RBSE Solutions for Class 10 Maths Chapter 16 Surface Area and Volume Miscellaneous Exercise 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 Miscellaneous Exercise.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 16 |

Chapter Name |
Surface Area and Volume |

Exercise |
Miscellaneous Exercise |

Number of Questions Solved |
14 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 10 Maths Chapter 16 Surface Area and Volume Miscellaneous Exercise

**Multiple Choice Questions**

**RBSE Class 10 Maths Chapter 16 Miscellaneous Question 1.**

Total surface area of cube (RBSESolutions.com) is 486 cm^{2}. Its side will be :

(A) 6 cm

(B) 8 cm

(C) 9 cm

(D) 7 cm

Solution :

Total surface area of cube = 486 cm^{2}

6a^{2} = 2 m

a^{2} = \(\frac { 7 }{ 5 }\) = 81

a^{2} = \(\sqrt { 81 }\) = 9 cm

∴ side of the cube is 9 cm

Hence, option (C) is correct.

**RBSE Solutions For Class 10 Maths Chapter 16 Miscellaneous Question 2.**

Length, breadth and height of a cuboid are 9 m, 2 m and 1 m respectively. Surface (RBSESolutions.com) area of cuboid is:

(A) 12 m^{2}

(B) 1 m^{2}

(C) 21 m^{2}

(D) 22 m^{2}

Solution :

Given,

l = 9 m

b = 2 m

h = 1 m

Surface area of cuboid = 2(l + b)h

= 2(9 + 2) × 1

=2 × 11 × 1

22 m^{2}

Hence, option (D) is correct.

**RBSE Solutions For Class 10 Maths Chapter 16 Question 3.**

If diameter of a sphere is 6 cm, (RBSESolutions.com) them volume is:

(A) 16π cm^{3}

(B) 20π cm^{3}

(C) 36π cm^{3}

(D) 30π cm^{3}

Solution :

Given,

Diameter of sphere = 6 cm

Radius (r) = \(\frac { 6 }{ 2 }\) = 3 cm

Volume V = \(\frac { 4 }{ 3 }\)πr^{3}

= \(\frac { 4 }{ 3 }\) × π × 3 × 3 × 3

36π cm^{3}

Hence, option (C) is correct.

**Class 10 Maths RBSE Solution Chapter 16 Question 4.**

The radius of a cylinder is 14 cm and its (RBSESolutions.com) height is 10 cm. Curved surface area of cylinder is:

(A) 881 cm^{2}

(B) 880 cm^{2}

(C) 888 cm^{2}

(D) 890 cm^{2}

Solution :

Given,

Radius of cylinder (r) = 14 cm

Height (h) = 10 cm

Curved Surface area = 2πrh

= 2 × \(\frac { 22 }{ 7 }\) × 14 × 10

= 880 cm^{2}

Hence, option (B) is correct.

**RBSE Solutions For Class 10 Maths Chapter 16.1 Question 5.**

The volume and height of a cone are 308 cm^{3 }and 6 cm respectively. Its (RBSESolutions.com) radius will be:

(A) 7 cm

(B) 8 cm

(C) 6 cm

(D) None of these

Solution :

Given.

Height of cone = 6 cm

Volume = 308 cm^{3}

Hence, option (A) is correct.

**RBSE Solutions For Class 10 Maths Chapter 16 In Hindi Question 6.**

The diameter of a metallic hemisphere is 42 cm. Find the cost of (RBSESolutions.com) polishing its total surface at the rate of paise 20 per cm^{2}.

Solution :

Diameter of hemisphere = 42 cm

Radius (r) = \(\frac { 42 }{ 2 }\) = 21 cm

Total surface area of hemisphere = 3πr^{2}

= 3 × \(\frac { 22 }{ 7 }\) × 21 × 21

= 3 × 22 × 3 × 21

= 4158 cm^{2}

∵ Cost of polishing 1 cm^{2} paise 20 = ₹ 0.20

∴ Cost of polishing 4158 cm^{2} = 4158 × 0.20 = ₹ 831.60

**RBSE Class 10 Maths Chapter 16 Question 7.**

A cone, a hemisphere and a cylinder have same base and (RBSESolutions.com) equal to their heights. Find the ratio among their volumes.

Solution :

Given, a cone, hemisphere and a cylinder have same base, their base and heights are equal.

Hence, the ratio of their volumes = 1 : 2 : 3

**Chapter 16 Maths Class 10 RBSE Question 8.**

Left side of a solid is cylindrical and right side is conical. If (RBSESolutions.com) diameter and length of cylindrical portion are 14 cm and 40 cm respectively and diameter and length of the conical portion are 14 cm and 12 cm respectively. Find the volume of the solid.

Solution :

Given,

Diameter of cylindrical portion = 14 cm

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

Height (h) = 40 cm

Volume of cylindrical portion V_{1} = πr^{2}h

= \(\frac { 7 }{ 5 }\) × 7 × 7 × 40 = 6160 cm^{3}

Again, given that diameter of conical portion = 14 cm.

∴ Its radius(R) = \(\frac { 14 }{ 2 }\) = 7 cm

∴ Height (H) = 12 cm

Volume of conical portion V_{2} = \(\frac { 1 }{ 3 }\)πR^{2}H

= \(\frac { 1 }{ 3 }\) × \(\frac { 22 }{ 7 }\) × 7 × 7 × 12 = 616 cm^{3}

∴ Volume of the solid = Volume cylinder + Volume of cone

= 6160 + 616 = 6776 cm^{3}

**RBSE Class 10 Maths Chapter 4 Miscellaneous Exercise Question 9.**

By melting a metallic solid sphere with radius 9 cm, some (RBSESolutions.com) cones are recasted. If the radius and height of cones recasted are 3 cm and 6 cm

respectively, then find the number of cones recasted.

Solution :

Given,

Radius of metallic sphere (r) = 9 cm

∴ Volume of metalic sphere = \(\frac { 4 }{ 3 }\)πr^{3} = \(\frac { 4 }{ 3 }\)π × (r)^{3}

And radius of cone recasted (R) = 3 cm

And its height (H) = 6 cm

∴ Volume of cone recasted = \(\frac { 1 }{ 3 }\)πR^{2}H

= \(\frac { 1 }{ 3 }\) × π × 3 × 3 × 6

Let the number of cones recasted be n. Then According to the question

Volume of metallic sphere = n × volume of a cone

Hence 54 cones be recasted

**Ch 16 Maths Class 10 RBSE Question 10.**

The population of a village is 4000, where each person needs 150 liter of water (RBSESolutions.com) per day. There is a water tank with dimensions 20 m × 15 m × 6 m in the village for how long will the water of the tank be sufficient?

Solution :

Population of the village = 4000

Each person needs water = 150 lit per day

Quantity of water that 4000 people need = 4000 × 150

= \(\frac { 4000\times 150 }{ 1000 } \) m^{3}

= 600 m^{3}

Volume of the tank = 20 × 15 × 6 m^{3} = 1800 m^{3}

Number of days for the tank has water in it for the village = \(\frac { 1800 }{ 600 }\) = 3

Hence, the water in the tank is sufficient for 3 days.

**Class 10 Maths Chapter 16 RBSE Question 11.**

Three spheres with radius 6 cm, 8 cm and 10 cm respectively are melted and a (RBSESolutions.com) large sphere is recasted. Find the radius of this sphere.

Solution :

Given

Radius of I^{st} sphere (r_{1}) = 6 cm

Radius of II^{nd} sphere (r_{2}) = 8 cm

Radius of III^{rd} sphere (r_{3}) 10 cm

Let the radius of the sphere recasted be R.

Since, by melting three given sphere into larger sphere is recasted

∴ Volume of sphere recasted = sum of volumes of three spheres

Hence radius of sphere recasted = 12 cm

**RBSE Class 10 Math Chapter 16 Question 12.**

The radius of a conical vessel is 10 cm and its height is 18 cm. It is completely (RBSESolutions.com) filled with water. The water is pored into another cylindrical

vessel with radius 5 cm. Find the height of water in this vessel.

Solution :

Given,

Radius of conical vessel (R) = 10 cm

and its height = 18 cm

Volume of conical vessel = \(\frac { 1 }{ 3 }\)r^{2}h

= \(\frac { 1 }{ 3 }\) × π × (10)^{2} × 18

= \(\frac { 1 }{ 3 }\) × π × 100 × 18

π × 100 × 6

= 600 π cm^{3}.

Let the height of water in cylindrical (RBSESolutions.com) vessel be H and its radius = 5 cm.

Now, according to question.

Volume of cylindrical vessel = Volume of water in the conical vessel.

πR^{2}H = 600 π

π(5)^{2}H = 600 π

H = \(\frac { 600\pi }{ 25\pi } \) = 24 cm

Hence, height of water in cylindrical vessel = 24 cm

**RBSE Solution Class 10 Maths Chapter 16 Question 13.**

A candle with diameter 2.8 cm is formed by melting wax cuboid with (RBSESolutions.com) dimensions 11 cm × 3.5 cm × 2.5 cm. Find the length of candle.

Solution :

Given,

Length of cuboid (l) = 11 cm.

Breadth (b) = 3.5 cm

Height (h) = 2.5 cm

Volume of wax-cuboid = l × b × h

=11 × 3.5 × 2.5 = 96.25 cm^{3}

Diameter (RBSESolutions.com) of candle = 2.8 cm

Radius = \(\frac { 2.8 }{ 2 }\) = 1.4 cm

Let the height of the candle of h

Volume of the candle πr^{2}h.

= \(\frac { 7 }{ 5 }\) × (1.4)^{2} × h

According to the question Volume of wax cuboid = Volume of candle

Hence, the length of candle = 15.625 cm.

**Class 10 Maths RBSE Solution Chapter 16 In Hindi Question 14.**

Diameter of a metallic sphere is 6 cm. By melting it an another (RBSESolutions.com) circular wire ¡s recasted. If the length of wire is 36 m, find its radius.

Solution :

Given,

Diameter of sphere = 6 cm

Radius of sphere (r) = \(\frac { 6 }{ 2 }\) = 3 cm

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

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

Let the radius of the wire be R.

According to question

Length of wire = 36 m = 3600 cm.

∴ Volume of wire = πR^{2}h.

π × R^{2} × 3600 cm^{3}

Again according to (RBSESolutions.com) question

Volume of sphere = volume of wire

Hence radius of the wire = 0.1 cm.

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