. = 2πrh

= 2 × 22/7 × 3.5 × 10

= 220 cm^{2}

Hemisphere:

Radius (r) = 3.5 cm

C.S.A. = 2πr^{2}

= 2 × 22/7 × 3.5 × 3.5

= 77 cm^{2}

2 × C.S.A. = 2 × 77 = 154 cm^{2}

∴ T.S.A. = 220 + 154 = 374 cm^{2}.

An iron pillar consists of a Cylindrical portion of 2.8 m. height and 20 cm. in diameter and a cone of 42 cm. height surmounting it. Find the weight of the pillar if 1 cm

Answer:

Volume of the iron pillar = Volume of the cylinder + Volume of the cone

Cylinder:

Radius = d/2 = 20/2 = 10 cm

Height = 2.8 m = 280 cm

Volume = πr^{2}h

=22/7 × 10 × 10 × 280

= 88000 cm^{3}

Cone:

Radius ‘r’ =d/2=20/2= 10 cm

height ‘h’ = 42 cm

Volume = 13πr^{2}h

=1/3 × 22/7 × 10 × 10 × 42

= 4400 cm^{3}

∴ Total volume = 88000 + 4400 = 92400 cm^{3}

∴ Total weight of the pillar at a weight of 7.5 g per 1 cm^{3} = 92400 × 7.5

= 693000 gms

=693000/1000 kg

= 693 kg.

A toy is made in the form of hemisphere surmounted by a right cone whose circular base is joined with the plane surface of the hemisphere. The radius of the base of the cone is 7 cm. and its volume is 3/2 of the hemisphere. Calculate the height of the cone and the surface area of the toy correct to 2 places of decimal.

(Take π = 3 1/7)

Answer:

Given r = 7 cm and

Volume of the cone =3/2 volume of the hemisphere

13πr^{2}h = 3/2 × 2/3 × πr^{3}

∴ h = 3r

= 3 × 7 = 21 cm

Surface area of the toy = C.S.A. of the cone + C.S.A. of hemisphere

Cone:

Radius (r) = 7 cm

Height (h) = 21 cm

= 22.135 cm.

∴ C.S.A. = πrl

=22/7 × 7 × 22.135 = 486.990 cm^{2}

Hemisphere:

Radius (r) = 7 cm

C.S.A. = 2πr^{2}

= 2 × 22/7 × 7 × 7

= 308 cm^{2}

C.S.A. of the toy = 486.990 + 308 = 794.990 cm^{2}

Find the volume of the largest right circular cone that can be cut out of a cube whose edge is 7 cm.

Answer:

Radius of the cone with the largest volume that can be cut out from a cube of edge 7 cm = 7/2 cm

Height of the cone = edge of the cube = 7 cm

∴ Volume of the cone V = 13πr^{2}h

= 1/3 × 22/7 × 7/2 × 7/2 × 7

= 89.83 cm^{3}.

A cylindrical tub of radius 5 cm and length 9.8 cm is full of water. A solid in the form of right circular cone mounted on a hemisphere is immersed into the tub. The radius of the hemi¬sphere is 3.5 cm and height of cone outside the hemisphere is 5 cm. Find the volume of water left in the tub. (Take π = 22/7)

Answer:

The tub is in the shape of a cylinder, thus

Radius of the cylinder (r) = 5 cm

Length of the cylinder (h) = 9.8 cm

Volume of the cylinder (V) = πr^{2}h

=22/7 × 5 × 5 × 9.8

Volume of the tub = 770 cm^{3}.

Radius of the hemisphere (r) = 3.5 cm

Volume of the hemisphere =2/3πr^{3}

=2/3 × 22/7 × 3.5 × 3.5 × 3.5

=22×12.25/3

=269.5/3

Radius of the cone (r) = 3.5 cm

Height of the Cone (h) = 5 cm

Volume of the cone V =1/3πr^{2}h

= 1/3 × 22/7 × 3.5 × 3.5 × 5

=192.5/3

Volume of the solid = Volume of the hemisphere + Volume of the cone

=269.5/3 + 192.5/3 = 462/3 = 154 cm^{3}

Now, when the solid is immersed in the tub, it replaces the water whose volume is equal to volume of the solid itself.

Thus the volume of the water replaced = 154 cm^{3}.

∴ Volume of the water left in the tub = Volume of the tub - Volume of the solid = 770 - 154 = 616 cm^{3}.

In the adjacent figure, the height of a solid cylinder is 10 cm and diameter 7 cm. Two equal conical holes of radius 3 cm and height 4 cm are cut off as shown in the figure. Find the volume of the remaining solid.

Answer:

Volume of the remaining solid = Volume of the given solid - Total volume of the two conical holes

Radius of the given cylinder (r) =d/2=7/2= 3.5 cm

Height of the cylinder (h) = 10 cm

Volume of the cylinder (V) = πr^{2}h

=22/7 × 3.5 × 3.5 × 10

=2695/7

= 385 cm^{3}.

Radius of each conical hole, ‘r’ = 3 cm

Height of the conical hole, h = 4 cm

Volume of each conical hole,

V =1/3πr^{2}h =1/3 × 22/7 × 3 × 3 × 4

=792/21

=264/7

Total volume of two conical holes = 2 × 264/7 = 528/7 cm^{3}

Hence, the remaining volume of the solid

Spherical marbles of diameter 1.4 cm. are dropped into a cylindrical beaker of diameter 7 cm., which contains some water. Find the number of marbles that should be dropped into the beaker, so that water level rises by 5.6 cm.

Answer:

Rise in the water level is seen in cylindrical shape of Radius = Beaker radius

= d/2 = 7/2 = 3.5 cm

Height ‘h’ of the rise = 5.6 cm.

∴ Volume of the ‘water rise’ = πr^{2}h

=22/7 × 3.5 × 3.5 × 5.6

=22×12.25×5.6/7

= 215.6

Volume of each marble dropped =4/3 πr^{3}

Where radius r = d/2 = 1.4/2 = 0.7 cm

∴ V = 4/3 × 22/7 × 0.7 × 0.7 × 0.7

= 1.4373 cm^{3}

∴ Volume of the ‘rise’ = Total volume of the marbles.

Let the number of marbles be ‘n’ then n × volume of each marble = volume of the rise.

n × 1.4373 = 215.6

=215.6/1.4373

∴ Number of marbles = 150.

A pen stand is made of wood in the shape of cuboid with three conical depressions to hold the pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depression is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Answer:

Volume of the wood in the pen stand = Volume of cuboid - Total volume of three depressions.

Length of the cuboid (l) = 15 cm

Breadth of the cuboid (b) = 10 cm

Height of the cuboid (h) = 3.5 cm

Volume of the cuboid (V) = lbh = 15 × 10 × 3.5 = 525 cm^{3}.

Radius of each depression (r) = 0.5 cm

Height / depth (h) = 1.4 cm

Volume of each depressions V =1/3πr^{2}h

=1/3 × 22/7 × 0.5 × 0.5 × 1.4

=7.7/3×7 = 1.1/3 cm^{3}

Total volume of the three depressions = 3 × 1.1/3

= 1.1 cm^{3}

∴ Volume of the wood = 525 - 1.1 = 523.9 cm^{3}

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.

Answer:

Given, sphere converted into cylinder.

Hence volume of the sphere = volume of the cylinder.

Sphere:

Radius, r = 4.2 cm

Volume V = 4/3πr^{3}

=4/3 × 22/7 × 4.2 × 4.2 × 4.2

= 4 × 22 × 0.2 × 4.2 × 4.2

= 4 x 22 x 0.2 x 4.2 x 4.2

= 310.464

Cylinder:

Radius, r = 6 cm

Height h = h say

Volume = πr^{2}h

=22/7 × 6 × 6 × h

=22×36/7 h

=792/7 h

Hence, 792/7 h = 310.464

h =310.464×7/792 = 2.744cm

!! π can be cancelled on both sides i.e., sphere = cylinder

Three metallic spheres of radii 6 cm., 8 cm. and 10 cm. respectively are melted together to form a single solid sphere. Find the radius of the resulting sphere.

Answer:

Given : Radii of the three spheres r_{1}= 6 cm r_{2}= 8 cm r_{3}= 10 cm

These three are melted to form a single sphere.

Let the radius of the resulting sphere be ‘r’.

Then volume of the resultant sphere = sum of the volumes of the three small spheres.

∴ 1728 = (2 × 2 × 3) × (2 × 2 × 3) × (2 × 2 × 3)

r^{3} = 12 × 12 × 12

r^{3} = 12^{3}

∴ r = 12

Thus the radius of the resultant sphere = 12 cm

A 20 m deep well with diameter 7 m. is dug and the earth got by digging is evenly spread out to form a rectangu¬lar platform of base 22 m. × 14 m. Find the height of the platform.

Answer:

Volume of earth taken out = πr^{2}h

=22/7 × 7/2 × 7/2 × 20

= 770 m

Let height of plot form = H m.

∴ 22 × 14 × H = 22/7 × 7/2 × 7/2 × 20

H =35/14 = 5/2 = 2 1/2 m

∴ The height of the plat form is 2 1/2 m

A well of diameter 14 m. is dug 15 m. deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 7 m to form an embankment. Find the height of the embankment

Answer:

Volume of the well = Volume of the embank

Volume of the cylinder = Volume of the embank

Cylinder :

Radius r = d/2 = 14/2 = 7 cm

Height/depth, h = 15 m

Volume V = πr^{2}h

=22/7 × 7 × 7 × 15

= 22 × 7 × 15

= 2310 m^{3}

Let the height of the embank = h m

Inner radius ‘r’ = Radius of well = 7 m

Outer radius, R = inner radius + width

= 7m + 7m = 14 m

Area of the base of the embank = (Area of outer circle) ? (Area of inner circle)

= πR^{2} - πr^{2}

= π(R^{2} - r^{2})

=22/7(14^{2}?7^{2})

=22/7× (14+7) × (14-7)

=22/7× 21 × 7

= 462 m^{2}

∴ Volume of the embank = Base area × height

= 462 × h = 462 h m^{3}

∴ 462 h m^{3}= 2310 m^{3}

h =2310/462= 5 m.

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.

Answer:

Let the number of cones that can be filled with the ice-cream be ‘n’.

Then total volume of all the cones with a hemi spherical top = Volume of the ice-cream

Ice-cream cone = Cone + Hemisphere = πr^{2}h

Cone:

Radius = d/2 = 6/2 = 3 cm

Height, h = 12 cm

Volume V = 1/3πr^{2}h

= 1/3 × 22/7 × 3 × 3 × 12

= 22/7 × 36

= 792/7

Hemisphere:

Radius = d/2 = 6/2 = 3 cm

Volume V = 23πr^{3}

= 2/3 × 22/7 × 3 × 3 × 3

= 44×9/7

= 396/7

∴ Volume of each cone with ice-cream = 792/7 + 396/7 = 1188/7 cm^{3}

Cylinder:

Radius = d/2 = 12/2 = 6 cm

Height, h = 15 cm

Volume V = πr^{2}h

= 22/7 × 6 × 6 × 15

= 22×36×15/7

= 11880/7

∴ 11880/7 = n × 11880/7

⇒ n = 11880/7 × 7/1188 = 10

∴ n = 10.

How many silver coins, 1.75 cm in diameter and thickness 2 mm., need to be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?

Answer:

Let the number of silver coins needed to melt = n

Then total volume of n coins = volume of the cuboid

n × πr^{2}h = lbh [∵ The shape of the coin is a cylinder and V = πr^{2}h]

∴ 400 silver coins are needed.

A vessel is in the form of an inverted cone. Its height is 8 cm. and the radius of its top is 5 cm. It is filled with water up to the rim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, 1/4 of the water flows out. Find the number of lead shots dropped into the vessel.

Answer:

Let the number of lead shots dropped = n

Then total volume of n lead shots = 14 volume of the conical vessel.

Lead shots:

Radius, r = 0.5 cm

Volume V = 4/3πr^{3}

= 4/3 × 22/7 × 0.5 × 0.5 × 0.5

Total volume of n - shots

= n × 4/3 × 22/7 × 0.125

Cone:

Radius, r = 5 cm;

Height, h = 8 cm

Volume, V = 1/3 πr^{2}h

= 1/3 × 22/7 × 5 × 5 × 8

= 1/3 × 22/7 × 200

∴ Number of lead shots = 100.

A solid metallic sphere of diameter 28 cm is melted and recast into a number of smaller cones, each of diameter 4 d2 cm and height 3 cm. Find the number of cones so formed.

Answer:

Let the no. of small cones = n Then,

total volume of n cones = Volume of sphere Diameter = 28 cm.

Cones:

Radius r = d/2

Height, h = 3 cm

Total volume of n-cones = n . 154/9 cmPDF Download

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