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

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.

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.

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.

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.

Question 5.
The volume and height of a cone are 308 cm³ 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 cm³

Hence, option (A) is correct.

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

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

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

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

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.

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 (r₁) = 6 cm
Radius of II^nd sphere (r₂) = 8 cm
Radius of III^rd 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

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

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.

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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