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      Class 9 Maths

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      • Class 09
      • Class 9 Maths
      CoursesClass 09MathsClass 9 Maths
      • 1.Number System
        14
        • Lecture1.1
          Numbers and its Classification, Rational Number and its Representation on the Number Line 50 min
        • Lecture1.2
          Decimal Representation of Rational Number and Conversion of Decimal Number into Rational Number 01 hour
        • Lecture1.3
          Representation of Decimal Number on Number Line, Finding Irrational Numbers Between Given Two Numbers 46 min
        • Lecture1.4
          Representing Irrational Number on the Number Line 41 min
        • Lecture1.5
          Rationalisation 54 min
        • Lecture1.6
          Introduction to Exponents, Laws of Exponents, BODMAS Rule 41 min
        • Lecture1.7
          Integral Exponents of Real Numbers 57 min
        • Lecture1.8
          Integral Exponents of Real Numbers and Solving Equation for x 01 hour
        • Lecture1.9
          Rational Exponents of Real Numbers 52 min
        • Lecture1.10
          Examples Based on Rational Exponents of Real Numbers 30 min
        • Lecture1.11
          Chapter Notes – Number System
        • Lecture1.12
          NCERT Solutions – Number System Exercise 1.1 – 1.6
        • Lecture1.13
          R D Sharma Solutions Number System
        • Lecture1.14
          Revision Notes – Number System
      • 2.Polynomials
        10
        • Lecture2.1
          Introduction to Polynomials and Some Basics Questions 58 min
        • Lecture2.2
          Value of Polynomial, Zeroes of Polynomial and Relationship between Zeroes and Coefficient of Linear Equation 01 hour
        • Lecture2.3
          Division of Polynomials and Division Algorithm and Remainder Theorem 51 min
        • Lecture2.4
          Factorising by Spliting Method 01 hour
        • Lecture2.5
          Factorisation By Different Methods and Factor Theorem 27 min
        • Lecture2.6
          Factorisation of Polynomial by Using Algebric Identities 01 hour
        • Lecture2.7
          Chapter Notes – Polynomials
        • Lecture2.8
          NCERT Solutions – Polynomials Exercise 2.1 – 2.5
        • Lecture2.9
          R S Aggarwal Solutions Polynomials
        • Lecture2.10
          Revision Notes Polynomials
      • 3.Coordinate Geometry
        8
        • Lecture3.1
          Representation of Position on Number Line, Finding the location by Using the Axis 01 hour
        • Lecture3.2
          Quadrant, Coordinates of Point on Axis, Distance from Axis 01 hour
        • Lecture3.3
          Finding the Area of Triangle and Rectangle and Mirror Image of Point On Axis 57 min
        • Lecture3.4
          Chapter Notes – Coordinate Geometry
        • Lecture3.5
          NCERT Solutions – Coordinate Geometry Exercise 3.1 – 3.3
        • Lecture3.6
          R D Sharma Solutions Coordinate Geometry
        • Lecture3.7
          R S Aggarwal Solutions Coordinate Geometry
        • Lecture3.8
          Revision Notes Coordinate Geometry
      • 4.Linear Equations
        8
        • Lecture4.1
          Introduction of Linear Equation in Two Variables and Graphical Representation 01 hour
        • Lecture4.2
          Example to find the value of Constant and Graph of Linear Equation in Two Variables 57 min
        • Lecture4.3
          Equations of lines Parallel to the x-axis and y-axis 46 min
        • Lecture4.4
          Chapter Notes – Linear Equations
        • Lecture4.5
          NCERT Solutions – Linear Equations Exercise 4.1 – 4.4
        • Lecture4.6
          R D Sharma Solutions Linear Equations
        • Lecture4.7
          R S Aggarwal Solutions Linear Equations
        • Lecture4.8
          Revision Notes Linear Equations
      • 5.Euclid's Geometry
        7
        • Lecture5.1
          Euclid’s Definitions and Euclid’s Axioms 32 min
        • Lecture5.2
          Eulcid’s Postulates and Playfair’s Axiom 37 min
        • Lecture5.3
          Chapter Notes – Euclid’s Geometry
        • Lecture5.4
          NCERT Solutions – Euclid’s Geometry Exercise 5.1 5.2
        • Lecture5.5
          R D Sharma Solutions Euclid’s Geometry
        • Lecture5.6
          R S Aggarwal Solutions Euclid’s Geometry
        • Lecture5.7
          Revision Notes Euclid’s Geometry
      • 6.Lines and Angles
        10
        • Lecture6.1
          Introduction to Lines and Angles, Types of Angle and Axioms 01 hour
        • Lecture6.2
          Vertically opposite Angles and Angle bisector 46 min
        • Lecture6.3
          Angle made by a transversal with two Lines and Parallel Lines 01 hour
        • Lecture6.4
          Lines Parallel to same Line; Exterior angle of a triangle; Angle Sum property of a Triangle 52 min
        • Lecture6.5
          Angle Sum Property of Triangle Including Interior and Exterior Angles 01 hour
        • Lecture6.6
          Chapter Notes – Lines and Angles
        • Lecture6.7
          NCERT Solutions – Lines and Angles Exercise 6.1 – 6.3
        • Lecture6.8
          R D Sharma Solutions Lines and Angles
        • Lecture6.9
          R S Aggarwal Solutions Lines and Angles
        • Lecture6.10
          Revision Notes Lines and Angles
      • 7.Triangles
        11
        • Lecture7.1
          Introduction, Types of Triangle, Altitude & Median of Triangle, Congruences, SAS criterion of Congruence 42 min
        • Lecture7.2
          Questions Based on SAS Criterion 01 hour
        • Lecture7.3
          ASA Criterion of Congruence and Its based Questions 42 min
        • Lecture7.4
          Questions Based on AAS and SSS Criterion 29 min
        • Lecture7.5
          RHS Criterion of Congruence and Its based Questions, Questions Based on Multiple Criterion 01 hour
        • Lecture7.6
          Inequalities of Triangle 42 min
        • Lecture7.7
          Inequalities of Triangle cont., Distance Between a line and The Point, Miscellaneous Questions 48 min
        • Lecture7.8
          Chapter Notes – Triangles
        • Lecture7.9
          NCERT Solutions – Triangles Exercise 7.1 – 7.5
        • Lecture7.10
          R D Sharma Solutions Triangles
        • Lecture7.11
          Revision Notes Triangles
      • 8.Quadrilaterals
        13
        • Lecture8.1
          Congruence of Triangles- SAS Congruence Criteria, ASA Congruence Criteria, SSS Congruence Criteria, RHS Congruence Criteria; Types of Quadrilateral- Parallelogram, Rectangle, Rhombus, Square, Trapezium, Kite 01 hour
        • Lecture8.2
          Theorem 1; Properties of parallelogram- Theorem 2, Theorem 3; 01 hour
        • Lecture8.3
          Theorem 4; Theorem 5; 24 min
        • Lecture8.4
          Conditions for a Quadrilateral to be a parallelogram- Theorem 6 01 hour
        • Lecture8.5
          Theorem 8 01 hour
        • Lecture8.6
          Theorem 9 51 min
        • Lecture8.7
          Theorem 9(Mid-point Theorem); 01 hour
        • Lecture8.8
          Theorem 10 27 min
        • Lecture8.9
          Chapter Notes – Quadrilaterals
        • Lecture8.10
          NCERT Solutions – Quadrilaterals Exercise 8.1 8.2
        • Lecture8.11
          R D Sharma Solutions Quadrilaterals
        • Lecture8.12
          R S Aggarwal Solutions Quadrilaterals
        • Lecture8.13
          Revision Notes Quadrilaterals
      • 9.Area of Parallelogram
        11
        • Lecture9.1
          Parallelograms on the Same Base and between the Same Parallels; A Diagonal of a parallelogram divides it into two triangles of Equal Area 57 min
        • Lecture9.2
          Questions based on parallelogram and Triangle are on same base & parallel lines and Area of Triangle 01 hour
        • Lecture9.3
          Triangles on the same Base and between the same Parallels 56 min
        • Lecture9.4
          Miscellaneous Questions Based on Triangles on the same Base and between the same Parallels 54 min
        • Lecture9.5
          Sums Based on Median of a Triangle 52 min
        • Lecture9.6
          Two Triangles on the same Base and equal Area lie b/w the same parallel 01 hour
        • Lecture9.7
          Summary of All Concepts and Miscellaneous Questions 56 min
        • Lecture9.8
          Chapter Notes – Area of Parallelogram
        • Lecture9.9
          NCERT Solutions – Area of Parallelogram Exercise 9.1 – 9.4
        • Lecture9.10
          R D Sharma Solutions Area of Parallelogram
        • Lecture9.11
          Revision Notes Area of Parallelogram
      • 10.Constructions
        7
        • Lecture10.1
          Construction of Perpendicular Bisector of a Line Segment, Different Angles and Angle bisector 42 min
        • Lecture10.2
          Construction of Different types of Triangle 43 min
        • Lecture10.3
          Chapter Notes – Constructions
        • Lecture10.4
          NCERT Solutions – Constructions Exercise
        • Lecture10.5
          R D Sharma Solutions Constructions
        • Lecture10.6
          R S Aggarwal Solutions Constructions
        • Lecture10.7
          Revision Notes Constructions
      • 11.Circles
        11
        • Lecture11.1
          Introduction, Important Terms related to circle, Theorem-1, 2, 3 and 4 01 hour
        • Lecture11.2
          Problems Solving 48 min
        • Lecture11.3
          Problems Based on Parallel Chords, Common Chord of Circles 01 hour
        • Lecture11.4
          Problems Based on Concentric Circles, Triangle in Circle 41 min
        • Lecture11.5
          Equal Chords and Their Distance, Theorem-5, 6, 7, 8 01 hour
        • Lecture11.6
          Circle Through Points, Theorem-9, Arc of a Circle, Theorem-10 01 hour
        • Lecture11.7
          Chapter Notes – Circles
        • Lecture11.8
          NCERT Solutions – Circles Exercise
        • Lecture11.9
          R D Sharma Solutions Circles
        • Lecture11.10
          R S Aggarwal Solutions Circles
        • Lecture11.11
          Revision Notes Circles
      • 12.Heron's Formula
        8
        • Lecture12.1
          Introduction of Heron’s Formula; Area of Different Triangles 01 hour
        • Lecture12.2
          Area of Triangles Using Heron’s Formula (Type-1 level-1 and 2) 53 min
        • Lecture12.3
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-1) 32 min
        • Lecture12.4
          Area of Quadrilaterals Using Heron’s Formula (Type-2 level-2) 38 min
        • Lecture12.5
          Chapter Notes – Heron’s Formula
        • Lecture12.6
          NCERT Solutions – Heron’s Formula Exercise 12.1 12.2
        • Lecture12.7
          R D Sharma Solutions Heron’s Formula
        • Lecture12.8
          Revision Notes Heron’s Formula
      • 13.Surface Area and Volume
        16
        • Lecture13.1
          Introduction, Surface Area of Cube, Cuboid, Cylinder, Hollow cylinder, Cone, Sphere 40 min
        • Lecture13.2
          Type-1: Surface Area and Volume of Cube and Cuboid 27 min
        • Lecture13.3
          Level-2, Question based on Cube and Cuboid, Level-1 Questions based on Cylinder 28 min
        • Lecture13.4
          Level-2 Questions based on Cylinder, Type-3 Surface Area and Volume of Right Circular Cone (Level-1) 25 min
        • Lecture13.5
          Level-2 Questions based on Right Circular Cone, Level-1& 2 Questions Based on Sphere & Hemisphere 27 min
        • Lecture13.6
          Type-5 Short Answer Types Questions, 12 min
        • Lecture13.7
          Chapter Notes – Surface Area and Volume of a Cuboid and Cube
        • Lecture13.8
          Chapter Notes – Surface Area and Volume of A Right Circular Cylinder
        • Lecture13.9
          Chapter Notes – Surface Area ad Volume of A Right Circular Cone
        • Lecture13.10
          Chapter Notes – Surface Area ad Volume of A Sphere
        • Lecture13.11
          NCERT Solutions – Surface Area and Volume Exercise 13.1 – 13.9
        • Lecture13.12
          R D Sharma Solutions Surface Areas And Volumes of A cuboid and A cube
        • Lecture13.13
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cylinder
        • Lecture13.14
          R D Sharma Solutions Surface Areas And Volumes of A Right Circular Cone
        • Lecture13.15
          R D Sharma Solutions Surface Areas And Volumes of A Sphere
        • Lecture13.16
          Revision Notes Surface Areas And Volumes
      • 14.Statistics
        15
        • Lecture14.1
          Data; Presentation of Data; Understanding Basic terms; 01 hour
        • Lecture14.2
          Questions; 52 min
        • Lecture14.3
          Bar Graph, Histogram, Frequency Polygon; 01 hour
        • Lecture14.4
          Histogram for Grouped Frequency Distribution (Class interval in Inclusive form), Frequency Polygon, Measures of Central Tendency, 01 hour
        • Lecture14.5
          Examples Based on Mean of Ungrouped Data 56 min
        • Lecture14.6
          Direct Method of Finding Mean of Ungrouped Data, Median of Cont. Freq. Distribution (Grouped Data) 01 hour
        • Lecture14.7
          Mean; Mean for an ungrouped freq. Dist. Table 25 min
        • Lecture14.8
          Chapter Notes – Statistics – Tabular Representation of Statistical Data
        • Lecture14.9
          Chapter Notes – Statistics – Graphical Representation of Statistical Data
        • Lecture14.10
          Chapter Notes – Statistics – Measures of Central Tendency
        • Lecture14.11
          NCERT Solutions – Statistics Exercise 14.1 – 14.4
        • Lecture14.12
          R D Sharma Solutions Tabular Representation of Statistical Data
        • Lecture14.13
          R D Sharma Solutions Graphical Representation of Statistical Data
        • Lecture14.14
          R S Aggarwal Solutions Statistics
        • Lecture14.15
          Revision Notes Statistics
      • 15.Probability
        8
        • Lecture15.1
          Activities; Understanding term Probability; Experiments; 44 min
        • Lecture15.2
          Event; Probability; Empirical Probability- Tossing a coin; 31 min
        • Lecture15.3
          Rolling a dice; Miscellaneous Example; 27 min
        • Lecture15.4
          Chapter Notes – Probability
        • Lecture15.5
          NCERT Solutions – Probability Exercise 15.1
        • Lecture15.6
          R D Sharma Solutions Probability
        • Lecture15.7
          R S Aggarwal Solutions Probability
        • Lecture15.8
          Revision Notes Probability

        NCERT Solutions – Surface Area and Volume Exercise 13.1 – 13.9

        Q.1     A plastic box 1. 5 m long , 1.25 m wide and 65 cm deep is to be made. It is to be open at the top. Ignoring the thickness of the plastic sheet, determine :
        (i) The area of the sheet required for making the box.
        (ii) The cost of sheet for it, if a sheet measuring 1 m2 costs Rs 20.
        Sol.

        We have, Length l = 1.5 m  Breadth b = 1.25 m  and  depth = Height. h = 65 cm = .65 m
        (i) Since the plastic box is open at the top. Therefore , Plastic sheet required for making such a box.
        =[2(ℓ+b)×h+ℓb]m2
        =[2(1.5+1.25)×.65+1.5×1.25]m2
        =[2×2.75×.65+1.875]m2
        =(3.575+1.875)m2=5.45m2

        (ii) Cost of 1m2 of sheet = Rs. 20
        Therefore Total cost of 5.45m2ofsheet=Rs(5.45×20)=Rs.109


        Q.2     The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs. 7.50 per m2.
        Sol.

        Here, Length l = 5m , Breadth b = 4 m  and  Height h = 3m
        Area of four walls including ceiling =[2(ℓ+b)×h+ℓb]m2
                                                             =[2(5+4)×3+5×4]m2
                                                             =[2×9×3+20]m2
                                                             =(54+20)m2=74m2
        Cost of white washing is Rs 7.50 per square metre.
        Therefore Cost of white washing = Rs (74 × 7.50) = Rs 555


        Q.3     The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs 10 per m2 is Rs 15000, find the height of the hall.
        Sol.

        Cost of painting the four walls = Rs 15000
        Rate of painting is Rs 10 per m2
        Therefore Area of four walls =(1500010)m2=1500m2
        ⇒ 2(ℓ+b)h=1500
        ⇒ Perimeter × Height = 1500
        ⇒ 250× Height = 1500
        ⇒ Height=1500250=6
        Hence , the height of the hall = 6 metres.


        Q.4     The paint in a certain container is sufficient to paint an area equal to 9.375 m2. How many bricks of dimensions 22.5 cm ×10 cm × 7.5 cm can be painted out of this container?
        Sol.

        Length l = 22.5 cm , Breadth b = 10 cm  and  Height h = 7.5 cm
        Surface area of one brick =2(ℓb+bh+hℓ)
                                                =2(22.5100×10100+10100×7.5100+7.5100×22.5100)m2
                                                =2×1100×1100(22.5×10+10×7.5+7.5×22.5)m2
                                                              =15000×(225+75+168.75)m2
                                                              =15000×468.75m2=0.09375m2
        Area for which the paint is just sufficient is 9.375 m2
        Therefore Number of bricks that can be painted with the available paint =9.3750.09375=100 bricks


        Q.5     A cubical box has each edge 10 cm and another cuboidal box is 12.5 cm long, 10 cm wide and 8 cm high.
                      (i) Which box has the greater lateral surface area and by how much ?
                      (ii) Which box has the smaller total surface area and by how much ?
        Sol.

        (i) Lateral surface area of cubical box of edge 10cm=4×102cm2=400cm2
        Lateral surface area of cuboid box =2(ℓ+b)×h
                                                             =2×(12.5+10)×8cm2
                                                                               =2×22.5×8cm2
                                                                               =360cm2
        Thus, lateral surface area of the cubical box is greater and is more by (400 – 360) cm2 i.e., 40cm2

        (ii) Total surface area of cubical box of edge 10 cm
        =6×102cm2=600cm2
        Total surface area of cuboidal box =2(ℓb+bh+hℓ)
                                                                              =2(12.5×10+10×8+8×12.5)cm2
                                                                              =2(125+80+100)cm2
                                                                              =(2×305)cm2=610cm2
        Thus, total surface area of cubical box is smaller by 10cm2


        Q.6      A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.
                      (i) What is the area of the glass?
                      (ii) How much of tape is needed for all the 12 edges?
        Sol.

        Here, l = 30 cm , b = 25 cm and h = 25 cm
        (i) Area of the glass = Total surface area =2(ℓb+bh+hℓ)
                                                                                           =2(30×25+25×25+25×30)cm2
                                                                                           =2(750+625+750)cm2
                                                                                           =(2×2125)cm2=4250cm2

        (ii) Tap needed for all the 12 edges =Thesumofalltheedges
                                                                                =4(ℓ+b+h)=4(30+25+25)cm
                                                                                =4×80cm=320cm


        Q.7    Shanti Sweets Stall was placing an order for making cardboard boxes for packing their sweets. Two sizes of boxes were required. The bigger of dimensions 25 cm × 20 cm × 5 cm and the smaller of dimensions 15 cm × 12 cm × 5 cm . For all the overlaps, 5 % of the total surface area is required extra. If the cost of the cardboard is Rs 4 for 1000 cm2, Find the cost of cardboard required for supplying 250 boxes of each kind.
        Sol.

        In case of bigger box :
        l = 25 cm , b = 20 cm and h = 5 cm
        Totalsurfacearea=2(ℓb+bh+hℓ)
                                          =2(25×20+20×5+5×25)cm2
                                          =2(500+100+125)cm2
                                          =(2×725)cm2
                                          =1450cm2
        In case of smaller box :
        l = 15 cm , b = 12 cm and h = 5 cm
        Total surface area =2(ℓb+bh+hℓ)
                                             =2(15×12+12×5+5×15)cm2
                                             =2(180+60+75)cm2
                                             =(2×315)cm2=630cm2
        Total surface area of 250 boxes of each type
        =250(1450+630)cm2
        =(250×2080)cm2=520000cm2
        Cardboard required (i.e., including 5% extra for overlaps etc.)
        =(520000×105100)cm2
        =546000cm2
        Cost of 1000cm2 of cardboard
        =Rs4
        Therefore Total cost of cardboard
        =Rs(5460001000×4)=Rs2184


        Q.8      Parveen wanted to make a temporary shelter for her car, by making a box- like structure with tarpaulin that covers all the four sides and the top of the car (with the front face as a flap which can be rolled up). Assuming that the stitching margins are very small, and therefore negligible, how much tarpaulin would be required to make the shelter of height 2.5 m, with base dimensions 4 m × 3 m?
        Sol.

        Dimensions of the box- like structure are l = 4m, b = 3m and h = 2.5 m.
        Since there is no tarpaulin for the floor.
        Therefore Tarpaulin required =[2(ℓ+b)×h+ℓb]m2
                                                                   =[2(4+3)×2.5+4×3]m2
                                                                   =[2×7×2.5+12]m2
                                                                   =[35×12]m2=47m2

         

        Assume π=227, unless stated otherwise.
        Q.1     The curved surface area of a right circular cylinder of height 14 cm is 88 cm2. Find the diameter of the base of the cylinder.
        Sol.

        Let r be the radius of the base and h = 14 cm be the height of the cylinder. Then,
        Curved surface area of cylinder =2πrh
        ⇒ 88=2×227×r×14
        ⇒ r=88×72×22×14=1
        Therefore Diameter of the base = 2r =2×1=2cm


        Q.2    It is required to make a closed cylindrical tank of height 1m and base diameter 140 cm from a metal sheet. How many square meters of the sheet are required for the same?
        Sol.

        Let r be the radius of the base and h be the height of the cylinder.
        Base diameter = 140 cm, radius of base = 1402 = 70 cm = .70 m
        Height = 1m
        Metal sheet required to make a closed cylindrical tank
        = Its total surface area
        =2πr(h+r) =2×227×0.7(1+0.70)m2
        [Because h = 1 m, r = 1402 cm = 70 cm = 0.70 cm]
        =2×22×0.1×1.70m2
        =7.48m2
        Hence, the sheet required =7.48m2

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        Q.3      A metal pipe is 77 cm long. The inner diameter of a cross- section is 4 cm, the outer diameter being 4.4 cm (see figure). Find its.
                    
        (i) Inner curved surface area,
                      (ii) Outer curved surface area,
                   
        (iii) Total surface area.
        surface

        Sol.

        We have, R = external radius =4.42cm=2.2cm
        r = internal radius =42cm=2cm
        h = length of the pipe = 77 cm
        (i)
         Inner curved surface =2πrhcm2
                                                 =2×227×2×77cm2
                                                 =968cm2

        (ii) Outer curved surface =2πRhcm2
                                                    =2×227×2.2×77cm2
                                                    =1064.8cm2

        (iii) Total surface area of a pipe
        = Inner curved surface area + outer curved surface area + areas of two bases
        = 2πrh+2πRh+2π(R2−r2)
        = [968+1064.8+2×227(4.84−4)] cm2
        = (2032.8+447×0.84)cm2
        =(2032.8+5.28)cm2=2038.08cm2


        Q.4     The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.
        Sol.

        The length of the roller is 120 cm i.e., h = 1.2 m and ,
        radius of the cylinder (i.e., roller) =842cm=42cm=0.42m.
        Distance covered by roller in one revolution
        = Its curved surface area = 2πrh
        = (2×227×0.42×1.2)m2
        = 3.168m2
        Area of the playground = Distance covered by roller in 500 revolution.
                                                = (500×3.168)m2=1584m2
        Hence , the area of playground is 1584m2

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        Q.5     A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting the curve surface of the pillar at the rate of Rs 12.50 per m2.
        Sol.

        Let r be the radius of the base and h be the height of the pillar.
        Therefore r=502 cm = 25 cm = .25 m and h = 3.5 m.
        Curved surface = 2πrh
                                  =(2×227×0.25×3.5)m2=5.5m2
        Cost of painting the curved surface @ Rs. 12. 50 per m2
        =(5.5×12.5) =  Rs. 68.75


        Q.6      Curved surface area of a right circular cylinder is 4.4 m2. If the radius of the base of the cylinder 0.7 m, Find its height.
        Sol.

        Let r be the radius of the base and h be the height of the cylinder. Then,
        Curved surface area = 4.4 m2
        ⇒ 2πrh=4.4
        ⇒ 2×227×0.7×h=4.4 [since r = 0.7]
        ⇒ h=(4.4×72×22×0.7)m=1m
        Thus, the height of the cylinder = 1 metre.

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        Q.7     The inner diameter of a circular well is 3.5 m. It is 10 m deep. Find.
                    
        (i) Its inner curved surface area.
                    
        (ii) The cost of plastering this curved surface at the rate of Rs 40 per m2
        Sol.

        (i) Let r be the radius of the face and h be depth of the well. Then,
        Curved surface = 2πrh
                                   = (2×227×3.52×10)m2=110m2

        (ii) Cost of plastering is Rs 40 per m2
        Therefore , cost of plastering the curved surface = Rs (110 × 40) = Rs 4400.


        Q.8     In a hot water heating system, there is a cylindrical pipe of length 28 m and diameter 5 cm. Find the total radiating surface in the system.
        Sol.

        Total radiating surface in the system
        = Curved surface area of the pipe = 2πrh
        [where r=52cm=2.5cm=2.5100m=0.025mandh=28m]
        =(2×227×0.025×28)m2=4.4m2


        Q.9     Find
                    (i) the lateral or curved surface area of cylindrical petrol storage tank that is 4.2 m in diameter and 4.5 m high.

                    (ii) How much steel was actually used, if 112 of the steel actually used was wasted in making the closed tank.

        Sol.

        (i) Here, r=(4.22)m=2.1mandh=4.5m
        Lateral surface area =2πrhm2
                                           =(2×227×2.1×4.5)m2
                                           =59.4m2

        (ii) Since 112 of the actual steel used was wasted,
        the area of the steel which has gone into the tank = (1−112) of x=1112ofx.
        Steel used = Letarl surface area + 2 x Area of base
                           = (2πrh+2πr2)cm2 = (59.4+2×227×2.1×2.1)cm2
                           = (59.4+27.72)cm=87.12cm2
        Therefore,   actual steel used =1112×x=87.12
        ⇒ x=87.12×1211=95.04m2
        Hence, the actual area of the steel used = 95.04m2

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        Q.10    In figure, you see the frame of a lampshade. It is to be covered with a decorative cloth. The frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. Find how much cloth is required for covering the lampshade.2Sol.

        Here , r=(202) cm = 10 cm and h = 30 cm + 2 × 2.5 cm (i.e., margin) = 35 cm
        Cloth required for covering the lampshade
        = Its curved surface area = 2πrh
        = (2×227×10×35)cm2
        = 2200cm2


        Q.11   The students of a Vidyalaya were asked to participate in a competition for making and decorating pen holders in the shape of a cylinder with a base, using cardboard. Each pen holder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors , how much cardboard was required to be bought for the competition?
        Sol.

        Cardboard required by each competitor
        = Curved surface area of one penholder + base area = 2πrh+πr2
        [ wherer=3cm,h=10.5cm]
        = [(2×227×3×10.5)+227×9]cm2
        = (198+28.28)cm2 = 226.28cm2(approx)
        Cardboard required for 35 competitors = (35×226.28)cm2
        = 7920cm2(approx)

         

         

        Assume π=227, unless stated otherwise.
        Q.1     Diameter of the base of a cone is 10.5 cm and its slant height is 10 cm. Find its curved surface area.

        Sol.

        Here,  radiusr=(10.52) cm = 5.25 cm  and  slant height( l) = 10 cm
        Curved surface area of the cone = (πrℓ)cm2
                                                                         =(227×5.25×10)cm2
                                                                         =165cm2


        Q.2     Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
        Sol.

        Here, radius r=(242) m = 12 m and slant height( l)= 21 m.
        Total surface area of the cone =(πrℓ+πr2)m2
                                                                     =πr(ℓ+r)m2
                                                                     =227×12×(21+12)m2
                                                                     =(227×12×33)m2
                                                                     =1244.57m2(approx)


        Q.3     Curved surface area of a cone is 308 cm2 and its slant height is 14 cm. Find
                   (i) radius of the base and
                   (ii) total surface area of the cone.

        Sol.

        (i) Curved surface of a cone = 308 cm2
        Slant height  ℓ=14cm
        Let r be the radius of the base .
        Therefore πrℓ=308
        ⇒ 227×r×14=308
        ⇒ r=308×722×14=7 cm
        Thus, the radius of the base = 7 cm

        (ii) Total surface area of the cone =πr(ℓ+r)cm2
                                                                            =227×7×(14+7)cm2
                                                                            =(22×21)cm2
                                                                            =462cm2


        Q.4      A conical tent is 10 m high and the radius of its base is 24 m. Find
                       (i) Slant height of the tent.
                       (ii) Cost of the canvas required to make the tent, if the cost of 1m2 canvas is Rs 70.
        Sol.

        (i) Here r = 24 m , h = 10 m
        Let l be the slant height of the cone. Then,
        ℓ2=h2+r2
        ⇒ ℓ=h2+r2−−−−−−√
        ⇒ ℓ=242+102−−−−−−−−√
        =576+100−−−−−−−−√
        =676−−−√=26m

        (ii) Canvas required to make the conical tent = Curved surface of the cone
        πrℓ=(227×24×26)m2
        Rate of canvas per 1m2 is Rs 70
        Therefore cost of canvas =(227×24×26×70) = Rs 137280


        Q.5     What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm. (Use π=3.14)
        Sol.

        Let r m  be the radius , h  m be the height and l m  be the slant height of the tent. Then ,
        r = 6 m , h = 8m
        ⇒ ℓ=r2+h2−−−−−−√=62+82−−−−−−√=36+64−−−−−−√
                   =100−−−√=10m
        Area of the canvas used for the tent = curved surface area of the tent
                                                                                =πrℓ=(3.14×6×10)m2=188.4m2
        Now, this area is bought in the form of a rectangle of width 3m.
        Therefore  length of tarpaulin required  =AreaoftarpaulinrequiredWidthoftarpaulin
                                                                                       =(188.43)m=62.8m
        The extra material required for stitching margins and cutting =20cm=0.2m
        So, the total length of tarpaulin required =(62.8+0.2)m=63m


        Q.6     The slant height and base diameter of a conical tomb are 25 m and 14 m respectively. Find the cost of white – washing its curved surface at the rate of Rs 210 per 100 m2.
        Sol.

        Here , slant height (l) = 25 m and radius r=(142)m=7m
        Curved surface area = πrℓm2
                                        =(227×25×7)m2
                                                   =550m2
        Rate of white- washing is Rs 210 per 100m2
        Therefore , cost of white – washing the tomb =(550×210100)=Rs1155


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

        Let r cm be the radius, h cm be the height and l cm be the slant height of the joker’s cap. Then ,
        r = 7 cm , h = 24 cm
        ℓ=h2+r2−−−−−−√=242+72−−−−−−−√
                   =576+49−−−−−−−√
                   =625−−−√=25cm
        Sheet required for one cap = Curved surface of the cone
                                                               =πrℓcm2
                                                               =(227×7×25)cm2
                                                               =550cm2
        Sheet required for 10 such caps = 5500cm2


        Q.8      A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1m. If the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m2, what will be the cost of painting all these cones? (Use π=3.14 and take 1.04−−−−√=1.02).
        Sol.

        Let r  be the radius , h  be the height and l  be the slant height of a cone. Then ,
        r=(402)cm = 20 cm = .2 m,
        and h = 1m.
        Therefore ℓ=r2+h2−−−−−−√=0.04+1−−−−−−−√=1.04−−−−√=1.02
        Curved surface of 1 cone =πrℓm2
                                                             =(3.14×.2×1.02)m2
        Curved surface of such 50 cones =(50×3.14×.2×1.02)m2
        Cost of painting @ Rs. 12 per m2 =(50×3.14×.2×1.02×12)
                                                                                                =384.68(approx)

         

         

        Q.1     Find the surface area of a sphere of radius :
                    (i) 10. 5 cm       (ii) 5.6 cm       (iii) 14 cm
        Sol.

        (i) We have :
        r = radius of the sphere = 10.5 cm
        Surface area =4πr2
                            =(4×227×10.5×10.5)cm2
                                   =1386cm2

        (ii) We have :
        r = radius of the sphere = 5.6 cm
        Surface area =4πr2 =(4×227×5.6×5.6)cm2
                                                                          =394.24cm2

        (iii) We have :
        r = radius of the sphere = 14 cm
        Surface area =(4×227×14×14)cm2
                            =2464cm2


        Q.2     Find the surface area of a sphere of diameter :
                   (i) 14 cm              (ii) 21 cm                        (iii) 3.5 m
        Sol.

        (i) Here , r=(142)cm=7cm
        Surface area = 4πr2 =(4×227×7×7)cm2
                                                        =616cm2

        (ii) Here , r=(212)cm=10.5cm
        Therefore Surface area = 4πr2 =(4×227×10.5×10.5)cm2
                                                                        =1386cm2

        (iii) Here , r=(3.52)cm=1.75m
                                      =(4×227×1.75×1.75)m2
                               =38.5m2


        Q.3     Find the total surface area of a hemisphere of radius 10 cm (Use π=3.14)
        Sol.

        Here , r = 10 cm
        Total surface area of hemisphere =3πr2 =(3×3.14×10×10)cm2
                                                                                                                = 942cm2


        Q.4     The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.
        Sol.

        Let r1andr2 be the radius of balloons in the two cases.
        Here , r1=7cmandr2=14cm
        Therefore , ratio of their surface area =4πr124πr22=r21r22
                                                                                   =7×714×14=14
        Thus, the required ratio of their surface areas = 1 : 4


        Q.5     A hemispherical bowl made of brass has inner diameter 10.5 cm. Find the cost of tin- plating it on the inside at the rate of Rs 16 per 100 cm2
        Sol.

        Here r=(10.52)cm=5.25cm
        Curved surface area of the hemisphere =2πr2=(2×227×5.25×5.25)cm2
                                                                   =173.25cm2
        Rate of tin – plating is Rs 16 per 100 cm2.
        Therefore Cost of tin – plating the hemisphere =(173.25×16100)
                                                                              =Rs.27.72


        Q.6      Find the radius of a sphere whose surface area is 154cm2
        Sol.

        Let r be the radius of the sphere.
        Surface area = 154cm2
        ⇒ 4πr2=154
        ⇒ 4×227×r2=154
        ⇒ r2=154×74×22=12.25
        ⇒ r=12.25−−−−√=3.5
        Thus, the radius of the sphere is 3.5 cm


        Q.7     The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas.
        Sol.

        Let the diameter of earth be R and that of the moon will be R4
        The radii of moon and earth are R8andR2 respectively.
        Ratio of their surface area =4π(R8)24π(R2)2=16414
                                                             =164×41=116i.e.,1:16


        Q.8     A hemispherical bowl is made of steel , 0.25 cm thick. The inner radius of the bowl is 5 cm. Find the outer curved surface area of the bowl.
        Sol.

        Inner radius r = 5 cm
        Thickness of steel = 0.25 cm
        Therefore , outer radius R = (r + 0.25) cm
                                                = (5 + 0.25) cm  = 5.25
        Therefore Outer curved surface = 2πR2
                                                                         =(2×227×5.25×5.25)cm2
                                                                          =173.25cm2


        Q.9 A right circular cylinder just encloses a sphere of radius r (see figure). Find

        3
        (i) Surface area of the sphere,
        (ii) Curved surface area of the cylinder,
        (iii) Ratio of the areas obtained in (i) and (ii).
        Sol.

        (i) The radius of the sphere r, so its surface area = 4πr2

        (ii) Since the right circular cylinder just encloses a sphere of radius r. So the radius of cylinder = r and its height = 2r
        Curved surface of cylinder =2πrh
        2πr(2r)         [Since r = r , h = r]
        = 4πr2

        (iii) Ratio of area = 4πr2:4πr2=1:1

         

         

        Q.1     A matchbox measures 4 cm × 2.5 cm × 1.5 cm . What will be the volume of a packet containing 12 such boxes?
        Sol.

        Here , l = 4 cm, b = 2.5 cm and h = 1.5 cm
        Therefore Volume of one matchbox =ℓ×b×hcm3
                                                                      =(4×2.5×1.5)cm3=15cm3
        Therefore Volume of a packet containing 12 such boxes =(12×15)cm3=180cm3


        Q.2     A cuboid water tank is 6 m long, 5 m wide and 4.5 m deep. How many liters of water can it hold? (1m3=1000ℓ).
        Sol. 

        Here , l = 6 m , b = 5 m and h = 4.5 m
        Therefore Volume of the tank =ℓbhm3
                                                           =(6×5×4.5)m3=135m3
        Therefore , the tank can hold = 135 × 1000 litres          [Since 1m3=1000litres]
                                                          = 135000 litres of water.


        Q.3    A cuboidal vessel is 10 m long and 8 m wide. How high must it be made to hold 380 cubic meters of a liquid?
        Sol.

        Here, Length = 10 m , Breadth = 8 m and Volume = 380m3
        Volume of cuboid = Length x Breadth x Height
        Height=VolumeofcuboidLength×Breadth 
                           =38010×8m
                           =4.75m


        Q.4     Find the cost of digging a cuboidal pit 8 m long, 6 m broad and 3m deep at the rate of Rs 30 per m3.
        Sol.

        Here, l = 8 m, b = 6 m and h = 3 m
        Volume of the pit =ℓbhm3
                                       = (8×6×3)m3=144m3
        Rate of digging is Rs 30 per m3
        Therefore Cost of digging the pit = Rs (144 × 30) = Rs 4320.


        Q.5    The capacity of a cuboidal tank is 50000 litres of water. Find the breadth of the tank, if its length and depth are respectively 2.5 m and 10 m.
        Sol.

        Here, length = 2.5 m, depth = 10 m and volume = 50000 litres
        =(50000×11000)m3[1m3=1000litres]
        =50m3
        Breadth=VolumeofcuboidLength×Depth=(5025×10)m=2m


        Q.6     A village having a population of 4000, requires 150 litres of water per head per day. It has a tank measuring 20 m × 15 m ×6 m . For how many days will the water of this tank last?
        Sol.

        Here , l = 20 m , b = 15 m and h = 6 m
        Therefore Capacity of the tank =ℓbhm3
                                                             =(20×15×6)m3=1800m3
        Water requirement per person per day =150litres
        Water required for 4000 person per day =(4000×150)ℓ
                                                                             =(4000×1501000)m3=600m3
        Number of days the water will last =CapacityoftankTotalwaterrequiredperday
                                                                   =(1800600)=3
        Thus, the water will last for 30 days.


        Q.7     A godown measures 60 m × 25 m × 10 m. Find the maximum number of wooden crates each measuring 1.5 m × 1.25 m × 0.5 m that can be stored in the godown.
        Sol. 

        Volume of the godown =(60×25×10)m3 =15000m3
        Volume of 1 crates =(1.5×1.25×0.5)m3 =0.9375m3
        Number of crates that can be stored in the godown
        =VolumeofthegodownVolumof1crate
        =150000.9375=16000


        Q.8     A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas.
        Sol. 

        Let V1 be volume of the cube of edge 12 cm . So , length = breadth = height = 12 cm
        Volume of the cube =(12×12×12)cm3
        and V2 = Volume of the cube cut out of the first one
               =18×V1=(18×12×12×12)cm3
               =(6×6×6)cm3
        Therefore , side of the new cube = 6 cm
        Ratio of their surface areas =6(side)26(side)2=6×12×126×6×6
                                                       =41i.e.,4:1


        Q.9     A river 3 m deep and 40 m wide is flowing at the rate of 2 km per hour. How much water will fall into the sea in a minute?
        Sol. 

        Since the water flows at the rate of 2 km per hour, the water from 2 km of river flows into the sea in one hour.
        Therefore The volume of water flowing into the sea in one hour = Volume of the cuboid
                                                                                                                    =ℓ×b×hm3
                                                                                                                    =(2000×40×3)m3
        Therefore , the volume of water flowing into the sea in one minute
        =(2000×40×360)m3
        =4000m3

         

         

        Assume π=227, unless stated otherwise.
        Q.1     The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. How many liters of water can it hold? (1000cm3=1ℓ)

        Sol.

        Let r cm be the radius of the base and h cm be the height of the cylinder.
        Circumference of the base = 132 cm
        ⇒ 2πr=132
        ⇒ 2×227×r=132
        ⇒ r=(132×72×22)cm=21cm
        Volume of the cylinder = πr2hcm3
                                               =(227×21×21×25)cm3
                                               =34650cm3
        Therefore Vessel can hold =(346501000)litres
        i.e., 34.65 litres of water.


        Q.2    The inner diameter of a cylindrical wooden pipe is 24 cm and its outer diameter is 28 cm. The length of the pipe is 35 cm. Find the mass of the pipe, if 1cm3 of wood has a mass of 0.6 g.
        Sol.

        We have h = Height of the cylindrical pipe = 35 cm
        R = External radius =(282)cm=14cm
        r = Internal radius =(242)cm=12cm
        Volume of the wood used in making the pipe = Volume the external cylinder – Volume of the internal cylinder
        πR2h−πr2h=π(R2−r2)h
                                      =227×(142−122)×35cm3
                                      =227×26×2×35cm3
                                      =5720cm3
        Weight of 1cm3=0.6g
        Therefore Weight of 5720 cm3=(5720×0.6)g
                                                                   =(5720×0.61000)kg     [1 kg = 1000 g]
                                                                   =3.432kg


        Q.3     A soft drink is available in two packs – (i) a tin can with a rectangular base of length 5 cm and width 4 cm, having a height of 15 cm and (ii) a plastic cylinder with circular base of diameter 7 cm and height 10 cm. Which container has greater capacity and by how much ?
        Sol.

        (i) Capacity of tin can =ℓbhcm3
                                              =(5×4×15)cm3
                                              =300cm3
        (ii) Capacity of plastic cylinder =πr2hcm3
                                                               =(227×72×72×10)cm
                                                               =385cm3
        Thus , the plastic cylinder has greater capacity by (385 – 300) = 85 cm3


        Q.4     If the lateral surface of a cylinder is 94.2cm2 and its height is 5 cm , then find (i) radius of its base (ii) its volume (Use π=3.14)
        Sol.

        (i) Let r be the radius of the base and h be the height of the cylinder. Then ,
        Lateral surface =94.2cm2
        ⇒ 2πrh=94.2
        ⇒ 2 × 3.14 × r × 5 = 94.2
        ⇒ r=94.22×3.14×5=3
        Thus , the radius of its base = 3 cm

        (ii) Volume of the cylinder =πr2h
                                                       =(3.14×32×5)cm3
                                                       =141.3cm3


        Q.5     It costs Rs 2200 to paint the inner curved surface of a cylindrical vessel 10 m deep. If the cost of painting is at the rate of Rs 20 per m2, Find.
                      (i) Inner curved surface area of the vessel,
                      (ii) Radius of the base,
                      (iii) Capacity of the vessel.
        Sol.

        (i) Inner curved surface area of the vessel
        =TotalcostofpaintingRateofpainting
        =(220020)m2=110m2

        (ii) Let r be the radius of the base and h be the height of the cylindrical vessel.
        Therefore 2πrh=110
        ⇒ 2×227×r×10=110
        ⇒ r×110×72×22×10=74=1.75
        Thus, the radius of the base = 1.75 m

        (iii) Capacity of the vessel = πr2h
                                                      =(227×74×74×10)m3
                                                      =96.25m3


        Q.6     The capacity of a closed cylindrical vessel height 1 m is 15.4 litres. How many square metres metal sheet would be needed to make it?
        Sol.

        Capacity of a closed cylindrical vessel = 15. 4 litres
                                                                          =(15.4×11000)m3=0.0154m3 [1 m3= 1000 liters]
        Let r be the radius of the base and h be the height of the vessel. Then ,
        Volume=πr2h=πr2×1=πr2                       [Since h = 1m]
        Therefore πr2=0.0154
        ⇒ 227×r2=0.0154
        ⇒ r2=0.0154×722=0.0049
        ⇒ r=0.0049−−−−−√=0.07
        Thus, the radius of the base of vessel = 0.07 m
        Metal sheet needed to make the vessel = Total surface area of the vessel
                                                                          =2πrh+2πr2=2πr(h+r)
                                                                          =2×227×0.07×(1+0.07)m2
                                                                          =44×0.01×1.07m2=0.4708m2


        Q.7      A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is 7 mm and the diameter of the graphite is 1 mm. If the length of the pencil is 14 cm, find the volume of the wood and that of the graphite.
        Sol.

        Diameter of the graphite cylinder = 1 mm = 110 cm
        Therefore Radius=120cm
        Length of the graphite = 14 cm
        Volume of the graphite cylinder =πr2h
                                                               =(227×120×120×14)cm3=0.11cm3
        Diameter of the pencil = 7mm = 710cm=0.11cm3
        Therefore Radius of the pencil = 720cm
        and , length of the pencil = 14 cm
        Therefore Volume of the pencil =πr2h
                                                               =(227×720×720×14)cm3
                                                               =5.39cm3
        Volume of wood = Volume of the pencil – Volume of the graphite
                                     =(5.39−0.11)cm3
                                     =5.28cm3


        Q.8     A patient in a hospital is given soup daily in a cylindrical bowl of diameter 7 cm. If the bowl is filled with soup to a height of 4 cm, how much soup the hospital has to prepare daily to serve 250 patients?
        Sol.

        Diameter of the cylindrical bowl = 7 cm
        Therefore Radius = 72cm
        Height of serving bowl = 4 cm
        Therefore Soup saved in on serving = Volume of the bowl
                                                                    =πr2h
                                                 =(227×72×72×4)cm3
                                                                     =1.54cm3
        Soup served to 250 patients = (250×1.54)cm3
                                                        =38500cm3i.e.,38.5ℓ
        Hence , the hospital has to prepare 38.5 l soup daily to serve 250 patients.

         

         

        Assume π=227, unless stated otherwise.
        Q.1      Find the volume of the right circular cone with

                       (i) Radius 6 cm , height 7 cm
                       (ii) Radius 3.5 cm , height 12 cm
        Sol.

        (i) Here , r = 6 cm and h = 7 cm
        Volume of the cone =13πr2h
                                          =(13×227×6×6×7)cm3
                                          =264cm3
        (ii) Here ,  r = 3.5 cm and h = 12 cm
        Volume of the cone =13πr2h
                                           =(13×227×3.5×3.5×12)cm3
                                           =154cm3.


        Q.2      Find the capacity in litres of a conical vessel with
                       (i) Radius 7 cm , slant height 25 cm
                       (ii) Height 12 cm, slant height 13 cm
        Sol.

        (i) Here , r = 7 cm and l = 25 cm
        Let the height of the cone be h cm. Then ,
        h2=ℓ2−r2=252−72
                  =625−49=576
        ⇒ h=576−−−√=24cm
        Volume of the conical vessel =13πr2h
                                                          =(13×227×7×7×24)cm3=1232cm3
        Therefore , capacity of the vessel in litres =(12321000)ℓ=1.232ℓ

        (ii) Here , h = 12 cm and l = 13 cm
        Let the radius of the base of the cone be r cm . Then,
        r2=ℓ2−h2=132−122
                 =169−144=25
        ⇒ r=25−−√=5cm
        Volume of the conical vessel =13πr2h
                                                          =(13×227×5×5×12)cm3=22007cm3
        Therefore Capacity of the vessel in litres  =(22007×11000)ℓ=1135ℓ


        Q.3    The height of a cone is 15 cm. If its volume is 1570 cm3, find the radius of the base. (Use π=3.14)
        Sol. 

        Here , h = 15 cm and volume = 1570 cm3
        Let the radius of the base of cone be r cm.
        Therefore Volume = 1570cm3
        ⇒ 13πr2h=1570
        ⇒ 13×3.14×r2×15=1570
        ⇒ r2=15703.14×5=100
        ⇒ r=100−−−√=10
        Thus , the radius of the base of cone is 10 cm.


        Q.4      If the volume of a right circular cone of height 9 cm is 48πcm3, find the diameter of its base.
        Sol. 

        Here , h = 9 cm and volume = 48πcm3,
        Let the radius of the base of the cone be r cm
        Therefore Volume =48πcm3,
        ⇒ 13πr2h=48π
        ⇒ 13×r2×9=48
        ⇒ 3r2=48
        ⇒ r2=483=16
        ⇒ r=16−−√=4
        Thus, the diameter of the base of the cone = 2X4=8 cm


        Q.5      A conical pit of top diameter 3.5 m is 12 m deep. What is its capacity in kilolitres?
        Sol.

        Diameter of the top of the conical pit = 3.5 m
        Therefore , radius =(3.52)m=1.75m
        Depth of the pit i.e., height = 12 m
        Volume =13πr2h
                      =(13×227×1.75×1.75×12)m3
                      =38.5m3 [1m3 = 1 kilolitres]
        Therefore Capacity of pit = 38.5 kilolitres.


        Q.6      The volume of a right circular cone is 9856 cm3. If the diameter of the base is 28 cm, find
                     
        (i) Height of the cone.
                        (ii) Slant height of the cone,
                        (iii) Curved surface area of the cone.
        Sol.

        (i) Diameter of the base of the cone = 28 cm
        Therefore  , radius = r = (282)cm=14cm
        Volume of the cone = 9856cm3
        Let the height of the cone be h cm.
        Now,    volume =9856cm3
        ⇒ 13πr2h=9856
        ⇒ 13×227×14×14×h=9856
        ⇒ h=9856×3×722×14×14=48 cm
        Thus , the height of the cone = 48 cm

        (ii) Here , r = 14 m and h = 48 cm
        Let l be the slant height of the cone. Then,
        ℓ2=h2+r2=482+142
                =2304+196=2500
        ⇒ ℓ=2500−−−−√=50
        Thus, the slant height of the cone = 50 cm.

        (iii) Here , r = 14 m and l = 50 cm.
        Curved surface area = πrℓ
                                          =(227×14×50)cm2
                                          =2200cm2


        Q.7     A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. Find the volume of the solid so obtained.
        Sol.         On revolving the right Δ ABC about the side AB ( = 12 cm) , we get a cone as shown in the figure.

        4
        Volume of solid so obtained =13πr2h                     
                                                         =(13×π×25×12)cm3
                                                         =100πcm3


        Q.8      If the triangle ABC in the question 7 above is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in question 7 and 8.
        Sol.

        On revolving the right Δ ABC about the side BC( = 5 cm), we get a cone as shown in the figure,
        Volume of solid so obtained =13πr2h
                                                         =13×π×12×12×5cm3   [h = 5 cm , r = 12 cm]
                                                          = 240πcm3
        5
        Therefore Ratio of their volumes = 100π:240π (i.e., of Q. 7 and Q. 8)  = 5 : 12


        Q. 9    A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3m. Find its volume. The heap is to be covered by canvas to protect it from rain. Find the area of the canvas required.
        Sol.

        Diameter of the base of the cone = 10.5 m
        Therefore , radius = r = (10.52)m=5.25m
        Height of the cone = 3m
        Therefore Volume of the cone (heap) =13πr2h
                                                                        =(13×227×5.25×5.25×3)m3
                                                                        =86.625m3
        To find the slant height l :
        We have, ℓ2=h2+r2=32+(5.25)2
                                     =9+27.5625=36.5625
        ⇒ ℓ=36.5625−−−−−−√=6.0467(approx)
        Canvas required to protect wheat from rain = Curve surface area
                                                                                 =πrℓ=(227×5.25×6.0467)m2
                                                                                 =99.77m2(approx)

         

        ( Assume π=227, unless stated otherwise. )
        Q.1     Find the volume of a sphere whose radius is

                      (i) 7 cm               (ii) 0.63 cm
        Sol.

        (i) We have : r = radius of the sphere = 7 cm
        Therefore, Volume of the sphere =43πr3
                                                                =(43×227×7×7×7)cm3
                                                                =43123cm3=143713cm3
        (ii) We have : r = radius of the sphere = 0.63 m
        Therefore, Volume of the sphere =43πr3
                                                                =(43×227×0.63×0.63×0.63)m3
                                                                =1.05m3(approx)


        Q.2      Find the amount of water displaced by a solid spherical ball of diameter.
                       (i) 28 cm                  (ii) 0.21 m
        Sol.

        (i) Diameter of the spherical ball = 28 cm
        Therefore, Radius =(282)cm=14cm
        Amount of water displaced by the spherical ball
        =Itsvolume=43πr3
                                   =(43×227×14×14×14)cm3
                                   =344963cm3=1149823cm3

        (ii) Diameter of the spherical ball = 0.21 m
        Therefore, Radius =(0.212)m=0.105m
        Amount of water displaced by the spherical ball
        =Itsvolume=43πr3
                                    =(43×227×0.105×0.105×0.105)m3
                                    =0.004851m3

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        Q.3     The diameter of a metallic ball is 4.2 cm. What is mass of the ball, if the density of the metal is 8.9 g per cm3?
        Sol.

        Diameter of the ball = 4.2 cm
        Therefore ,     Radius =(4.22)cm=2.1cm
        Volume of the ball =43πr3
                                        =(43×227×2.1×2.1×2.1)cm3
                                        =38.808cm3
        Density of the metal is 8.9g per cm3
        Therefore Mass of the ball = (38.808 × 8.9) g = 345.3912 g


        Q.4      The diameter of the moon is approximately one – fourth the diameter of the earth. What fraction of the volume of the earth is the volume of the moon?
        Sol.

        Let the diameter of the moon be r. Then, the radius of the moon =r2
        According to the question, diameter of the earth is 4r, so its radius =4r2=2r.
        V1 = The volume of the moon =43π(r2)3
                                                                            =43πr3×18
        ⇒ 8V1=43πr3 … (1)
        and, V2 = The volume of the earth =43π(2r)3
                                     =43πr3×8
        ⇒ V28=43πr3 … (2)
        From (1) and (2) , we have
        8V1=V28 ⇒ V1=164V2
        Hence, the volume of the moon is 164 of the volume of the earth.

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        Q.5      How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
        Sol.

        Diameter of a hemispherical bowl = 10.5 cm
        Therefore , its radius =(10.52)cm=5.25cm
        Volume of the bowl =23πr3
                                          =(23×227×5.25×5.25×5.25)cm3
                                          =303.1875cm3
        Hence, the hemispherical bowl can hold 303 l (approx.) of milk.


        Q.6     A hemispherical tank is made up of an iron sheet 1 cm thick. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
        Sol.

        Let R cm and r cm be respectively the external and internal radii of the hemispherical vessel. Then,
        R = 1.01 (as thickness = 1 cm = .01 m)
        and r = 1 m.
        Volume of iron used = External volume – Internal volume
                                            =23πR3−23πr3
                                            =23π(R3−r3)
                                     =23×227×[(1.01)3−(1)3]m3
                                            =4421×(1.030301−1)m3
                                            =(4421×0.030301)m3
                                            =0.06348m3(approx)

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        Q.7      Find the volume of a sphere whose surface area is 154 cm2.
        Sol.

        Let r cm be the radius of the sphere.
        So, surface area = 154cm2
        ⇒ 4πr2=154
        ⇒ 4×227×r2=154
        ⇒ r2=154×74×22=12.25
        ⇒ r=12.25−−−−√=3.5cm
        Now , Volume =43πr3
                                 =(43×227×3.5×3.5×3.5)cm3
                                 =5393cm3=17923cm3


        Q.8      A dome of a building is in the form of a hemisphere. From inside it was white – washed at the cost of Rs 498.96. If the cost of white- washing is Rs.2.00 per square metre, find the
                       (i) Inside surface area of the dome,
                       (ii) Volume of the air inside the dome.
        Sol.

        (i) Inside surface area of the dome =TotalcostofwhitewashingRateofwhitewashing
                                                                     = (498.962.00)m2=249.48m2

        (ii) Let r be the radius of the dome.
        Therefore Surface area =2πr2
        ⇒ 2×227×r2=249.48
        ⇒ r2=249.48×72×22=39.69
        ⇒ r=39.69−−−−√=6.3m
        Volume of the air inside the dome = Volume of the dome
                                                                 =23πr3=23×227×6.3×6.3×6.3m3
                                                                   =523.9m3(approx)

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        Q.9     Twenty seven solid iron spheres , each of radius r and surface area S are melted to form a sphere with surface area S’. Find the
                        (i) Radius r’ of the new sphere,
                        (ii) Ratio of S and S’.
        Sol.

        (i) Volume of 27 solid sphere of radius r=27×43πr3 … (1)
        Volume of the new sphere of radius r′=43πr3 … (2)
        According to the problem, we have
        43πr′3=27×43πr3
        ⇒ r′3=27r3=(3r)3
        Therefore r′=3r

        (ii) Required ratio =SS′=4πr24πr′2=r2(3r)2
                                           =r29r2=19=1:9


        Q.10    A capsule of medicine is in the shape of a sphere of diameter 3.5 mm. How much medicine (inmm3) is needed to fill this capsule?
        Sol.

        Diameter of the spherical capsule = 3.5 mm
        Radius =3.52mm
                     =1.75mm
        Medicine needed for its filling = Volume of spherical capsule
                                                             =43πr3
                                                             =(43×227×1.75×1.75×1.75)mm3
                                                             =22.46mm3(approx)

         

        Q.1     A wooden bookshelf has external dimensions as follows : Height = 110 cm , Depth = 25 cm, Breadth = 85 cm (see figure). The thickness of the plank is 5 cm everywhere. The external faces are to be polished and the inner faces are to be painted . If the rate  of polishing is 20 paise per cm2 and the rate of painting is 10 paise per cm2, find the total expenses required for polishing and painting the surface of the bookshelf.

        6
        Sol.

        Area to be polished  = (110 × 85 + 2 × 85 × 25 + 2 × 25 × 110 + 4 × 75 × 5 + 2 × 110 × 5)cm2
                                            = (9350 + 4250 + 5500 + 1500 + 1100)cm2
                                            = 21700 cm2
        Cost of polishing @ 20 paise per cm2
        =(21700×20100)=4340
        Area to be painted = (6 × 75 × 20 + 2 × 90 × 20 + 75 × 90)cm2
                                        = (9000 + 3600 + 6750) cm2 = 19350 cm2
        Cost of painting @ 10 paise per cm2 =(19350×10100)=Rs1935
        Therefore total expenses =Rs (4340 + 1935) =Rs  6275


        Q.2     The front compound wall of a house is decorated by wooden spheres of diameter 21 cm, placed on small supports as shown in figure. Eight such spheres are used for this purpose, and are to be painted silver. Each support is a cylinder of radius 1.5 cm and height 7 cm and is to be painted black. Find the cost of paint required if silver paint costs 25 paise per cm2 and black paint costs 5 paise per cm2.

        7
        Sol.

        Clearly, we have to subtract the part of the sphere that is resting on the sphere while calculating the cost of silver paint.
        Surface area to be silver painted = 8 (Curved surface area of the sphere – area of circle on which sphere is resting).
        =8(4πR2−πr2)cm2        [where R=212cm,r=1.5cm]
        =8π(4×4414−2.25)cm2
        =8π(441−2.25)cm2
        =8π(438.75)cm2
        Therefore Cost of silver paint @ 25 paise per cm2
        =Rs(8×227×438.75×25100)
        =Rs(193057)
        =Rs2757.86(approx)
        Surface area to be black painted = 8 × curved area of cylinder.
        =8×2πrh
        =8×2×227×1.5×7cm2
        =528cm2
        Cost of black paint @ 5 paise per cm2

        8
        =Rs(528×5100)
        =Rs26.40

        Total cost of painting =Rs (2757.86 + 26.40) =Rs 2784.26 (approx)


        Q.3     The diameter of a sphere is decreased by 25%, By what percent does its curved surface area decrease?
        Sol.

        Let d be the diameter of the sphere . Then it’s surface area.
        =4π(d2)2=πd2
        On decreasing its diameter by 25% ,the new diameter,
        d1=(75100×d)=3d4
        Therefore New surface area =4π(d12)2
        =4π(12×3d4)2
        =4π9d264
        =πd2916
        Decrease in surface area = πd2−πd2.916
        =πd2(1−916)
        =πd2(716)
        Therefore Percentage decrease in surface area = (DecreaseinsurfaceareaInitialsurfacearea×100)%
        =(πd2×716×1πd2×100)%
        =(70016)%=43.75%

        Prev Chapter Notes – Surface Area ad Volume of A Sphere
        Next R D Sharma Solutions Surface Areas And Volumes of A cuboid and A cube

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