RBSE Solutions for Class 10 Maths Chapter 16 Surface Area and Volume Additional Questions is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 16 Surface Area and Volume Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 16 |

Chapter Name |
Surface Area and Volume |

Exercise |
Additional Questions |

Number of Questions Solved |
80 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 16 Surface Area and Volume Additional Questions

**Multiple Choice Questions
**Question 1.

A pencil, pointed at one and has the (RBSESolutions.com) assembling as: [NCERT Exemplar Problem]

(A) A cone and a cylinder

(B) A frustum and a cylinder

(C) A hemisphere and a cylinder

(D) Two cylinders

Solution :

Option (A) is correct.

Question 2.

A Surahi is a combination of: [NCERT Exemplar Problem]

(A) A sphere and a cylinder

(B) A hemisphere and a cylinder

(C) Two hemispheres

(D) A cylinder and a cone

Solution :

(A) is the correct option.

Question 3.

A hollow cube with Internal cube 22 cm is filled with the (RBSESolutions.com) spherical marbles with diameter 0.5 cm. and assumed that 1/8 part of the cube can not be filled up. Number of marbles incorporated into cube will be : [NCERT Exemplar Problem]

(A) 142296

(B) 142396

(C) 142496

(D) 142596

Solution :

Given,

Internal side of cube = 22 cm

And diameter of spherical mable = 0.5 cm

Internal volume of cube = 22 × 22 × 22 cm^{3}

When \(\frac { 1 }{ 8 }\) part of hollow cube can (RBSESolutions.com) not be filled by the marbles.

∴ The part of hollow cube filled with marbles = 1 – \(\frac { 1 }{ 8 }\) = \(\frac { 7 }{ 8 }\)

The volume of the part of the hollow cube filled with marbles = \(\frac { 7 }{ 8 }\) × 22 × 22 × 22

From equation (i) and (ii), we get

Hence, option (A) is correct.

Question 4.

If the ratio of radii of two cones is 3 : 1 and ratio of (RBSESolutions.com) their height is 1 : 3, then the ratio of their volumes is :

(A) 2 : 1

(C) 1 : 3

(B) 3 : 1

(D) 1 : 2

Solution :

Let radii of two cones are r_{1} and r_{2}

respectively heights be h_{1} and h_{2}.

∴ According to the question

Ratio of volumes = 3 : 1

Hence, option (B) is correct.

Question 5.

A largest cone is cut off from a cube with (RBSESolutions.com) core 14 cm. The volume of cone is :

(A) 766.18 cm^{3}

(B) 817.54 cm^{3}

(C) 1232 cm^{3}

(D)718.66 cm^{3}

Solution :

A Largest cone is cut off from a cube (RBSESolutions.com) with its core 14 cm.

∴ Height of cone = 14 cm

And diameter of base = 14 cm

∴ Base radius of cone = \(\frac { 14 }{ 2 }\) = 7 cm

Volume of cone = \(\frac { 1 }{ 3 }\)πr^{2}h

= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (7)^{2} × 14

= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × 14

= \(\frac { 1 }{ 3 }\) × 22 × 7 × 14

= \(\frac { 22\times 98 }{ 3 } \)

= \(\frac { 2156 }{ 3 }\)

= 718.66 cm^{3}

Hence, option (D) is correct.

Question 6.

The ratio of volumes of two spheres is 64 : 27. The ratio (RBSESolutions.com) of their surface areas will be : [NCERT Exemplar Problem]

(A) 3 : 4

(B) 4 : 3

(C) 9 : 16

(D)16 : 9

Solution :

Given,

Volume of I^{st} sphere : Volume of II^{nd} sphere = 64 : 27

Hence, option (B) is correct.

Question 7.

Having melted a hollow spherical shell with internal and external (RBSESolutions.com) radii 4 cm and 8 cm, a cone with base diameter 8 cm is recasted.

Height of recasted cone is : [NCERT Exemplar Problem]

(A) 12 cm

(B) 14 cm

(C) 15 cm

(D) 18 cm

Solution :

Given,

Internal diameter of spherical shell = 4 cm.

∴ Internal radius (r_{1}) = \(\frac { 4 }{ 2 }\) = 2 cm

External diameter of spherical shell = 8 cm

∴ External radius (r_{2}) = \(\frac { 8 }{ 2 }\) = 4 cm

Base diameter of cone = 8 cm

∴ Radius of cone = \(\frac { 8 }{ 2 }\) = 4 cm

Let height of cone be h.

Conical shape is recastecd by melting the spherical cell

∴ Volume of spherical shell = Volume of cone recasted

Hence, option (B) is correct.

Question 8.

12 spheres of some volumes are recasted by melting a cylindrical (RBSESolutions.com) solid with base diameter 2 cm and height 16 cm. The diameter of each sphere recasted is : [NCERT Exemplar Problem]

(A) 4 cm

(B) 3 cm

(C) 2 cm

(D) 6 cm

Solution :

Given

Height of metallic cylinder (h) = 16 cm

And diameter of its base = 2 cm

∴ Base radius (r_{1}) = \(\frac { 2 }{ 2 }\) = 1 cm

Let radius of each sphere recasted be r_{2} since, by melting the solid cylinder, 12 solid spheres of same volumes are recasted.

∴ Volume of solid cylinder = Volume of 12 spheres

∴ Diameter of solid sphere = 1 × 2 = 2 cm

Hence, option (C) is correct.

Question 9.

If a solid is reformed into another shape, its (RBSESolutions.com) volume will be : [NCERT Exemplar Problem]

(A) increased

(B) decreased

(C) remain same

(D) doubled

Solution :

Option (C) is correct.

Question 10.

A spherical solid is melted and recasted into a cone. If radius of spherical (RBSESolutions.com) solid is r cm and the height of cone is also r cm, then radius of the cone recasted is :

(A) 2r

(B) 3r

(C) r

(D) 4r

Solution :

Given

Radius of sphere = r

Height of cone = r

Let the radius of cone be x.

According to question,

Volume of spherical solid = Volume of cone

∴ Base radius of cone recasted = 2r.

Hence, option (A) is correct.

Question 11.

A metallic cone is turned into the cylindrical shape with same (RBSESolutions.com) radius. If the height of cylinder is 5 cm, the height of cone is :

(A) 10 cm

(B) 15 cm

(C) 18 cm

(D)24 cm

Solution :

Let the common radius of the cone and cylinder be r.

Let height of cone be h_{1}.

And the height of cylinder h_{2} = 5 cm.

According to the problem

Volume of cone = Volume of cylinder

⇒ \(\frac { 1 }{ 3 }\)πr^{2}h_{1} = πr^{2}h_{2}

⇒ \(\frac { { h }_{ 1 } }{ 3 }\) = h_{2} = 5

⇒ h_{1} = 3 × 5

h_{1} = 15

∴ Height of cone = 15 cm

Hence, option (B) is correct.

Question 12.

A copper wire of diameter 5 mm is evenly wrapped on a (RBSESolutions.com) cylinder of length 15 cm to cover its whole surface. The number of wrap will be :

(A) 75

(B) 50

(C) 45

(D) 30

Solution :

Given,

Diameter of wire = 5 mm = \(\frac { 5 }{ 10 }\) cm

length of the cylinder = 15 cm

∴ Number of wrap

∴ Number of wrap = 30.

Hence option (D) is correct.

Question 13.

If base radius of cone is 21 cm and slant height is 35 cm, the (RBSESolutions.com) total surface area of cone will be :

(A) 3096 cm^{2}

(B) 3696 cm^{2}

(C) 5696 cm^{2}

(D)None of these

Solution :

Given,

Base radius of cone (r) = 21 cm

Slant Height of cone (l) = 35 cm

Total surface area = πr(l + r)

= \(\frac { 22 }{ 7 }\) × 21 × (35 + 21)

= 22 × 3 × 56

= 3696 cm^{2}

Hence, option (D) is correct.

Question 14.

The sides of right angled triangle are 3 cm and 4 cm. If the triangle is (RBSESolutions.com) revolved round the hypotenuse. The radius of the obtained double cone will be:

(A) 2.4 cm

(B) 2.8 cm

(C) 5 cm

(D) 6 cm

Solution:

Let in ∆ABC,

∠B = 90°, AB = 4 cm, BC = 3 cm

Hence, option (A) is correct.

Question 15.

The sides of three cubes are 3 cm, 4 m and 5 cm respectively. The (RBSESolutions.com) side of the cube which recasted by the melting three cubes is:

(A) 6 cm

(B) 5 cm

(C) 7 cm

(D) 4 cm

Solution :

Volume of I^{st} cube = (3)^{3} = 27 cm^{3}

Volume of II^{nd} cube (4)^{3} = 64 cm^{3}

Volume of III^{rd} cube = (5)^{3} = 125 cm^{3}

Sum of the volumes of these three cubes = 27 + 64 + 125 = 216 cm^{3}

Now cube is recasted with these three cubes.

∴ Volume of the cubes recasted

= Sum of the volumes of three cubes.

(core)^{3} = 216

core = \(\sqrt [ 3 ]{ 216 }\)

= \(\sqrt [ 3 ]{ 6\times 6\times 6 } \) = 6 cm

Hence, the option (A) is correct.

Question 16.

Having melted a cuboid of dimensions 18 cm × 12 cm × 9 cm, the cubes (RBSESolutions.com) with side 3 cm are recasted. The number of cubes recasted is:

(A) 60

(B) 55

(C) 69

(D) 72

Solution :

Given,

Dimensions of cuboid = 18 cm × 12 cm × 9 cm

∴ Volume = 18 × 12 × 9

= 18 × 108

= 1944 cm^{3}

Volume of cube with side 3 cm = (3)^{3} = 27 cm^{3}

Hence, option (D) is correct.

Question 17.

The base radius of cylinder is 7 cm and its height Is 3 cm. Measurement (RBSESolutions.com) of Its curved surface area is:

(A) 132 cm^{2}

(B) 220 cm^{2}

(C) 1232 cm^{2}

(D) 440 cm^{2}

Solution :

Given,

r = 7 cm

h = 3 cm

∴ Curved surface area of cylinder = 2πrh

= 2 × \(\frac { 22 }{ 7 }\) × 7 × 3

= 132 cm^{2}.

Question 18.

The diameter of a sphere Is d, the (RBSESolutions.com) volume of the sphere is :

(A) \(\frac { 1 }{ 3 }\)πd^{3} cm^{3}

(B) \(\frac { 1 }{ 24 }\)πd^{3} cm^{3}

(C) \(\frac { 4 }{ 3 }\)πd^{3} cm^{3}

(D) \(\frac { 1 }{ 6 }\)πd^{3} cm^{3}

Solution :

Given,

Diameter of sphere = d

∴ radius = \(\frac { d }{ 2 }\)

Volume of sphere = \(\frac { 4 }{ 3 }\)πr^{3}

= \(\frac { 4 }{ 3 }\)π\({ \left( \frac { d }{ 2 } \right) }^{ 2 }\)

= \(\frac { 1 }{ 6 }\)πd^{3} cm^{3}

Hence option (D) is correct.

Question 19.

In a cylindrical vessel whose height is equal to its (RBSESolutions.com) diameter, a sphere of same diameter is put. Volume of this sphere is part of cylinder.

(A) one third

(B) two third

(C) double

(D) thrice

Solution :

Let the radius of the base of cylinder be r.

∴ Diameter = 2r

And height of cylinder is (RBSESolutions.com) equal to its diameter.

∴ Height = 2r

Radius of sphere = r

∴ Volume of sphere = \(\frac { 2 }{ 3 }\) Volume of Cylinder

Hence option (B) is correct.

Question 20.

The upper part of a cone has been separated by (RBSESolutions.com) cutting it at half of its height. If the Volume of whole cone is V, the volume of remaining

part of cone is :

(A) \(\frac { V }{ 2 }\)

(B) \(\frac { 3V }{ 2 }\)

(C) \(\frac { 5V }{ 6 }\)

(D) \(\frac { 7V }{ 8 }\)

Solution :

Volume of whole cone = \(\frac { 1 }{ 3 }\)πr^{2}h

Volume of upper smaller cone = \(\frac { 1 }{ 3 }\)π\({ \left( \frac { r }{ 2 } \right) }^{ 2 }\) × \(\frac { h }{ 2 }\)

∴ Volume of remaining part of cone = V – \(\frac { V }{ 8 }\)

= \(\frac { 7V }{ 8 }\)

Hence, option (D) is correct.

**Very Short Answer Type Questions**

Question 1.

Side of a cube ¡s 6.5 cm. Find its (RBSESolutions.com) total surface area.

Solution :

Given,

Side (a) = 6.5 cm

Total surface area = 6a^{2}

= 6 × (6.5)^{2}

= 6 × 6.5 × 6.5

= 253.5 cm^{2}

Question 2.

The dimensions of a under ground water (RBSESolutions.com) tank are 48 m, 36 m and 28 m. Find its volume.

Solution :

Given:

Length of water tank (l) = 48 m

Breath (b) = 36 m

Height (h) = 28 m

∴ Volume of the water tank = l × b × h

= 48 × 36 × 28

= 48384 m^{3}.

Hence, volume of water tank = 48384 m^{3}

Question 3.

The length and breadth of a tank are 4 m and 3 m respectively How (RBSESolutions.com) many cubic meter should be poured into it so that the depth of water in the take becomes 2 m?

Solution :

Given,

Length of the tank (l) = 4 m

Breadth (b) = 3 m

Depth of water into tank (h) = 2 m

Volume of the water poured into tank = l × b × h

= 4 × 3 × 2

= 24 m^{3}

Hence, 24 m^{3} water should be poured.

Question 4.

Total surface area of a cube is 1014 m^{2}. Find the (RBSESolutions.com) length of its side.

Solution :

Let the side of cube be a

∴ Total surface area of cube = 1014 cm^{2}

⇒ 6a^{2} = 1014

⇒ a^{2} = \(\frac { 1014 }{ 6 }\) = 169

⇒ a = \(\sqrt { 169 }\) = 13 m

Hence, the side of cube is 13 m.

Question 5.

The base area of a right circular cylinder is 63 π cm^{2}. If its height is 10 cm. If its (RBSESolutions.com) height is 10 cm, find the volume.

Solution :

Base area of cylinder = 63 π cm^{2}

Volume of cylinder = base area × height

= 63 π × 10

= 630 π cm^{3}

Hence, the volume of cylinder = 630 π cm^{3}

Question 6.

The base diameter and height of right circular cylinder are 10 cm and 14 cm respectively. Find its curved surface area.

Solution :

Given

Height (h) = 14 cm

and Base diameter = 10 cm

∴ Base radius (r) = \(\frac { 10 }{ 2 }\) = 5 cm

Curved surface area = 2πrh

= 2 × \(\frac { 22 }{ 7 }\) × 5 × 14

= 440 cm^{2}

Question 7.

The base radius of a right cylindrical cup is 3 cm and its height 8 cm. The cup is (RBSESolutions.com) filled with water by the half of its height. Find the volume of water in the cup.

Solution :

Given,

Base radius (r) = 3 cm

Height (h) = 8 cm

Volume of right cylindrical cup = πr^{2}h

= π × (3)^{2} × 8

= 72 π cm^{3}

Volume of water in the cup = \(\frac { 72\pi }{ 2 } \)

= 36 π cm^{3}

Question 8.

The ratio of the base radii of two right cylinders in 1 : 2. If ratio of (RBSESolutions.com) their volumes is 5 : 12, find the ratio of their heights.

Solution :

Let base radius of I^{st} cylinder be r_{1} = 1

And base radius of II^{nd} cylinder be r_{2} = 2

Volume of Ist cylinder V_{1} = 5

Volume of IInd cylinder V_{1} = 12

∴ h_{1} : h_{2} = 5 : 3

Question 9.

The volume of a right cylinder is 2160 cm^{3}. If the base diameter of (RBSESolutions.com) cylinder is 24 cm, then find its height.

Solution :

Given V = 2160 cm^{3}

Diameter (d) = 24 cm

∴ radius (r) = \(\frac { 24 }{ 2 }\) = 12 cm

∵ V = πr^{2}h = 2160

Question 10.

The length of cylindrical hollow brass pipe is 21 cm. Its external and internal (RBSESolutions.com) diameters are 10 cm and 6 m respectively. Find the volume of brass used to make the pipe.

Solution :

Given h = 21 cm

Length of the pipe (h) = 10 cm

External radius (r_{1}) = \(\frac { 10 }{ 2 }\) = 5 cm

Internal radius r_{2}= \(\frac { 6 }{ 2 }\) = 3 cm

Volume of brass used to make the pipe

Question 11.

The circumference of the base of a right cone is 44 cm. If its vertical (RBSESolutions.com) height is 24 cm, find its slant height.

Solution :

Given,

Vertical height of cone (h) = 24 cm

Circumference of base = 44 cm

⇒ 2πr = 44

2 × \(\frac { 22 }{ 7 }\) × r = 44

r = 7 cm

Question 12.

A conical tent contains 528 m^{3} of air inside it. If the vertical height of (RBSESolutions.com) the tent is 14 m, find the diameter of tent.

Solution :

Given,

Vertical height of tent = 14 m

Volume of the tent = Volume of the air inside tent

= 528 m^{3}

Hence, diameter of tent = 2 × 6 = 12 meter.

Question 13.

The vertical height of a conical tent is 12 m and radius of its circular (RBSESolutions.com) floor is 14 m. If one person needs 112 m^{3} of air to breath, then how many person can be stand inside in the tent?

Solution :

Given :

Height of conical tent (h) = 12 m

Radius of circular floor (r) = 14 m.

Essential air for one man = 112 m^{3}

Total number of men inside the tent

∴ Total number of men inside the tent = 22.

Question 14.

The height of a right cone is equal to its base radius and (RBSESolutions.com) volume is 9π cm^{3}. Find the height of given cone.

Solution :

Let radius of cone be r and height be h.

Volume of cone = \(\frac { 1 }{ 3 }\)πr^{2}h

⇒ 9π = \(\frac { 1 }{ 3 }\) × (h)^{2} × h × π

⇒ h^{3} = 27

h = 3

Hence, the height of cone is 3 cm.

Question 15.

How many times the volume of cone will be increased if its base radius is made double for the same height?

Solution :

Let the radius of cone be r and height be h also let the volume be V_{1},

Then V_{1} = \(\frac { 1 }{ 3 }\)πr^{2}h

Now, when radius is 2r and height is h, then the new volume

V_{2} = \(\frac { 1 }{ 3 }\)π(2r)^{2}h

Hence, the volume will be increased by 4 times.

Question 16.

The diameter of a cricket ball is 14 m. Find the (RBSESolutions.com) surface area and volume.

Solution :

Given :

Diameter of cricket ball = 14 cm

Radius = \(\frac { 14 }{ 2 }\) = 7 cm

Surface area of ball = 4πr^{2}

= 4 × \(\frac { 22 }{ 7 }\) × 7 × 7

= 4 × 22 × 7

= 616 cm^{2}

Volume of the ball = \(\frac { 4 }{ 3 }\)πr^{3}

= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (7)^{3}

= 1437.33 cm^{3}

Question 17.

Find the surface area and the volume of a ball whose radius is 14 cm.

Solution :

Given, r = 14 cm

Surface area of ball = 4πr^{2}

= 4 × \(\frac { 22 }{ 7 }\) × 14 × 14

= 2464 cm^{2}

Volume of the ball = \(\frac { 4 }{ 3 }\)πr^{3}

= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (14)^{3}

= 40245.33 cm^{3}

Question 18.

A right circular cone and hemisphere have same base and (RBSESolutions.com) volumes. Find the ratio of the height of the cone and the hemisphere.

Solution :

Let height of cone be h.

And radius of hemisphere (r) = height of the hemisphere (H)

Volume of cone = Volume of hemisphere

\(\frac { 1 }{ 3 }\)πr^{2}h = \(\frac { 2 }{ 3 }\)πr^{2} × H

⇒ h = 2H

⇒ \(\frac { h }{ H }\) = \(\frac { 2 }{ 1 }\)

⇒ h : H = 2 : 1

Question 19.

If the radius of a solid hemisphere is 3.5 cm. Find its volume.

Solution :

Radius of hemisphere (r) = 3.5 cm

Volume of hemisphere = \(\frac { 2 }{ 3 }\)πr^{3}

= \(\frac { 2 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (3.5)^{3}

= 89.83 cm^{3}

Question 20.

If the radius of hemisphere is 4 cm, then (RBSESolutions.com) find its curved surface area.

Solution :

Given,

Radius of hemisphere (r) = 4 cm

Curved surface area = 2πr^{2}

= 2 × π × 4^{2}

= 32π cm^{2}

**Short Answer Type Questions**

Question 1.

The external and internal radii of a hollow right circular cylinder are 5 cm and 4 cm respectively. If the volume of metal used to make it is 198 cm^{3}, find the height of cylinder.

Solution :

Given External radius of hollow cylinder r_{1} = 5 cm

Internal radius (r_{1}) = 4 cm.

Volume of hollow cylinder = 198 cm^{3}

Hence height of cylinder = 7 cm

Question 2.

A conical vessel can be completely filled with 8 liter (RBSESolutions.com) of water. Find the total surface area of vessel.

Solution :

Quantity of water in the vessel = 8 liter

∴ Volume of the vessel = 8 l

= 8 × 1000 cm^{3} [∵ 1 litre = 1000 cm^{3}]

Let the side of conical vessel be = x

∴ Volume of the vessel = x^{3} cm^{3}

∴ 8000 = x^{3}

x = \(\sqrt [ 3 ]{ 8000 }\)

= (20 × 20 × 20)^{1/3}

= [(20^{3})]^{1/3}

= 20 cm

∴ Total surface area of conical vessel

= 6x^{2} = 6 × (20)^{2}

= 6 × 400

= 2400 cm^{2}

Hence, total surface area of conical vessel = 2400 cm^{2}

Question 3.

Two hemispheres of same base are fixed on the two circular ends (RBSESolutions.com) of a right cylinder. Total length of solid is 19 cm and diameter is 7 cm. Find its total surface area. \(\left( \pi =\frac { 22 }{ 7 } \right) \)

Solution :

Given

Diameter of solid = 7 cm

Hence, common radius of the cylinder and hemisphere = \(\frac { 7 }{ 2 }\) cm

And the total height of solid = 19 cm

∴ Height of the cylinder (h) 19 – 2 × \(\frac { 7 }{ 2 }\) = 12 cm

Total surface area of solid curved surface area of cylinder + 2 × curved surface area of hemisphere

Hence, total surface area of solid = 418 cm^{2}

Question 4.

Height and base radius of a conical (RBSESolutions.com) tent are 10 m and 24 m respectively. Find

(i) Slant height of the tent

(ii) Cost the canvas used in the tent at the rate of ₹ 70/m^{2}.

Solution:

Given,

(i) Base radius of tent (r) = 24 m and height (h) = 10 m

Hence, slant height of tent (l) = 26 meter.

(ii) Curved surface area of the conical tent = πrl

= \(\frac { 22 }{ 7 }\) × 24 × 26

= \(\frac { 13,728 }{ 7 }\) m^{2}

∴ Cost of canvas = Area of canvas × rate of per m^{2} canvas

= \(\frac { 13,728 }{ 7 }\) × 70 = ₹ 1,37,280

Hence, the cost of canvas used = ₹ 1,37,280.

Question 5.

A sphere with radius 8 cm is melted and recasted into small (RBSESolutions.com) spheres of equal radius 1 cm each. Find the number of spheres recasted.

Solution :

Given

Radius of given sphere (r) = 8 cm

∴ Its volume = \(\frac { 4 }{ 3 }\)πr^{3}

= \(\frac { 4 }{ 3 }\) × π × 8 × 8 × 8 cm^{3}

Radius of the sphere recasted = 1 cm.

Its volume = \(\frac { 4 }{ 3 }\)π(1)^{3}

= \(\frac { 4 }{ 3 }\)π cm^{3}

No. of spheres recasted

Hence number of spheres recasted = 512

Question 6.

A copper rod of radius 1 cm and length 2 cm, is converted into 18 meter (RBSESolutions.com) long circular wire. Find the thickness of the wire. [Higher secondary board Raj. 2013, 15]

Solution :

Given

Radius of copper rod (r) = 1 cm

Length of the copper rod (l) = 2 cm

∴ Volume of copper rod = πr^{2}h = π × (1)^{2} × 2

= 2 π cm^{3}

the Length of the wire (l_{1}) = 18 meter

= 1800 cm

If the radius of transverse section of wire be R

The volume of wire = πR^{2}h = πR^{2} × 1800

R = \(\frac { 1 }{ 30 }\) cm

Diameter of wire = 2R = 2 × \(\frac { 1 }{ 30 }\) = \(\frac { 1 }{ 15 }\) cm

Hence, the thickness of the wire = diameter = \(\frac { 1 }{ 15 }\) cm = 0.67 cm(approx)

Question 7.

A metallic sphere of radius 4.2 cm is melted and recasted into a cylinder (RBSESolutions.com) with its radius 7 cm. Find the height of the cylinder recasted. [Higher secondary board Raj. 2013, 15]

Solution : Given

The radius of metallic sphere (r) = 4.2 cm

And the radius of cylinder reasted (R) = 7 cm

∵ Volume of sphere = Volume of cylinder recasted

Hence, Height of cylinder = 6.048 cm

Question 8.

How many silver coins with diameter 1.75 cm and thickness 2 mm will have (RBSESolutions.com) to melted to recast a cuboid with dimensions 5.5 cm × 10 cm × 3.5 cm.

Solution:

Let the number of silver coins to melt be n

The diameter of each coin = 7.5 cm

∴ Radius (r) = \(\frac { 1.75 }{ 2 }\) cm

= \(\frac { 175 }{ 200 }\) cm = \(\frac { 7 }{ 8 }\) cm

The thickness of each coin (h) = 2 mm

= \(\frac { 2 }{ 10 }\) cm = \(\frac { 1 }{ 5 }\) cm

∴ Volume of each coin = πr^{2}h

∴ Volume of n coins = \(\frac { 77 }{ 160 }\)n cm^{3}

Volume of cuboid = 5.5 × 10 × 3.5 = 192.5 cm^{3}

Since the cuboid is recasted by melting the n silver coins.

⇒ Volume of n coins = Volume of Cuboid

⇒ \(\frac { 77 }{ 160 }\)n = 192.5

⇒ n = \(\frac { 192.5\times 160 }{ 77 } \) = 400

Hence, 400 silver coins will be melted.

Question 9.

A hemisphere with internal radius 9 cm ¡s completely filled with water. This water is (RBSESolutions.com) radius into a cylindrical vessel of internal poured 6 cm. Find the height of water in this vessel. [CBSE 2012]

Solution :

Given,

Internal radius of hemisphere (R) = 9 cm

Internal radius of cylindrical vessel (r) = 6 cm

Let the height of water in the cylindrical vessel since, water from the hemispheric bowl is poured into the cylindrical vessel.

∴ Volume of cylindrical vessel = Volume of hemispheric bowl.

⇒ h = \(\frac { 27 }{ 2 }\)

⇒ h = 13.5 cm

Hence, the height of water into cylindrical vessel = 13.5 cm

Question 10.

A hemispheric tank filled with water is made empty by a pipe at the rate of 5 liter per second. If (RBSESolutions.com) diameter of tank is 3.5 m, then how among time will be taken to make half of the filled tank empty. [Higher secondary board Raj. 2014]

Solution :

Given,

Diameter of hemispheric tank = 3.5 m

∴ Radius (r) = \(\frac { 3.5 }{ 2 }\) m

The volume of hemispheric tank = \(\frac { 2 }{ 3 }\)πr^{3}

Hence, time taken to empty half tank = 10.69 minutes.

Question 11.

By joining the two corresponding faces of two cubes of volume 27 cm^{3} each, another (RBSESolutions.com) solid is formed. Find the surface area of the obtained cuboid. [Higher secondary education board Raj. 2014]

Solution :

Let side of the cube be a

Volume of cube = 27 cm^{3}

27 cm^{3} = a^{3}

a^{3} = 27

a = 3 cm

Now, the length of cuboid form = a + a = 3 + 3 = 6 cm

Breadth = a = 3 cm

Height = a = 3 cm

∴ Surface area of cuboid

= 2 (l × b + b × h + h + l)

= 2 (3 × 3 + 3 × 6 + 6 × 3)

= 2(9 + 18 + 18)

= 2(45) = 90 cm^{2}

Question 12.

If the volume and surface area of a cuboid with (RBSESolutions.com) dimensions a, b and c are V and S respectively, then prove that :

Solution :

Question 13.

The curved surface area of a cone with (RBSESolutions.com) height 24 cm is 550 cm^{2}. Find its volume.

Solution :

Given:

Height of the cone (h) =24 cm

Let the radius of cone be r.

Question 14.

The diameter of a roller is 84 cm and its length ¡s 120 cm. If it takes 500 (RBSESolutions.com) revolutions to make a field plane, then find the cost of making the field plane at the rate of paise 30 per meter^{2}.

Solution :

Diameter of roller = 84 cm

∴ Radius = \(\frac { 84 }{ 2 }\) = 42 cm

Length of the roller (h) = 120 cm

Curved surface area of the roller = 2πrh

=2 × \(\frac { 22 }{ 7 }\) × 42 × 120

= 31680 cm^{2}

= 3.1680 m^{2}

∴ The area covered by the roller in one revolution = 3.168 m^{2}

∴ area covered in 500 revolutions

= 500 × 3.168

= 1584 m^{2}

∴ The cost of planing the field at the rate 30 paise per m^{2} = 0.30 × 1584 = ₹ 475.20

Question 15.

The depth of metallic conical tank is 4 m and diameter of Its circular base is 6 m. Find (RBSESolutions.com) the area of metal sheet required to make this tank. (take, π = 3.14)

Solution :

Given,

Base diameter of conical tank = 6 m

∴ Radius = \(\frac { 6 }{ 2 }\) = 3 m

Height of tank = 4 m

Let the slant height of tank be l

∴ Area of metal.sheet required = Surface area of conical tank

= πrl

= 3.14 × 3 × 5

= 47.10 m^{3}

Question 16.

The flow of water in a canal of breadth 30 m and depth 12 dm is 20 km/h. If 9 cm (RBSESolutions.com) height of water is needed in fields, then how much area of a field will be watered with in 30 minutes.

Solution :

Given,

Breadth of canal = 30 dm = 3 m

Depth of the canal = 12 dm = 1.2 m

The length water flowing in one hour = 20 km = 20,000 m

The length of water flowing in 30 minutes = \(\frac { 20000 }{ 2 }\) = 10000 m.

The volume of the water flow in 30 minutes = 10000 × 3 × 1.2

Let the area of water be A.

And the height of the water in the filed = 9 cm = \(\frac { 9 }{ 100 }\) m

Volume of water required = A × \(\frac { 9 }{ 100 }\)

Hence, the area of watered = 4,00,000 m^{2}

Question 17.

A bag contains 2.8 m^{3} grains inside it. How many bags full of grains are needed to (RBSESolutions.com) fill a cylindrical drum with radius 4.2 m and height 5 m.

Solution :

Given

Volume of grain inside the bag = 2.8 m^{3}

Radius of the drum = 4.2 m

Height of the drum = 5 m

∴ Volume of the drum =πr^{2}h

= \(\frac { 22 }{ 7 }\) × (2.1)^{2} × 5

The number of bag full of grains required

Question 18.

There are two right circular cones X and Y. Radius of X is three time that of radius of Y and (RBSESolutions.com) the volume of Y Is half of the volume of X. Find the ratio of the heights of X and Y.

Solution :

Let the height of cones X and Y be H and h respectively.

And let radius of cone Y be r.

∴ radius of cone X = 3r

∴ Volume of X = \(\frac { 1 }{ 4 }\) × π × 3r × 3r × H

and Volume of Y = \(\frac { 1 }{ 3 }\) × π × r × r × H

According to question

Volume of Y = \(\frac { 1 }{ 2 }\) Volume of X

Question 19.

The diameter of a metallic ball is 4.2 cm. If density of this (RBSESolutions.com) metal is 8.9 gram/cm^{3}, then find the mass of the ball.

Solution:

Given :

Diameter of ball = 4.2 cm

Radius(r) = \(\frac { 4.2 }{ 2 }\) = 21 cm

Volume of the ball = \(\frac { 4 }{ 3 }\)πr^{3}

= \(\frac { 4 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (2.1)^{3}

= \(\frac { 38808 }{ 1000 }\)

= 38.808 cm^{3}

∵ mass of 1 cm^{3} metal = 8.9 gram

∵ mass of 38.808 metal = 38.808 × 8.9 = 345.3912 gram

Question 20.

The height and base radius of cone are 8.4 cm and 2.1 cm respectively. A sphere is recasted after (RBSESolutions.com) melting. Find the radius of the sphere recasted.

Solution :

Given,

Height of cone (h) = 8.4 cm

radius (r) = 2.1 cm

Then volume of cone (V) = \(\frac { 1 }{ 3 }\)πr^{2}h

= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × (2.1)^{2} × 8.4

Let the radius of sphere recasted be R, then

According to question

Volume of sphere = volume of cone

Hence, the radius of sphere = 2.1 cm

**Long Answers Type Questions**

Question 1.

If the diameter of the transaction of wire is made less by 5%, then how much its (RBSESolutions.com) length will be increased while its volume remains same?

Solution :

Let the diameter be 2r.

radius = \(\frac { 2r }{ 2 }\) = r

After becomes less by 5%

Hence, the length will be increased by 10.8%

Question 2.

A solid toy ¡s in the form of a hemisphere surmounted by a right circular cone. If (RBSESolutions.com) radius of the cone and the total height of the toy are 7 cm and 31 cm respectively. Find the total surface are of the toy. \(\left( Take\pi =\frac { 22 }{ 7 } \right) \) [CBSE 2013]

Solution :

Given,

The common radius of conical part and hemispherical part of the toy (r) = 7 cm.

Total height of the toy = 31 cm

Hence, the height of conical part = 31 – 7 = 24 cm

Slant height of conical part l

Hence, total surface area of toy = 858 cm^{2}

Question 3.

A hollow cylinder is surmounted on a hemispherical vessel. Diameter of hemisphere is 14 cm (RBSESolutions.com) and total height of the vessel is 13 cm. Find the internal curved surface area of the vessel.

Solution :

Given

Common diameter of cylinder and hemisphere = 14 cm

∴ Radius of hemisphere = \(\frac { 14 }{ 2 }\) = 7 cm

Radius of cylinder = 7 cm

∵ Total height of vessel = 13 cm

∴ Height of cylinder (h) = (13 – 7) = 6 cm

Internal curved surface area of vessel = Internal surface area of cylinder + internal surface area of hemisphere

= 2πrh + 2πr^{2} = 2πr(h + r)

= 2 × \(\frac { 22 }{ 7 }\) × 7 × (6 + 7)

= 44 × 13 = 572 cm^{2}.

Hence, the total surface area of the vessel = 572 cm^{2}

Question 4.

A sphere of diameter 5 cm Is poured into a partially filled cylindrical vessel. The diameter of (RBSESolutions.com) vessel is 10 cm. If the sphere completely poured into water, then how much the water level ¡n the vessel will be rise?

Solution :

Given

Diameter of cylindrical vessel = 10 cm

∴ Radius (R) = \(\frac { 10 }{ 2 }\) = 5 cm

Diameter of sphere = 5 cm

∴ Radius = \(\frac { 5 }{ 2 }\)

When the sphere is poured into the vessel the water level rises by h.

According to the question

Volume of sphere = Volume of water risen in the vessel

⇒ \(\frac { 4 }{ 3 }\)πr^{3} = πR^{2}h

⇒ \(\frac { 4 }{ 3 }\)π × (2.5)^{3} = π × (5)^{2} × h

Hence, on pouring the sphere water level will be risen by \(\frac { 5 }{ 6 }\) cm

Question 5.

The flowing water through a cylindrical pipe with internal diameter 7 cm at (RBSESolutions.com) the speed of 192.5 liter per minute is collecting in a vessel. Find the speed of water in km/h \(\left( \pi =\frac { 22 }{ 7 } \right) \) [CBSE 2013]

Solution :

Given,

Diameter of cylindrical pipe = 7 cm

∴ Radius (r) = \(\frac { 7 }{ 2 }\) = 3.5 cm = 0.035 m and the volume of flowing water in one minute = 192.5 liter

∴ The volume of flowing water in one hour = 192.5 × 60 liter

= \(\frac { 192.5\times 60 }{ 1000 } \) m^{3}

= 11.55 m^{3} …….(i)

Let the speed of the water in pipe be x km./h

∴ Length of flowing water in one hour (h) = x m

∴ Volume of water flowing in one hour = πr^{2}h = \(\frac { 22 }{ 7 }\) × (0.035)^{2} × x

= 0.00385 × x m^{3} …….(ii)

From equation (i) and (ii) we get

0.00385 × x = 11.55

⇒ x = \(\frac { 11.55 }{ 0.00385 }\)

⇒ x = \(\frac { 1155000 }{ 385 }\)

⇒ x = 3000 m

= \(\frac { 3000 }{ 1000 }\) km = 3 km

Hence the speed of water in the pipe = 3 km/h

Question 6.

The length and breadth of a rectangular plot are 30 m and 20 m respectively. Out of (RBSESolutions.com) the plot a pit of dimension 15 m × 10 m × 5 m is dig. The earth digout from the pit is spread in the plot evenly. Find the height of earth in the plot.

Solution :

Given Length of the plot (l) = 30 m

Breadth (b) = 20 m

Let the height of the earth spreaded into plot be h.

∴ Volume of the earth spreaded into plot = l × b × h

= 30 × 20 × h m^{3}

Length of pit = 15 m

breadth = 10 m

And depth = 5 m

∴ Volume of the earth digout = 15 × 10 × 5 m^{3}

Now, according to question

Volume of the earth spreaded into plot = Volume of earth digout

Hence the height of the earth spreaded in to plot = 1.25 m

Question 7.

A well with diameter 3 m is digout to the depth of 14 m. The earth digout is (RBSESolutions.com) spreaded out 4 m round the well evenly in the form of circular ring to make a plateform. Find the height of the plateform.

Solution :

Given:

∵ Diameter of well = 3 m

Radius of well (r) = \(\frac { 3 }{ 2 }\) m

And the depth (h) = 14 m

∴ The volume of the earth digout from the well

= πr^{2} = \(\frac { 22 }{ 7 }\) × \(\frac { 3 }{ 2 }\) × \(\frac { 3 }{ 2 }\) × 14

= \(\frac { 22\times 3\times 3 }{ 2 }\) = 99 m^{3}

∵ Since, radius of the well is \(\frac { 3 }{ 2 }\) m and a ring is formed round the well. The ring is 4 m broad.

∴ radius of the well with ring r_{1} = \(\frac { 3 }{ 2 }\) + 4 = \(\frac { 11 }{ 2 }\) m

and the radius of the well r_{2} = \(\frac { 3 }{ 2 }\) m

Let the height of the circular ring be h

Then the volume of the earth of plateform = 88 × h m^{3}

Now volume of the earth of plateform = volume of earth digout.

88 × h = 99

∴ h = \(\frac { 99 }{ 8 }\) = \(\frac { 9 }{ 8 }\) = 1.125 m

Hence the height of plateform = 1.125 m.

Question 8.

A solid metallic sphere with diameter 8 cm is melted and recasted into (RBSESolutions.com) a circular wire. If the length of wire is 12 m, then find the thickness of the wire. [CBSE – 2013]

Solution : Given,

Diameter of metallic sphere = 8 cm

∴ Radius (R) = \(\frac { 8 }{ 2 }\) = 4 cm

The length of wire recasted (h) = 12 m = 1200 cm

Let the radius of wire recasted be r since, by melting the metallic sphere, the wire is recasted.

∴ Volume of cylindrical wire = Volume of sphere

Hence the thickness of the wire = \(\frac { 8 }{ 15 }\) cm

Question 9.

A hemispherical tank ¡s completely filled with water which is (RBSESolutions.com) making empty at the rate of \(\frac { 25 }{ 7 }\) liter per second with the help of a pipe. If the base diameter of the tank is 3 m, then how long will it take to empty half of the tank. [CBSE – 2012]

Solution :

Given:

Diameter of hemispheric tank = 3 m

∴ its radius (r) = \(\frac { 3 }{ 2 }\) m

Volume of the hemispheric tank = \(\frac { 2 }{ 3 }\)πr^{3}

Hence, time taken to empty the half tank = 16.5 minutes.

Question 10.

An iron cuboid with dimensions 4.4 m × 2.6 m × 1 m is melted (RBSESolutions.com) and recasted it into a hollow circular pipe. If internal radius and thickness of the pipe are 30 cm and 5 cm respectively. Find its length. [NCERT Exemplar Problem]

Solution :

Given:

Internal radius of pipe (r) = 30 cm

Thickness of the pipe = 5 cm

External radius of the pipe R =30 + 5 = 35 cm

Let the length of the cylindrical pipe be h.

Volume of iron cuboid = l × b × h

= 4.4 × 2.6 × 1 m^{3}

= 4.4 × 2.6 × 1 × 1000000 cm^{3}

∴ The cuboid is melted and recasted into a hollow cylindrical pipe.

∴ Volume of hollow cylindrical pipe = Volume of cuboid

Hence the length of cylindrical pipe = 112 m.

Question 11.

By melting 504 cones with diameter 3.5 cm and height 3 cm each, (RBSESolutions.com) a metallic sphere is recasted. Find the diameter of sphere. Also

find the surface area of the sphere. \(\left( take\pi =\frac { 22 }{ 7 } \right) \) [CBSE 2015]

Solution:

Given,

Diameter of cone = 3.5 cm

∴ Radius (r) = \(\frac { 3.5 }{ 2 }\) cm

Height (h) = 3 cm

∴ Volume of each cone = \(\frac { 1 }{ 3 }\)πr^{2}h

= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × \({ \left( \frac { 3.5 }{ 2 } \right) }^{ 2 }\) × 3 = 9.625 cm^{3}

∴ Volume of 504 cones = 504 × 9.625 = 4851 cm^{3}

Now, Let the radius metallic sphere be R

According to the question

Volume of sphere = Volume of 504 cones

Diameter of sphere = 2R = 2 × 10.5 = 21 cm

∴ Surface area of sphere

= 4πR^{2} = 4 × \(\frac { 22 }{ 7 }\) × (10.5)^{2}

= 1386 cm^{2}

Question 12.

The length, breadth and height of wall are 4 m, 15 cm and 3 m (RBSESolutions.com) respectively. The brick with dimensions 20 cm × 10 cm × 8 cm will be used to build the wall. If the rate of the bricks is 120 per one thousand, then find the total cost of the bricks used.

Solution :

Given:

Length of wall (L) 4 m = 400 cm.

Breadth (B) = 15 cm

Height (H)= 3 m = 300 cm

∴ Volume of the wall L × B × H

= 400 × 15 × 300 cm^{3}

Length of one brick used (l) = 20 cm

Breadth (b) = 10 cm

And height (h) = 8 cm

Volume of one brick = l × b × h

= 20 × 10 × 8 cm^{3}

= \(\frac { 400\times 15\times 300 }{ 20\times 10\times 8 } \)

∵ cost of 1000 bricks = ₹ 120

∴ cost of 1 bricks = \(\frac { 120 }{ 1000 }\)

∴ cost of 1125 bricks = \(\frac { 120 }{ 1000 }\) × 1125

Hence, the number of bricks used in wall = 1125

And the cost the bricks used = ₹ 135

Question 13.

The length, height and thickness of a wall are 8 m, 4 m and 35 cm respectively. It contains (RBSESolutions.com) one gate of size 3 m × 1 m and two windows of size 1.20 m × 1 m. Find the cost of building the wall at the rate of ₹ 1500 per m^{3}.

Solution :

Given :

Length of wall (l) = 8 m

Thickness of wall (b) = 35 cm = 0.35 m

Height of wall (h) = 4 m

Volume of wall = l × b × h

= 8 × 4 × 0.35

= 11.2 m^{3}

Dimension of the gate in the wall = 2 m × 1 m × 0.35 ms [∵ Thickness of wall = 0.35 m]

∴ Volume of the gate in the wall = 2m × 1 m × 0.35 m = 0.7 m^{3}

Measures of 2 windows in the wall = 1.20 m × 1 m × 0.35 m

Volume of 2 windows in the wall =2 × 1.2 × 1 × 0.35 = 0.84 m^{3}

Volume of wall with out a gate and two windows

= 11.2 – (0.7 + 0.84)

= 9.66 m^{3}

∵ Cost of building 1 m^{3} wall = ₹ 1500

∴ Cost of building 9.66 m^{3} wall = 1500 × 9.66 = ₹ 14490

∴ Hence cost of building the wall = ₹ 14490

Question 14.

11 persons can sit in side a conical tent. Each man can needs 4 m^{2} and (RBSESolutions.com) to sit and 20 m^{3} space to breath. Find the height of the tent.

Solution :

Given

The land for a man to sit = 4 m^{2}

∴ Base area for 11 men to sit = 4 × 11 = 44 m^{2}

Base area = 44

⇒ πr^{2} = 44

⇒ \(\frac { 22 }{ 7 }\) × r^{2} = 44

r^{2} = 14

Each man needs the air = 20 m^{3}

Thus the volume of tent =20 × 11 = 220 m^{3}

Question 15.

The radius of a solid sphere is 3 cm. It is melted and recasted into (RBSESolutions.com) smaller spheres with diameter 0.6 cm. Find the number of spheres recasted.

Solution :

Given :

Radius of solid sphere (R) 3 cm

Diameter of the sphere recasted = 0.6 cm

its radius = \(\frac { 0.6 }{ 2 }\) =0.3 cm

Volume of solid sphere = \(\frac { 4 }{ 3 }\)π(0.3)^{3} cm^{3}

Volume of sphere recated = \(\frac { 4 }{ 3 }\)π(0.3)^{3} cm^{3}

Let the number spheres recasted be n.

∴ Volume of solid sphere n × volume of sphere recasted

Volume of solid sphere n × volume of sphere recasted.

n = \(\frac { 3\times 3\times 3 }{ 0.3\times 0.3\times 0.3 } \)

= 1000

Hence the number of spheres reasted = 1000

Question 16.

The two ends of solid cylinder are in the shape of hemisphere. Total (RBSESolutions.com) length of solid is 19 cm and diameter is 7 cm. Find the volume and surface area of the solid.

Solution :

Given

Total height of solid = 19 cm

diameter of the solid = 7 cm

∴ Its radius = \(\frac { 7 }{ 2 }\) cm

Hence, volume of solid = 641.67 cm^{3}

Surface area = 418 cm^{2}

Question 17.

A hemispherical tank ¡s made of 1 cm thick iron sheet. If its internal (RBSESolutions.com) radius is 1 m, Find the volume of the iron used to make the tank.

Solution :

For hemispherical tank

Internal radius (r) = 1 m = 100 cm.

External radius R = 100 + 1 = 101 cm

Volume of the iron sheet

Question 18.

A glass sphere with diameter 3 cm is melted and recasted into 3 smaller (RBSESolutions.com) spheres. If the diameters of two spheres recasted are 1.5 cm and 2 cm respectively, find the diameter of third sphere.

Solution :

Let the radius of third bail be r.

The volume of the given largest sphere = \(\frac { 4 }{ 3 }\)π\({ \left( \frac { 3 }{ 2 } \right) }^{ 3 }\)

The volume of the sphere of diameter 1.5 cm = \(\frac { 4 }{ 3 }\)π\({ \left( \frac { 1.5 }{ 2 } \right) }^{ 3 }\) cm^{3}

The volume of the sphere of diameter 2 cm = \(\frac { 4 }{ 3 }\)π\({ \left( \frac { 2 }{ 3 } \right) }^{ 3 }\)

The volume of the sphere of radius r = \(\frac { 4 }{ 3 }\)π(r)^{3}

Now according to question

Volume of the largest sphere = Sum of the volume of spheres recasted

The diameter of third sphere

= 2r = 2 × \(\frac { 5 }{ 4 }\)

= \(\frac { 5 }{ 2 }\) cm

= 2.5 cm

Question 19.

50 plates are put one by one on each other to form a (RBSESolutions.com) cylindrical shape. The radius and thickness of each plate are 7 cm and \(\frac { 1 }{ 2 }\) cm respectively. Find the total surface area and the volume of the cylinder formed in this way.

Solution :

The thickness of the plate = \(\frac { 1 }{ 2 }\) cm

∴ Thickness of the 50 plates = \(\frac { 1 }{ 2 }\) × 50 = 25 cm

Thus the height of cylinder formed = 25 cm

Radius of the cylinder = 7 cm

Total surface area = 2πr(r + h)

= 2 × \(\frac { 22 }{ 7 }\) × 7 × (7 + 25)

= 2 × \(\frac { 22 }{ 7 }\) × 7 × 32

= 1408 cm^{2}

Volume of cylinder = πr^{2}h

= \(\frac { 22 }{ 7 }\) × 7 × 7 × 25

= 3850 cm^{3}

Question 20.

Total surface area of a cylindrical solid is 231 cm^{3}. If its curved (RBSESolutions.com) surface area is one third of the total surface area, then find the volume of cylinder.

Solution : Given,

Total surface area = 231 cm^{2}

Curved surface area = \(\frac { 2 }{ 3 }\) × 231 = 154 cm^{2}

Let the base radius of cylinder be r.

curved surface area 2πrh = 154 …(i)

Total surface area

Hence, the volume of cylinder = 269.5 cm^{2}

We hope the given RBSE Solutions for Class 10 Maths Chapter 16 Surface Area and Volume Additional Questions will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 16 Surface Area and Volume Additional Questions, drop a comment below and we will get back to you at the earliest.

## Leave a Reply