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

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 cm2. Its side will be :
(A) 6 cm
(B) 8 cm
(C) 9 cm
(D) 7 cm
Solution :
Total surface area of cube = 486 cm2
6a2 = 2 m
a2 = $$\frac { 7 }{ 5 }$$ = 81
a2 = $$\sqrt { 81 }$$ = 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 m2
(B) 1 m2
(C) 21 m2
(D) 22 m2
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 m2
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 cm3
(B) 20π cm3
(C) 36π cm3
(D) 30π cm3
Solution :
Given,
Diameter of sphere = 6 cm
Radius (r) = $$\frac { 6 }{ 2 }$$ = 3 cm
Volume V = $$\frac { 4 }{ 3 }$$πr3
= $$\frac { 4 }{ 3 }$$ × π × 3 × 3 × 3
36π cm3
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 cm2
(B) 880 cm2
(C) 888 cm2
(D) 890 cm2
Solution :
Given,
Radius of cylinder (r) = 14 cm
Height (h) = 10 cm
Curved Surface area = 2πrh
= 2 × $$\frac { 22 }{ 7 }$$ × 14 × 10
= 880 cm2
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 cmand 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 cm3

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 cm2.
Solution :
Diameter of hemisphere = 42 cm
Radius (r) = $$\frac { 42 }{ 2 }$$ = 21 cm
Total surface area of hemisphere = 3πr2
= 3 × $$\frac { 22 }{ 7 }$$ × 21 × 21
= 3 × 22 × 3 × 21
= 4158 cm2
∵ Cost of polishing 1 cm2 paise 20 = ₹ 0.20
∴ Cost of polishing 4158 cm2 = 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 V1 = πr2h
= $$\frac { 7 }{ 5 }$$ × 7 × 7 × 40 = 6160 cm3
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 V2 = $$\frac { 1 }{ 3 }$$πR2H
= $$\frac { 1 }{ 3 }$$ × $$\frac { 22 }{ 7 }$$ × 7 × 7 × 12 = 616 cm3
∴ Volume of the solid = Volume cylinder + Volume of cone
= 6160 + 616 = 6776 cm3

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 = $$\frac { 4 }{ 3 }$$πr3 = $$\frac { 4 }{ 3 }$$π × (r)3
And radius of cone recasted (R) = 3 cm
And its height (H) = 6 cm
∴ Volume of cone recasted = $$\frac { 1 }{ 3 }$$πR2H
= $$\frac { 1 }{ 3 }$$ × 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
= $$\frac { 4000\times 150 }{ 1000 }$$ m3
= 600 m3
Volume of the tank = 20 × 15 × 6 m3 = 1800 m3
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 Ist sphere (r1) = 6 cm
Radius of IInd sphere (r2) = 8 cm
Radius of IIIrd sphere (r3) 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 = $$\frac { 1 }{ 3 }$$r2h
= $$\frac { 1 }{ 3 }$$ × π × (10)2 × 18
= $$\frac { 1 }{ 3 }$$ × π × 100 × 18
π × 100 × 6
= 600 π cm3.
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.
πR2H = 600 π
π(5)2H = 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 cm3
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 πr2h.
= $$\frac { 7 }{ 5 }$$ × (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 = $$\frac { 4 }{ 3 }$$πr3
= $$\frac { 4 }{ 3 }$$ × π × (3)3 cm3
Let the radius of the wire be R.
According to question
Length of wire = 36 m = 3600 cm.
∴ Volume of wire = πR2h.
π × R2 × 3600 cm3
Again according to (RBSESolutions.com) question
Volume of sphere = volume of wire

Hence radius of the wire = 0.1 cm.

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