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Mathematics Chapter 14 Statistics

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    NCERT Solution For Class 10 Mathematics

    Statistics Here is the CBSE Mathematics Chapter 14 for Class 10 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 10 Mathematics Statistics Chapter 14 NCERT Solutions for Class 10 Mathematics Statistics Chapter 14 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 10 Mathematics.

    Question 1
    CBSEENMA10008805
    Wired Faculty App

    2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.

    Solution

    Let ‘a’ be the length of a side of a cube. Then
    Volume of one cube = 64 cm3
    ⇒    n3 = 65 cm3
    ⇒    n = 4 cm
    On joining the cubes, a cuboid is formed.
    Then The length of the resulting cuboid (l) = 2
    The breadth of the resulting cuboid (b) = a cm
    The thickness of the resulting cuboid (h) = a cm
    Now,
    Surface area of the resulting cuboid
    = 2(lb + bh + hl)
    = 2 (2a. a + a. a + a. 2a)
    = 2 (2a2 + a2 + 2a2)
    = 2 (5a2) = 10a2 = 10 (4)2
    = 160 cm2.

    Question 2
    CBSEENMA10008806
    Wired Faculty App

    A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner curved surface area of the vessel.

    Solution

    Let r cm be the radius of the cylinder and h cm be the height of the cylinder, then
    r = 7 cm,
    and    h = (13–7) cm
    = 6 cm.
    Let r1 cm be the radius of the hemisphere, then
    r1 = 7 cm
    Now,
    the inner curved surface area of the vessel
    = C,S.A of hemisphere
    + C,S.A of cylinder
    = (2πr12 + 2 π rh) cm2
    = (2 π r2 + 2 π rh) cm2 [∵ r1 = r]
    = [2 π r (r + h)] cm2

    equals open square brackets open parentheses 2 space straight x space 22 over 7 space straight x space 7 close parentheses open parentheses 7 space plus space 6 close parentheses close square brackets space cm squared

    = (44 x 13) cm2
    = 572 cm2.

    Question 3
    CBSEENMA10008807
    Wired Faculty App

    A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

    Solution
    Let r cm be the radius, h cm be the height and l cm be the slant height of the cone, then

    r = 3.5 cm,
    h = (15.5 – 3.5) cm = 12 cm.
    Now,   l = square root of straight r squared plus straight h squared end root rightwards double arrow space space straight l space equals space square root of left parenthesis 3.5 right parenthesis squared plus left parenthesis 12 right parenthesis squared end root space space space space space space space equals space square root of 12.25 plus 144 end root space space space space space space space equals space square root of 156.25 end root space space space space space space space equals space 12.5 space cm. space space space space space space space space

     Let r1 cm be the radius of the hemisphere.
    Then, r1 = 3.5 cm    [∵ r = r1]
    Now,
    The total surface area of the toy
    = CSA of hemisphere
    + CSA of cone
    = 2 π r12 + πrl
    = 2π r2 + πrl    [ ∵ r1= r]
    = π r [2r + l]
    equals space 22 over 7 space straight x space 3.5 space left square bracket 2 space straight x space 3.5 space plus space 12.5 right square bracket equals space space left square bracket 11 space left parenthesis 7 space plus 12.5 right parenthesis right square bracket space cm squared equals space space left square bracket space 11 space straight x space 19.5 right square bracket space cm squared space equals space 214.5 space cm squared
    Question 4
    CBSEENMA10008808
    Wired Faculty App

    A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

    Solution

    Let a be the length of an edge of the cube. Then
    a = 7 cm
    Greatest diameter of the hemisphere
    = Length of an edge of the cube
    = 7 cm
    Now,
    Surface area of the cube
    = 6 (edge)2
    = 6 x 72
    = 6 x 49 = 294 cm2
    Let r be the radius of the hemisphere.
    Then,       r = 7 over 2 space cm
    Now,  
    Curbed surface area of hemisphere
    equals space 2 πr squared equals space open parentheses 2 space straight x space 22 over 7 straight x 7 over 2 straight x 7 over 2 close parentheses equals space 77 space cm squared
    And,  Base area = πr squared equals open parentheses 22 over 7 straight x 7 over 2 straight x 7 over 2 close parentheses space cm squared
                                  space space space equals space space 77 over 2 space cm squared
    Total surface area
    = Surface area of the cube + curved surface area of the hemisphere – base area of the hemisphere
    equals space open parentheses 294 space plus space 77 minus 77 over 2 close parentheses space cm squared equals space open parentheses 294 plus 77 over 2 close parentheses space cm squared equals space left parenthesis 294 plus 38.5 right parenthesis space cm squared equals space 332.5 space cm squared

    Question 5
    CBSEENMA10008809
    Wired Faculty App

    A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

    Solution

    It is given that,
    Edge of the cube = l
    Then, Surface area = 6 (edge)2 = 6l2
    Now, the greatest diameter of hemisphere
    = length of an edge of the cube
    = l
    So, curved surface area of the hemisphere
                      
                       equals space 2 space straight pi space open parentheses 1 half close parentheses squared equals space 2 space straight pi space straight l squared over 4 equals πl squared over 2
    And, Area of base of the hemisphere
                   
                        equals straight pi space open parentheses l over 2 close parentheses squared equals space straight pi l to the power of italic 2 over 4

    Required surface area
    = Surface area of the cubical wooden block – area of the base of the hemisphere + curved surface area of the hemisphere
    equals space 6 l to the power of italic 2 italic minus pi over italic 4 l to the power of italic 2 italic equals italic 6 l to the power of italic 2 italic plus pi over italic 4 l to the power of italic 2

    Question 6
    CBSEENMA10008810
    Wired Faculty App

    A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig. 13.10). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.
    Solution
    Let r mm be radius of the hemisphere, then
    straight r space equals space 5 over 2 equals space 2.5 space mm
    Let r mm be the radius and h mm be the height of the cylinder, then
    r1 = 2.5 mm
    and    h = (14 – 5) mm = 9 mm
    Now,
    Surface Area of Capsule
    = CSA of cylinder + CSA of two
    hemispherical ends = 2πr1h + 2(2πr2)
    = (2πr1h + 4πr2) mm2
    = (2πrh + 4πr2) mm2 [∵ r1 = r]
    = [2πr (h + 2r)]mm2
    equals space open square brackets open parentheses 2 space straight x space 22 over 7 space straight x space 25 close parentheses space left parenthesis 9 plus 5 right parenthesis close square brackets space mm squared space equals space open parentheses 44 over 7 space straight x space 25 space straight x space 14 close parentheses space mm squared equals space left parenthesis 44 space straight x space 2.5 space straight x space 2 right parenthesis space mm squared equals space 220 space mm squared
    Question 7
    CBSEENMA10008811
    Wired Faculty App

    A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m2. (Note that the base of the tent will not be covered with canvas.)
    Solution

    Let r m be the radius and l m be the slant height of the cone, then
    r = 2 m, and l = 2.8 m
    Let r1 m be the radius and h m be the height of the cylinder, then
    Now, r1 – 2 m and h = 2.1 m
    Area of the canvas used
    = CSA of cone + CSA of cylinder
    = πrl + 2 πr1h
    = πrl + 2πrh [r = r1]
    = π r (l + 2h) m2 
    equals space open square brackets 22 over 7 space straight x space 2 space open curly brackets 2.8 space plus space 2 space left parenthesis 2.1 right parenthesis close curly brackets close square brackets space straight m squared equals space open square brackets 44 over 7 left parenthesis 2.8 plus 4.2 close square brackets space straight m squared equals space open parentheses 44 over 7 space straight x space 7 close parentheses space straight m squared equals space 44 space cm squared

    and Cost of canvas of the tent
    = Rs. (500 x 44)
    = Rs. 22,000.

    Question 8
    CBSEENMA10008812
    Wired Faculty App

    From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
    Solution

    Let r cm be the radius and h cm be the height of the cylinder, then
    r = 0.7 cm and h = 2.4 cm.
    Let r1 cm be the radius, l cm be the slant height and h1 cm be the height of the cone, then
    r1 = 0.7 cm and h1 = 2.4 cm
    Now comma space space l italic space equals space square root of straight r subscript 1 superscript 2 plus straight h subscript 1 squared end root space space space space space space space space space space space equals space square root of left parenthesis 0.7 right parenthesis squared plus left parenthesis 2.4 right parenthesis squared end root space space space space space space space space space space space equals space square root of 0.49 plus 5.76 end root space space space space space space space space space space space equals square root of 6.25 end root space equals space 2.5 space space cm

    Now,
    Total surface area of the remaining solid
    = (C.S.A of cylinder) + (C.S.A of cone) + (area of upper base of the cylinder)
    = 2 π rh + πr2l + πr2
    = 2 π rh + π rl + π r2 [ ∵ r = r1]
    equals space open parentheses 2 space straight x space 22 over 7 space straight x space 0.7 space straight x space 2.4 space plus 22 over 7 straight x space 0.7 space straight x space 2.5 space plus space 22 over 7 straight x space 0.7 space straight x space 0.7 close parentheses space space cm squared equals space left parenthesis 1.56 space plus space 5.5 space plus 1.54 right parenthesis space cm squared equals space 17.6 space cm squared

     

    Question 9
    CBSEENMA10008813
    Wired Faculty App

    A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 13.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
    Solution

    Let r cm be the radius and h cm be the height of cylinder, then
    r = 3.5 cm
    and
    h = 10 cm.
    Let r1 be the radius of the hemisphere, then r1 = 3.5 cm.
    Now,
    Total Surface area of the article
    = Curved Surface Area of the cylinder + 2 (Curved Surface area of hemisphere)
    = 2 π rh + 2(2 π r12)
    = 2 π rh + 4 π r 2
    = 2 π rh + 4 π r2    [∵ r = r1]
    = [2 π r (h + 2r)] cm2
    equals space open square brackets open parentheses 2 space straight x space 22 over 7 straight x space 3.5 close parentheses left parenthesis 10 plus 1 space straight x space 3.5 close square brackets space cm squared equals space open parentheses 2 space straight x space 22 over 7 straight x space 3.5 space straight x space 17 close parentheses space cm squared equals space 374 space cm squared

    Question 10
    CBSEENMA10008814
    Wired Faculty App

    A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

    Solution

    Let r cm be the radius of the hemisphere, then r = 1 cm.
    Let R be the radius of the cone and h cm be the height. Then

    R = 1 cm and h = 1 cm
    [It is given that R = h)
    Now, Volume of solid
    = Volume of hemisphere + volume of cone
    equals space 2 over 3 straight pi space straight r cubed plus 1 third space πR squared straight h equals space 2 over 3 straight pi space straight r cubed plus 1 third space straight pi space straight r squared straight h space space space space space left square bracket because space straight r space equals space straight R right square bracket equals space space 2 over 3 straight pi space left parenthesis 1 right parenthesis cubed plus 1 third straight pi left parenthesis 1 right parenthesis squared left parenthesis 1 right parenthesis equals space space 2 over 3 straight pi plus 1 third straight pi space equals space fraction numerator 3 straight pi over denominator 3 end fraction equals straight pi space cm cubed

    Question 11
    CBSEENMA10008815
    Wired Faculty App

    Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)
    Solution

    Let r cm be the radius and h cm be the height of a cone, then
    r = 1.5 cm, and h = 2 cm,
    Now, Volume of conical part

    Let r cm be the radius and h cm be the height of the cylindrical part then r1 = 1.5 cm, h1= 8 cm
    Now, Volume of cylindrical part
    = πr12 h’
    = (π x 1.5 x 1.5 x 8) cm3 = 18 π cm3
    Hence, the volume of air contained in the model that Rachel made
    = Volume of two conical part + Volume of cylindrical part
    = (2 x 1.5 π + 18 π) cm3
    = (3 π + 18 π) cm3
    = (21 π) cm3
    equals open parentheses 21 space straight x space fraction numerator 2 minus 2 over denominator 7 end fraction close parentheses space cm cubed equals space 66 space cm cubed

    Question 12
    CBSEENMA10008816
    Wired Faculty App

    A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see Fig. 13.15).



    Fig. 13.17.
    Solution
    Let r cm be the radius of hemispherical part, then
    straight r space equals space fraction numerator 2.8 over denominator 2 end fraction space equals space 1.4 space cm

    Now,
    Volume of hemispherical part
    equals space open parentheses 2 over 3 straight pi space straight x space 1.4 space straight x space 1.4 space straight x space 1.4 close parentheses space cm cubed equals space open parentheses fraction numerator 5.488 over denominator 3 end fraction straight pi close parentheses space cm cubed equals space 1.892 space straight pi space cm cubed.
    Let R cm be the radius and H cm be the height of cylinndrical part, then
    straight R equals 28 over 2 equals space 1.4 space cm

    and    H = 5 cm – (2 x 1.4) cm = 2.2 cm

    Now,
    Volume of cylindrical part
    = πR2H = (πx1.4 x 1.4 x 2.2)cm3
    = 4.312 π cm3.
    Now,
    Volume of each Gulab Jamun
    = Volume of cylindrical + 2 (Volume of hemispherical part)
    = (4.312 π + 2(1.829) π] cm3
    = (4.312 π + 3.658 π) cm = 7.97 π
    equals space 7.97 space straight x space 22 over 7 space equals space space 25.05 space cm cubed

    Hence,
    Volume of syrup found in 45 Gulab Jamuns
    = 45 x 30% of volume of each Gulab Jamun
    equals space open square brackets 45 space straight x space 30 over 100 space straight x space 25 space.05 close square brackets space cm cubed equals space 338.175 space cm cubed space left parenthesis Approx. right parenthesis

     

    Question 13
    CBSEENMA10008817
    Wired Faculty App

    A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand (see the given Fig.).
    Solution

    Let l, b and h be respectively the length, breadth and height of a cuboid, then
    l = 15 cm
    b = 10 cm
    and    h = 3.5 cm
    Volume of cuboid
    = (l x b x h) cm3 = (15 x 10 x 3.5) cm3
    = 525 cm3
    Let r cm be the radius and h cm be the height of conical part, the
    r = 0.5 cm, h = 1.4 cm
    Now, 
       Volume = V o l u m e space space equals space 1 third space πr squared straight h equals space open parentheses 1 third straight x 22 over 7 straight x space 0.5 space straight x space 0.5 space straight x space 1.4 space close parentheses space cm cubed space equals space open parentheses fraction numerator 7.7 over denominator 21 end fraction close parentheses space cm cubed
    Hence, the volume of wood in the entire stand
    = Volume of cuboid - 4 (Volume of cone)
    equals space open parentheses 252 space minus space 4 space straight x space 737 over 21 close parentheses space cm cubed equals space left parenthesis 525 space minus 1.466 right parenthesis space cm cubed equals space 523.534 space cm cubed.

    Question 15
    CBSEENMA10008819
    Wired Faculty App

    A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm3 of iron has approximately 8g mass. (Use π = 3.14)
    Solution

    Let h cm be the height and r cm be the radius of the first cylinder. Then

    h = 220 cm and r = 24 over 2 = 12 cm
    Now, Volume = πr2 h
    = (π x 12 x 12 x 220) cm3
    = 31680 π
    Let H cm be the height and R cm be the radius of the second cylinder.  ThenH = 60 cm and R = 8 cm
    Now, Volume = πR squared straight H
    open parentheses straight pi space straight x space 8 space straight x space 8 space straight x space 60 close parentheses space c m cubed equals space 3840 space straight pi
    Total volume of the pole 
     = 31680 straight pi + 3840 straight pi
     = 35520 straight pi = (35520 x 3.14) cm3
     = 111532.8 cm3 
    Now, Required weight
        = open parentheses fraction numerator 111532.8 space straight x space 8 over denominator 1000 end fraction close parentheses space kg equals space 892.26 space kg 

    Question 16
    CBSEENMA10008820
    Wired Faculty App

    A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.
    Solution
    Let r cm be the radius and h cm be the height of a cone, then
                   r = 60 cm,  h = 120 cm
    Volume of cone  = 1 third space πr squared straight h
                            = open parentheses 1 third straight pi space straight x space 60 space straight x space 60 space straight x space 120 close parentheses space c m cubed equals space 144000 space straight pi space cm cubed space
    Let r1 cm be the radius of a hemisphere then
                   r1 = 60 cm
    Now,
        Volume of hemisphere = 2 over 3 πr cubed
                           equals space open parentheses 2 over 3 straight pi space straight x space 60 space straight x space 60 space straight x space 60 close parentheses space cm cubed equals space 144000 space straight pi space cm cubed

    Let R cm be the radius and H cm be the height of a cyclinder, then
    R = 60 cm, H = 180 cm
    Now,
    Volume of Cylinder = πR2H
    = (π x 60 x 60 x 180) cm3
    = (π x 64800) cm3
    = 648000 π cm3.
    Volume of Solid
    = Volume of cone + volume of hemisphere
    = (144000π + 144000π) cm3
    = 288000π cm3 Hence,
    Volume of water left in the cylinder
    = Volume of cylinder – Volume of solid
    = 1648000 π – 288000 π) cm3
    = 360000 π cm3
    equals space open parentheses 360000 space straight x space 22 over 7 close parentheses space cm cubed equals space 7920000 over 7000000 space equals space 1.131 space straight m cubed

    Question 17
    CBSEENMA10008821
    Wired Faculty App

    A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm3. Check whether she is correct, taking the above as the inside measurements and π = 3.14.

    Solution

    Let r cm be the radius of the spherical glass. Then
                             straight r space space equals space fraction numerator 8.5 over denominator 2 end fraction space cm
    Now,  
           Volume  = 4 over 3 space πr cubed
    equals space open parentheses 4 over 3 straight x space 3.14 space straight x space fraction numerator 8.5 over denominator 2 end fraction straight x fraction numerator 8.5 over denominator 2 end fraction fraction numerator 8.5 over denominator 2 end fraction space close parentheses space cm cubed equals space open parentheses fraction numerator 7713.41 over denominator 24 end fraction close parentheses space cm cubed equals space 321.4 space cm cubed
    Let R cm be the radius and h cm be the height of cylindrical part.
    Then,      R  = 2 over 2 = 1 cm,  h = 8 cm
    Volume = πR squared straight h
                = (3.14 x 1 x 1 x 8) cm3
                = (25.12) cm3
    Quantity of water = (321.4 + 25.12) cm3
                                      = 346.52 cm3
    Hence, Answer is not correct. 

     

    Sponsor Area

    Question 18
    CBSEENMA10008822
    Wired Faculty App

    A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.
    Solution

    Let r cm be the radius of the sphere. Then, r = 4.2 cm

    Now,  
    Volume = 4 over 3 space πr cubed
                
                equals open square brackets 4 over 3 space straight x space straight pi space straight x left parenthesis 4.2 right parenthesis space straight x space left parenthesis 4.2 right parenthesis space left parenthesis 4.2 right parenthesis close square brackets space cm cubed

    Let R cm be the radius and h cm be the height at the cylinder. Then
    R = 6 cm and h = ?
    Now, Volume = πR2h
    = (π x 6 x 6 x h) cm3
    ∵ Since sphere is recast into the shape of a cylinder. So, volume remains same
    bold italic i bold. bold italic e bold. bold space bold space bold space bold 4 over bold 3 bold space bold πr to the power of bold 3 bold space bold equals bold space bold πR to the power of bold 2 bold h
    rightwards double arrow space 4 over 3 straight pi space straight x space 4.2 space straight x space 4.2 space straight x space 4.2 space equals space straight pi space straight x space 6 space straight x space 6 space straight x space straight h space space space space space space space space space space space space space space straight h space equals space fraction numerator 4 space straight x space 4.2 space straight x space 4.2 space straight x space 4.2 over denominator 6 space straight x space 6 space straight x space 3 end fraction space space space space space space space space space space space space space space space space equals space fraction numerator 296.352 over denominator 108 end fraction equals space 2.74 space cm

    Question 19
    CBSEENMA10008823
    Wired Faculty App

    Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
    Solution

    Let r1 r2, and r3 be the radius of metallic spheres, then r1 = 6 cm, r2 = 8 cm, r3 = 10 cm.
    Let R cm be the radius of a single solid sphere.
    Since, three metallic spheres are formed from a single solid sphere, so their volumes are equal.
    rightwards double arrow space 4 over 3 πr subscript 1 cubed plus 4 over 3 πr subscript 2 cubed plus 4 over 3 πr subscript 3 cubed space equals space 4 over 3 πR cubed rightwards double arrow space 4 over 3 straight pi left parenthesis straight r subscript 1 cubed plus straight r subscript 2 cubed plus straight r subscript 3 cubed right parenthesis space equals space 4 over 3 straight pi left parenthesis straight R right parenthesis cubed rightwards double arrow space space straight r subscript 1 cubed plus straight r subscript 2 cubed plus straight r subscript 3 cubed equals straight R cubed rightwards double arrow space 6 cubed plus 8 cubed plus 10 cubed equals straight R cubed rightwards double arrow space space 216 space plus 512 space plus space 1000 space equals space straight R cubed rightwards double arrow space space straight R cubed space equals space 1728 rightwards double arrow space space space straight R space equals space 12
    Hence, radius of sphere = 12 cm

    Question 20
    CBSEENMA10008824
    Wired Faculty App

    A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.

    Solution
    r m be the radius and h m be the height of the will
    Then,        r = 7 over 2 m and h = 20 m
    Now, 
    Volume of earth dug out = πr squared straight h
                         equals space open parentheses 22 over 7 straight x 7 over 2 straight x 7 over 2 straight x 20 close parentheses space straight m cubed
         
                          = 770 m3
    Area of embankment
                               = 22 m x 14 m
                               = 308 m2
    therefore  Height of the embankment
                    equals space fraction numerator Volume space of space the space earth space dug space out over denominator Area space of space the space embankment end fraction equals 770 over 308 equals space 2.5 space m
                      
    Question 22
    CBSEENMA10008826
    Wired Faculty App

    A container shaped like a right circular cylinder having diameter 12 cm and height 15 cm is full of ice cream. The ice cream is to be filled into cones of height 12 cm and diameter 6 cm, having a hemispherical shape on the top. Find the number of such cones which can be filled with ice cream.
    Solution

    Let R cm be the radius and H cm be the height of a container, then

    R space equals space 12 over 2 = 6 cm and H = 15 cm
    Therefore, volume of cylindrical container
                      equals space πR squared straight H equals space left parenthesis straight pi space straight x space 6 space straight x space 6 space straight x space 15 right parenthesis space cm cubed equals space 540 space straight pi space cm cubed
    Let r1 cm be the radius and h cm be the height of a cone, then
     
                straight r subscript 1 equals 6 over 2 equals 3 space cm space and space straight h space equals space 12 space cm comma
    Therefore, Volume of conical part
                     equals space 1 third space straight pi space straight r subscript 1 squared straight h equals space open parentheses 1 third πx 3 space straight x space 3 space straight x space 12 close parentheses space cm cubed equals space 36 space straight pi space space cm cubed
    Let r2 cm  be the radius of hemispherical part then  straight r subscript 2 equals 3 space cm comma space space space left square bracket because space straight r subscript 1 equals straight r subscript 2 right square bracket
    Therefore, Volume of Hemispherical part
                   equals space 2 over 3 space straight pi space straight r subscript 1 squared straight h equals space open parentheses 2 over 3 straight pi space straight x space 3 space straight x space 3 space straight x space 3 close parentheses space cm cubed equals space 18 space straight pi space space cm cubed
    Now, Volume of ice-creme cone with hemispherical top
                           =  Volume of cone + Volume of Hemisphere
                           = (36 straight pi + 18 straight pi) cm3
                           = 54 straight pi cm3 
    Therefore,
      the required no. of such cones
                        equals space fraction numerator Volume space of space cylindrical space container over denominator Volume space of space cone space with space hemispherical space top end fraction equals space fraction numerator 540 straight pi over denominator 54 straight pi end fraction space equals space 10
    Question 23
    CBSEENMA10008827
    Wired Faculty App

    How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cm × 10 cm × 3.5 cm?
    Solution
    Let r cm be the radius and h cm be the thickness of the silver coin (cylindrical in shape). Then,

    straight r space equals space fraction numerator 1.75 over denominator 2 end fraction space cm space space and space straight h space equals space fraction numerator 2 space mm over denominator 10 end fraction space equals space 0.2 space cm
    Now,      Volume = πr squared straight h
        equals space straight pi open parentheses fraction numerator 1.75 over denominator 2 end fraction close parentheses open parentheses fraction numerator 1.75 over denominator 2 end fraction close parentheses space straight x space 0.2 space equals open parentheses fraction numerator straight pi space straight x space 0.6125 over denominator 4 end fraction close parentheses space cm cubed equals space 0.153125 space straight pi equals space 0.153125 space straight x space 22 over 7 space equals space fraction numerator 3.36875 over denominator 7 end fraction equals 0.478125 space cm cubed
    Let l, b and h are respectively the length, breadth and height of the cuboid. Then
    l = 5.5 cm, b = 10 cm and h = 3.5 cm
    Now, Volume = l x b x h
    = (5.5 x 10 x 3.5) cm3 = 192.5 cm3
    Hence, Required number of silver coins
    equals fraction numerator 192.5 over denominator 0.48125 end fraction equals 400
    Question 24
    CBSEENMA10008828
    Wired Faculty App

    A cylindrical bucket, 32 cm high and with radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the conical heap is 24 cm, find the radius and slant height of the heap.

    Solution

    Let r cm be the radius and l cm be the slant height of the bucket (cylindrical). Then
    r = 18 cm and h = 32
    Now, Volume = πr2h
    = π (18)2 (32)
    = (π x 18 x 18 x 32) cm3
    Let r cm be the radius and h cm be the height of the conical heap. Then
    r = ? and h = 24 cm
    Now, Volume  = 1 third space πr squared straight h
                    equals space 1 third space straight pi space straight x space straight r squared space straight x space 24 equals space left parenthesis 8 πr squared right parenthesis space cm cubed.
    Since sand of bucket is emptied on the ground and a conical heap of sand is formed. So, volume remains same
    i.e.      straight pi x 18 x 18 x 32 = 8 πr squared
    rightwards double arrow space space space space space space space space space space space space space straight r squared space equals space fraction numerator 18 space straight x space 18 space straight x space 32 over denominator 8 end fraction space space space space space space space space space space space space space space space space space space space equals space 10368 over 8 equals 1296 rightwards double arrow space space space space space straight r space equals space 36 space cm.

    Question 25
    CBSEENMA10008829
    Wired Faculty App

    Water in a canal, 6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much area will it irrigate in 30 minutes, if 8 cm of standing water is needed?

    Solution

    We have,
    Width of the canal = 6 m
    Depth of the canal = 1.5 m
    It is given that the water is flowing with velocity 10 km/hr. Therefore, length of the water column formed in 1 half hr (30 minutes) = 5  km = 5000 m.

    ∴ Volume of the water flowing in 30 minutes = volume of cuboid at of length 5000 m, width 6 m and depth 1.5 m.
    ⇒ Volume of the water flowing in 30 minutes
    = (5000 x 6 x 1.5) m3
    = 45000 m3

    Suppose x m2 area is irrigated in space space 1 half hr.
    Then,  x x 8 over 100 = 45000
    rightwards double arrow space space space space space space space space space space space space space space space space space space space space straight x space space equals space space fraction numerator 45000 space straight x space 100 over denominator 8 end fraction rightwards double arrow space space space space space space space space space space space space space space space space space space space space straight x space space equals space 562500
    Hence, the canal irrigates 562500 m2 area in 1 half hour.

    Question 26
    CBSEENMA10008830
    Wired Faculty App

    A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in her field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 3 km/h, in how much time will the tank be filled?
    Solution
    Let r m be the radius and h m be the heignt of the cylindrical tank.
    Then,  r = 10 over 2 space equals 5 space cm space and space straight h space equals space 2 space straight m
    Now, volume of cylindrical tank
                                     equals space space πr squared straight h equals space space space straight pi space straight x space 5 squared space straight x space 2 space equals space 50 space straight pi space straight m cubed

    Now, Volume of the water that flows through the pipe t hours
    = Volume of cylinder of radius of 10 cm and length (3000f m)
    equals space open square brackets straight pi space straight x open parentheses 1 over 10 close parentheses squared straight x space 3000 straight t close square brackets straight m cubed space equals space 30 πt space straight m cubed therefore space space space space 50 space straight pi space straight m cubed space space equals space 30 space straight pi space straight t space straight m cubed rightwards double arrow space space space space straight t space equals space 5 over 3 space hour space space space space space space space space space space
      = 1  hour  40 minutes

    Question 27
    CBSEENMA10008831
    Wired Faculty App

    A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
    Solution
    Let R and r be the radii of bigger and smaller ends of the frustum and h be its height. Then,
    straight R space equals space 4 over 2 equals 2 space cm comma space space straight r space equals space 2 over 2 equals 1 space cm space and space straight h space equals space 14 space cm Now comma space volume space equals space πh over 3 space left square bracket straight R squared plus straight R. straight r space plus space straight r squared right square bracket space space space space space space space space space space space space space space space space space space space equals space 22 over 7 space straight x space 14 space left square bracket left parenthesis 2 right parenthesis squared space plus space 2 space straight x space 1 space plus space left parenthesis 1 right parenthesis squared right square bracket space space space space space space space space space space space space
                     = 44 (4 + 2 + 1)
                     = 44 x 7 = 308 cm3 
    Question 28
    CBSEENMA10008832
    Wired Faculty App

    The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum. 
    Solution

    Let R and r be the radii of bigger and smaller ends of the frustum, and l be the slant height, then l = 4cm.
    Perimeter of bigger end = 18 cm
    ⇒    2 πR = 18
    ⇒    π R = 9 cm
    Perimeter of smaller end = 6 cm
    ⇒    2 πr = 6
    ⇒    πr = 3 cm
    Now,
    Curved Surface of the fustum
    = πRl + πrl
    = 1 (πR + πr)
    = 4 (9 + 3)
    = 4 x 12
    = 48 cm2.

    Question 29
    CBSEENMA10008833
    Wired Faculty App

    A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig. 13.24). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it. 
    Solution
    Let R and r be respectively the radii of bigger and smaller ends of the frustum and l be the slant height then.

    R = 10 cm; r = 4 cm and l = 15
    Now,
    Curved Surface Area
    = π Rl + πrl
    = πl (R + r)
    equals space 22 over 7 straight x space 15 space left parenthesis 10 space plus 4 right parenthesis space cm squared equals space open parentheses 22 over 7 straight x space 15 space straight x space 14 close parentheses space cm squared equals space 22 space straight x space 30 space equals space 660 space cm squared
    and area of the top of the cap
    equals space πr squared equals space 22 over 7 straight x space 16 equals space 352 over 7 space cm squared

    Thus,
    Total area of the material used for making the cap
    equals space open parentheses 660 plus 352 over 7 close parentheses space cm squared equals space 710 2 over 7 space cm squared

    Question 30
    CBSEENMA10008834
    Wired Faculty App

    A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of ` 20 per litre. Also find the cost of metal sheet used to make the container, if it costs ` 8 per 100 cm2 . (Take π = 3.14)
    Solution

    Let R and r be respectively the radii of bigger and smaller ends of the frustum, then
    R = 20 cm, r = 8 cm
    Let l and h be respectively the slant height and height of the frustum then
    h = 16 cm
    and space space space space space space l space equals space square root of straight h squared plus left parenthesis straight R space straight r right parenthesis squared end root space space space space space space space space space space space space space equals space square root of 16 squared plus left parenthesis 20 minus 8 right parenthesis squared end root space space space space space space space space space space space space space equals space square root of 256 space plus space 144 end root space space space space space space space space space space space space space equals space square root of 400 space equals space 20 space cm space
    Now,
    (i) Curved Surface area of frustum
    = πl (R + r)
    = 3.14 x 20 (20 + 8)
    = (3.14 x 20 x 28) cm2
    = 1758.4 cm2
    (ii) Total tin required
    = CSA. + area of base
    = 1758.4 + 3.14 x (8)2
    = 1758.4 + 200.96
    = 1959.36 cm2
    (iii) Cost of required tin
    equals space fraction numerator 1959.36 space straight x space 8 over denominator 100 end fraction space equals space space Rs. space 156.75
    (iv) Volume of frustum
    equals space πh over 3 space left square bracket space straight R squared plus straight R. space straight r space plus space straight r squared right square bracket equals space fraction numerator 22 space straight x space 16 over denominator 21 end fraction space left square bracket 400 space plus space 160 space plus space 64 right square bracket equals space fraction numerator 22 space straight x space 16 space over denominator 21 end fraction straight x space 624 equals space 10459.43 space cm cubed equals space 10.45943 space litres

    Now,
    Cost of the milk which can completely fill the container at the rate of 20 per litre
    = 10.46 x 20
    = Rs. 209.20
    Hence, cost of milk is Rs. 209 and cost of metal is Rs. 156.75.

    Question 31
    CBSEENMA10008835
    Wired Faculty App

    A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1 over 16 cm, find the length of the wire.
    Solution
    In fig. cone ABC is cut out by a plane parallel to the base FG. DEFG is the frustum so obtained. Let O be the centre of the base of the cone and O’ the centre of the base of the frustum.

    It is given that ∠BAC = 60° ∠OAC = 30°
    In right triangle AOC, tan 30 degree space equals space OC over OA
    rightwards double arrow space space Oc space space equals space OA space straight x space tan space 30 degree space space space space space space space space space space space space equals space 10 space straight x space fraction numerator 1 over denominator square root of 3 end fraction equals fraction numerator 10 over denominator square root of 3 end fraction space cm
    And, C =  equals fraction numerator 10 over denominator square root of 3 end fraction space cm. space space space space... left parenthesis straight i right parenthesis Since space space space increment AOC space minus space increment AO apostrophe straight F space space space space space space space space space space space space space space left parenthesis Using space AA space similar space condition right parenthesis therefore space space space fraction numerator AO over denominator AO apostrophe end fraction space equals space fraction numerator OC over denominator straight O apostrophe straight F end fraction rightwards double arrow space space space 10 over 20 equals fraction numerator begin display style fraction numerator 10 over denominator square root of 3 end fraction end style over denominator straight O apostrophe straight F end fraction rightwards double arrow space space space space straight O apostrophe straight F space equals space fraction numerator 20 over denominator square root of 3 end fraction space space space space space space space space space... left parenthesis ii right parenthesis
    Height of the frustum = P'O = 1 half space AO apostrophe space equals space 10 space cm
    therefore volume of the frustum = πh over 3 left square bracket straight R squared plus Rr plus straight r squared right square bracket
                                             equals fraction numerator straight pi space straight x space 10 over denominator 3 end fraction left square bracket OF squared plus straight O apostrophe straight F space straight x space PE space plus space PE squared right square bracket equals space fraction numerator straight pi space straight x space 10 over denominator 3 end fraction open square brackets open parentheses fraction numerator 20 over denominator square root of 3 end fraction close parentheses plus fraction numerator 20 over denominator square root of 3 end fraction straight x fraction numerator 10 over denominator square root of 3 end fraction plus open parentheses fraction numerator 10 over denominator square root of 3 end fraction close parentheses squared close square brackets space space space space space space space space space space space space space space space space space space left square bracket space Using space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis right square bracket equals space fraction numerator πx 10 over denominator 3 end fraction open square brackets fraction numerator 400 plus 200 plus 100 over denominator 3 end fraction close square brackets equals space fraction numerator straight pi space straight x space 10 space straight x space 700 over denominator 9 end fraction space cm cubed equals space fraction numerator straight pi space 7000 over denominator 9 end fraction space cm cubed space space space space space space space space space space space space space space space space space space space space space... left parenthesis iii right parenthesis space space space space space space space space space space space space space space space space space space space space
    Radius of the wire = 1 over 32 space cm
    Let h be length of the wire
    therefore         Volume of wire = straight p space straight x space open parentheses 1 over 32 close parentheses squared space straight x space straight h space space space space space space space space space space space space space space space space.... left parenthesis iv right parenthesis
    From (iii) and (iv), we get
    straight pi space straight x space open parentheses 1 over 32 close parentheses squared space straight x space straight h space equals space fraction numerator straight pi space straight x space 7000 over denominator 9 end fraction therefore space space space space space space space straight h space equals space fraction numerator straight pi space straight x space 7000 over denominator 9 end fraction space straight x space fraction numerator 32 space straight x space 32 space over denominator straight pi end fraction cm space space space space space space space space space space space space space space space space space space space space equals space 7168000 over 9 equals 796444.4 cm space left parenthesis approx right parenthesis space space space space space space space space space space space space space space space space space space space space space space space space equals space 7964.44 space straight m space space space space space space space space space space space space space space space space space space space space space space space space equals space 7964 space straight m. space left parenthesis approx right parenthesis space

    Question 32
    CBSEENMA10008836
    Wired Faculty App

    A copper wire, 3 mm in diameter, is wound about a cylinder whose length is 12 cm, and diameter 10 cm, so as to cover the curved surface of the cylinder. Find the length and mass of the wire, assuming the density of copper to be 8.88 g per cm3.
    Solution

    Length of the cylinder = 12 cm = 120 mm
    ∵ Number of rounds to cover 3 mm = 1
    ∴ Number of rounds to cover 120 mm
                      equals space 120 over 3 equals 40
    Let r cm be the radius of the cylinder, Then
                     equals space straight r space equals space 10 over 2 equals space 5 space cm
    therefore  Length of the wire in completing one round
                      = 2πr
                      = 2straight pi(5) = 10 straight pi cm
    therefore Length of the wire in completing the whole surface (40 rounds)
                      = 10 straight pi x 40 = 400 straight pi cm
    Radius of the copper wire = 3 over 2 space mm space equals space 3 over 20 space cm
    therefore Volume of wire = straight pi space open parentheses 3 over 20 close parentheses squared space left parenthesis 400 space straight pi right parenthesis space space space space space space space
                               =  9 straight pi squared space cm cubed
    therefore space space space Mass of wire  = 9 straight pi squared space straight x space 8.88
                               =  79.92 straight pi squared space 8 space straight m.
                                                                      

    Question 33
    CBSEENMA10008837
    Wired Faculty App

    A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.)
    Solution

    In right triangle CAB :
    BC2 = AB2 + AC2
    [Using Pythagoras theorem]
    ⇒ BC2 = 32 + 42
    ⇒ BC2 = 9 + 16 = 25
    ⇒    BC = 5 cm

    Now, in ΔAOB and ΔCAB :
    ∠AOB = ∠CAB (90°)
    ∠B = ∠B
    [Common]
    Therefore, by using A A similar condition
    increment AOB space minus space increment CAB OA over AC equals AB over CB rightwards double arrow space OA over 4 equals 3 over 5 rightwards double arrow OA equals 12 over 5 cm Similarly comma space space OB over AB equals AB over CB rightwards double arrow space space space space OB over 3 equals 3 over 5 rightwards double arrow OB equals 9 over 5 space cm therefore space space space space OC space equals space BC space minus space OB space space space space space space equals space 5 minus 9 over 5 equals fraction numerator 25 minus 9 over denominator 5 end fraction equals 16 over 5 space cm

    Now,
    Volume of the double cone so formed
    equals space 1 third straight pi open parentheses 12 over 5 close parentheses squared straight x 16 over 5 plus 1 third straight pi open parentheses 12 over 5 close parentheses squared straight x 9 over 5 equals space 1 third straight pi open parentheses 12 over 5 close parentheses squared open square brackets 16 over 5 plus 9 over 5 close square brackets equals 1 third straight pi open parentheses 12 over 5 close parentheses squared open square brackets 25 over 4 close square brackets equals open parentheses 1 third straight pi cross times 12 over 5 cross times 12 over 5 cross times 12 over 5 close parentheses space cm cubed equals space open parentheses 3600 over 375 straight pi close parentheses space cm cubed equals 9.6 space straight pi space cm cubed equals space left parenthesis 9.6 space straight x space 3.14 right parenthesis space cm cubed equals space 30.14 space cm cubed
    and Surface area of the double cone
    equals space straight pi space straight x space 12 over 5 cross times 3 plus straight pi cross times 12 over 5 cross times 4 equals straight pi cross times 12 over 5 left square bracket 3 plus 4 right square bracket equals space straight pi cross times 12 over 5 cross times 7 equals 84 over 5 straight pi equals space 84 over 5 cross times 3.14 equals 52.75 space cm squared.

    Question 34
    CBSEENMA10008838
    Wired Faculty App

    A cistern, internally measuring 150 cm × 120 cm × 110 cm, has 129600 cm3 of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water. How many bricks can be put in without overflowing the water, each brick being 22.5 cm × 7.5 cm × 6.5 cm?
    Solution

    Volume of water in the cistern
    = 129600 cm3
    Let l, b and h are the length, breadth and height of the cistern. Then
    l = 150 cm, b = 120 cm and h = 110 cm
    Now, Volume of cistern = l x b x h
    = 150 x 120 x 110 = 1980000 cm3
    ∴ Volume of cistern to be filled
    = (1980000 – 129600) cm3
    = 1850400 cm3
    Volume of one brick
    = (22.5 x 7.5 x 6.5) cm3
    = 1096.875 cm3
    Let the total number of bricks be x.
    then, water absored by x bricks
             equals space open parentheses straight x over 17 cross times 1096.875 close parentheses space cm cubed
    therefore  Volume of the water left in the cistem
            equals space open parentheses 129600 minus straight x over 17 straight x 1096.875 close parentheses space cm cubed

    Since, the cistern is filled upto the brim. Therefore,
    Volume of the cistern
    = Volume of the water left in the cistern + volume of the bricks
    rightwards double arrow space 1980000 space equals space open parentheses 129600 minus straight x over 17 straight x 1096.875 close parentheses space plus space straight x space cross times space 1096.875 rightwards double arrow space straight x space equals space 1792.410
    Hence, total no of bricks = 1792 (approx).

    Question 36
    CBSEENMA10008840
    Wired Faculty App

    An oil funnel made of tin sheet consists of a 10 cm long cylindrical portion attached to a frustum of a cone. If the total height is 22 cm, diameter of the cylindrical portion is 8 cm and the diameter of the top of the funnel is 18 cm, find the area of the tin sheet required to make the funnel (see the given Fig.
    Solution
    Let R and r be respectively the radii of the bigger and smaller ends of the frustum, then
    straight R space equals space 18 over 2 equals 9 space cm semicolon space space space straight r space equals space 8 over 2 equals 4 space cm
    Let l and h be respectively the slant height and height of frustum, then
    h = total height – height of cylindrical part
    = 22 cm – 10 cm
    = 12 cm
     a n d space space space space space space space space l italic space italic space italic space equals space space italic space square root of straight h squared plus left parenthesis straight R minus straight r right parenthesis squared end root italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space italic space equals space space space square root of left parenthesis 12 right parenthesis squared plus left parenthesis 9 minus 4 right parenthesis squared end root space space space space space space space space space space space space space space space space equals space space space space square root of 144 plus 25 end root space space space space space space space space space space space space space space space space equals space space space space square root of 169 space equals space 13 space cm

    Now,
    Curved surface area of frustum
    equals space rl space left parenthesis straight R space plus space straight r right parenthesis equals space 22 over 7 straight x space 13 space left parenthesis 9 plus 4 right parenthesis equals space space 22 over 7 straight x space 13 space straight x space 13 equals space 531.14 space cm squared

    Let r1 and h1 be respectively the radius and height of cylindrical part.
    Then, r1 = 4cm and h1 = 10cm
    Now,
    Curved Surface area of the cylinder
    = 2πr1h 1
    equals space open parentheses 2 space straight x space 22 over 7 straight x space 4 space straight x space 10 close parentheses space cm squared equals space 251.43 space cm squared

    Hence,
    Area of tin required
    = Curved surface area of frustum + curved surface area of cylinder
    = 531.14 + 251.43
    = 782.57 cm2.

     

    Question 37
    CBSEENMA10008841
    Wired Faculty App

    Derive the formula for the curved surface area and total surface area of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
    Solution

    Let h be the height, l the slant height and r1 and r2 the radii of the circular bases of the frustum ABB’ A’ shown in Fig. such that r1 > r2.
    Let the height of the cone VAB be h1 and its slant height be i.e., VO = h1 and VA = VB = l1
    ∴ VA’ = VA – AA’ = l1– l
    and VO’ = VO – OO’ = h1– h
    Here, ΔVOA ~ ΔVO‘A’
    therefore space space space fraction numerator VO over denominator VO apostrophe end fraction equals fraction numerator OA over denominator OA apostrophe end fraction equals fraction numerator VA over denominator VA apostrophe end fraction rightwards double arrow space space space fraction numerator straight h subscript 1 over denominator straight h subscript 1 minus straight h end fraction equals straight r over straight R equals fraction numerator straight l subscript 1 over denominator straight l subscript 1 minus 1 end fraction rightwards double arrow space space space space fraction numerator straight h subscript 1 minus straight h over denominator straight h subscript 1 end fraction equals straight F subscript 2 over straight F subscript 1 equals 1 minus 1 over straight l subscript 1 rightwards double arrow space space space straight h over straight h subscript 1 equals 1 minus straight r subscript 2 over straight r subscript 1 space and space straight l over straight l subscript 1 equals 1 minus straight r subscript 2 over straight r subscript 1 rightwards double arrow space space space space straight h over straight h subscript 1 equals fraction numerator straight r subscript 1 minus straight r subscript 2 over denominator straight r subscript 1 end fraction space and space straight l over straight l subscript 1 equals fraction numerator straight r subscript 1 minus straight r subscript 2 over denominator straight r subscript 1 end fraction rightwards double arrow space space space space space straight h subscript 1 equals fraction numerator hr subscript 1 over denominator straight r subscript 1 minus straight r subscript 2 end fraction space and space straight l subscript 1 equals fraction numerator lr subscript 1 over denominator straight r subscript 1 minus straight r subscript 2 end fraction space space space space... left parenthesis straight A right parenthesis

    Now,
    Height of the cone VA‘B’
    equals space straight h subscript 1 minus straight h equals fraction numerator hr subscript 1 over denominator straight r subscript 1 minus straight r subscript 2 end fraction minus straight h equals fraction numerator hr subscript 2 over denominator straight r subscript 1 minus straight r subscript 2 end fraction space space space space... left parenthesis straight B right parenthesis
    Slant height of the cone VA‘B’
    equals space straight l subscript 1 minus straight l space equals fraction numerator lr subscript 1 over denominator straight r subscript 1 minus straight r subscript 2 end fraction minus straight l equals fraction numerator lr subscript 2 over denominator straight r subscript 1 minus straight r subscript 2 end fraction space space space space space space.... left parenthesis straight C right parenthesis

    Let S denote the curved surface area of the frustum of cone. Then,
    S = Lateral (curved) surface area of cone VAB
    - Curved surface area of cone VA‘B’

    rightwards double arrow space space space space space space space space straight S space equals space πr subscript 1 straight l subscript 1 minus πr subscript 2 left parenthesis straight l subscript 1 minus straight l right parenthesis rightwards double arrow space space space space space space space space straight S space equals space πr subscript 1. fraction numerator lr subscript 1 over denominator straight r subscript 1 minus straight r subscript 2 end fraction minus πr subscript 1. fraction numerator lr subscript 2 over denominator straight r subscript 1 minus straight r subscript 2 end fraction
    [Using (A) and (C)]
    rightwards double arrow space space space space straight S space equals space straight pi space open parentheses fraction numerator straight r subscript 1 squared minus straight r subscript 2 squared over denominator straight r subscript 1 minus straight r subscript 2 end fraction close parentheses straight l rightwards double arrow space space space space space space space equals space straight pi space left parenthesis straight r subscript 1 plus straight r subscript 2 right parenthesis space straight l

    Curved surface area of the frustum
    = π(r1 + r2)l
    Total surface area of the frustum
    = Lateral (curved) surface area
    + Surface area of circular bases
    = π (r1 + r2) I + πr12 + πr22
    = π {(r1 + r2) l + r12 + r22}.

     

    Question 38
    CBSEENMA10008842
    Wired Faculty App

    Derive the formula for the volume of the frustum of a cone, given to you in Section 13.5, using the symbols as explained.
    Solution

    Let V be the volume of the frustum of cone. Then,
    V = Volume of cone V AB
    – Volume of cone VA ‘B’

    Thus, the volume of the frustum of the cone is given by
    bold rightwards double arrow bold space bold space bold V bold space bold equals bold space bold 1 over bold 3 bold pi bold space bold left parenthesis bold r subscript bold 1 to the power of bold 2 bold plus bold r subscript bold 1 bold r subscript bold 2 bold plus bold r subscript bold 2 to the power of bold 2 bold right parenthesis bold space bold h bold.

    Question 39
    CBSEENMA10008843
    Wired Faculty App

    The radius and height of a cylinder are in the ratio 5:7 and its volume is 550 cm3. Find its radius.

    Solution
    Let radius (r) = 5x and height (h) = 7x.
    Now. Volume = 550
    rightwards double arrow space 22 over 7 cross times space 5 straight x space cross times space 5 straight x space cross times space 7 straight x space equals space 550 rightwards double arrow space space space straight x cubed equals fraction numerator 550 cross times 7 over denominator 22 cross times 25 cross times 7 end fraction
    x = 1
    ∴ The required radius = 5x = 5 x 1 = 5cm.

    Sponsor Area

    Question 40
    CBSEENMA10008844
    Wired Faculty App

    The curved surface area of height circular cone is 12320 cm2. If the radius of the base is 56 cm, find its height.

    Solution

    Here, we have r = 56 cm
    and, Curved Surface Area = 12320
    rightwards double arrow space space 22 over 7 space straight x space 56 space straight x italic space straight l space equals space 12320 rightwards double arrow space space l space equals space 70 space cm Now comma Height space left parenthesis straight h right parenthesis space equals space square root of l to the power of italic 2 minus straight r squared end root equals space square root of open parentheses 70 close parentheses squared minus left parenthesis 56 right parenthesis squared end root equals space square root of 4900 minus 3136 end root equals square root of 1764 equals space 42 space space space
    Hence, height of one cone = 42 cm.

    Question 41
    CBSEENMA10008845
    Wired Faculty App

    A conical flask is full of mater. The flask was base m radius r and height n. the mater is poored into a cylindrical flask of base radius mr. find the height of water in the cylindrical flask.

    Solution

    Here, we have base radius of conical flask = r and height = h
    Then,
    Volume of conical flask = 1 third space πr squared straight h
    Again,
    Base radius of cylindrical flask = mr.
    Let height of cylindrical flask = h
    Then, volume = π (mr)2h1
    Now,
    Volume of Conical flask
    = volume of cylindrical flask
    rightwards double arrow space space space space 1 third πr squared straight h space equals space straight pi space left parenthesis mr right parenthesis squared straight h subscript 1 rightwards double arrow space space space space space straight h subscript 1 equals fraction numerator straight r squared straight h over denominator 3 straight m squared straight r squared end fraction equals fraction numerator straight h over denominator 3 straight m squared end fraction

    Question 42
    CBSEENMA10008846
    Wired Faculty App

    A cylinder, a cone and a hemisphere are of equal base and have the same height. What is the ratio of their volumes?

    Solution

    Let r be radius of a cylinder, a cone and a hemisphere respectively and h be the height of the cylinder, a cone and hemisphere respectively.
    Then, h = r
    V1 = Volume of a cylinder = πr squared straight h comma
    Now, V2 = Volume of a cone = 1 third πr squared straight h
    and V3 = Volume of a hemisphere = 2 over 3 πr cubed
    therefore space space space straight V subscript 1 colon space straight V subscript 2 space colon space straight V subscript 3 equals space πr squared straight h space colon space 1 third πr squared straight h space colon space 2 over 3 πr cubed rightwards double arrow space straight V subscript 1 colon space straight V subscript 2 space colon space straight V subscript 3 space equals space 3 straight h space colon space straight h space colon space 2 straight r rightwards double arrow space space straight V subscript 1 colon space straight V subscript 2 space colon space straight V subscript 3 equals space 3 space colon space 1 space colon space 2

    Question 43
    CBSEENMA10008847
    Wired Faculty App

    A cone and a hemisphere have equal bases and equal volumes. Find the ratio of their heights.

    Solution

    Let the radius of both cone and hemisphere be r cm.
    Let h be the height of the cone.
    It is given that,
    Volume of a Cone = Volume of a hemisphere
    rightwards double arrow space 1 third πr squared straight h equals 2 over 3 πr cubed rightwards double arrow straight h rightwards double arrow 2 straight r rightwards double arrow space straight h over straight r equals 2 over 1 rightwards double arrow space straight h space colon space straight r space equals space 2 space colon space 1 comma
    Hence, the ratio of the heights of a cone and hemisphere is 2 : 1.

    Question 45
    CBSEENMA10008849
    Wired Faculty App

    If two cones have their volumes in the ratio 3 :1 and their heights are in the ratio 1 : 3, then find the ratio of their radius.

    Solution
    Let the radii be r1 and r2 and heights be h1 and h1 Then
    straight V subscript 1 over straight V subscript 2 equals fraction numerator begin display style 1 third end style πr subscript 1 squared straight h subscript 1 over denominator begin display style 1 third end style πr subscript 2 squared straight h subscript 2 end fraction equals open parentheses r subscript 1 over r subscript 2 close parentheses squared space x space open parentheses h subscript 1 over h subscript 2 close parentheses rightwards double arrow space 3 over 1 equals open parentheses r subscript 1 over r subscript 2 close parentheses squared cross times open parentheses 1 third close parentheses rightwards double arrow open parentheses 3 over 1 close parentheses squared equals open parentheses r subscript 1 over r subscript 2 close parentheses squared rightwards double arrow space space space r subscript 1 over r subscript 2 equals 3 over 1
    Thus, the ratio of their radius is 3 :1.
    Question 46
    CBSEENMA10008850
    Wired Faculty App

    The radius of a solid hemispherical toy is 3.5 cm. Find the total surface area.

    Solution

    Here, we have
    radius of hemispherical toy (r) = 3.5 cm
    Now,
    Total surface area = 3πr2
    equals space open parentheses 3 space straight x space 22 over 7 straight x space 3.5 space straight x space 3.5 close parentheses space space cm squared equals space 115.5 space cm squared

    Question 48
    CBSEENMA10008852
    Wired Faculty App

    Determine the ratio of the volume of a cube to that of a sphere which will exactly fit inside the cube.

    Solution
    Let V1 and V2 be the volume of the cube and sphere respectively, then
    V1= (2r)3 and V2 = 4 over 3 comma space πr cubed
    therefore space space space space straight V subscript 1 over straight V subscript 2 equals fraction numerator left parenthesis 2 straight r right parenthesis cubed over denominator begin display style 4 over 3 end style πr cubed end fraction equals fraction numerator 8 straight r cubed over denominator begin display style 4 over 3 end style πr cubed end fraction equals 6 over straight pi Hence comma space straight V subscript 1 space colon space straight V subscript 2 equals 6 space colon space straight pi
    Question 49
    CBSEENMA10008853
    Wired Faculty App

    A cylinder and a cone are of the same base radius and of same height. Find the ratio of the volume of cylinder to that of the cone.

    Solution

    Let r be the base radius and h be the height.
    Then, volume of the cylinder i. e.,
    V1 = πr2h
    and volume of the cone i.e.,
    straight V subscript 2 equals 1 third πr squared straight h therefore space space straight V subscript 1 over straight V subscript 2 equals fraction numerator πr squared straight h over denominator 1 divided by 3 space πt squared straight h end fraction equals 3 over 1
    Hence, the required ratio is 3 : 1.

    Question 50
    CBSEENMA10008854
    Wired Faculty App

    A right circular cylinder and a right circular cone have the same base radius and same height. What is the ratio of the volume of the cylinder to that of cone?

    Solution

    Let V and ‘R’ be the radii of cylinder and cone respectively.
    And ‘h’ and ‘H’ be the heights of cylinder and cone respectively.
    Now volume of cylinder (v1) = πr squared straight h
    And, volume of cone (v2) = 1 third space πr squared straight h
    Now comma space space space space straight v subscript 1 over straight v subscript 2 equals fraction numerator πr squared straight h over denominator begin display style 1 third end style πR squared straight H end fraction space space space space space space space space space space space space space space space space equals space fraction numerator πr squared straight h over denominator begin display style 1 third end style πr squared straight h end fraction space left square bracket because space straight r space equals space straight R comma space straight h space equals space straight H right square bracket space space space space space space space space space space space space space space space space space equals space fraction numerator 1 over denominator begin display style 1 third end style end fraction equals 3 over 1

    Question 61
    CBSEENMA10008865
    Wired Faculty App

    Two cubes each of 10 cm edge are joined end to end. Find the surface area of the resulting cuboid.

    Solution

    Lei l, b and h are respectively the length, breadth and height of the cuboid, then
    l = 10 cm + 10 cm 20 cm
    b = 10 cm and h = 10 cm.
    Now,
    Surface Area of the cuboid
    = 2 (lb + bh + hl)
    = 2 (20 x 10+ 10 x 10+ 10 x 20) cm2
    = 2 (200 + 100 + 200) cm2
    = 1000 cm2.

    Question 62
    CBSEENMA10008866
    Wired Faculty App

    Three cubes each of side 5 cm are joined end to end. Find the Surface Area of the resulting cuboid.

    Solution

    Let l, b and h are respectively the length, breadth and height of the cuboid, then
    l = (5 + 5 + 5) cm = 15 cm
    when three cubes are joined we get a cuboid
    b = 5 cm and h = 5 cm.
    Now,
    The Surface Area of the resulting cuboid
    = 2 (lb + bh + hl) cm2
    = 2 (15 x 5 + 5 x 5 + 5 x 15) cm2
    = 2 (75 + 25 + 75) cm2
    = 2 (175) cm2
    = 350 cm2.

    Question 63
    CBSEENMA10008867
    Wired Faculty App

    Three equal cubes of side 4 cm are joined end to end. Find the surface area of the resulting cuboid.

    Solution

    Let l, b and h are respectively the length, breadth and height of the cuboid, then
    l = (4 + 4 + 4) cm = 12 cm
    when three cubes are joined we gel a cuboid.
    So, b = 4 cm and h = 4 cm.
    Now,
    The Surface Area of the resulting cuboid
    = 2 (lb + bh + hl) cm2
    = 2 (12 x 4 + 4 x 4 + 4 x 12) cm2
    = 2 (48 + 16 + 48) cm2
    = (2 x 112) cm2
    = 224 cm2.

    Question 64
    CBSEENMA10008868
    Wired Faculty App

    Three cubes of metal whose edges are 6 cm, 8 cm and 10 cm are melted and a new cube is made (i) Find the length of the edge of the new cube.
    (ii) Find the total surface area of the new cube.
    Solution

    Let the edges of the cubes of metal be a, b and c respectively, then
    a = 6 cm, b = 8 cm and c = 10 cm
    and thus respective volumes are,
    v1 = (6)3, V2 = (8)3 and V3 = (10)3
    Now, the volume of new cube
    = V1 + V2 + V3
    = 63 + 83 + 103
    = 1748 = 123 cm3.
    Let the length of the edge of the new cube be = a cm, then
    a3 = 123 ⇒ a = 12 cm.
    (ii) The total surface area of the new cube
    = 6a2 = 6 x 12 x 12
    = 864 cm2.

    Question 65
    CBSEENMA10008869
    Wired Faculty App

    Two cubes each of 12 cm edge are joined end to end. Find the surface area of the resulting cuboid.

    Solution

    Let l, b and h are respectively the length, breadth and height of the cuboid, then
    l = (12 + 12) cm = 24 cm
    when two cubes are joined we get a cuboid b = 12 cm and h = 12 cm.
    Now,
    The Surface Area of the resulting cuboid
    = 2 (lb + bh + hl) cm2
    = 2 (24 x 12 + 12 x 12 + 24 x 12) cm2
    = 2 (288 + 144 + 288) cm2
    = 2 (720) cm2
    = 1440 cm2.
    Problems Based on Conversion of Solids from Shape to Another

    Question 66
    CBSEENMA10008870
    Wired Faculty App

    A cone of height 24 cm and radius of base 6 cm is made up of modelling clay. A child reshapes it in the form of a sphere. Find the radius of the sphere.

    Solution

    Let the radius and height of a cone be 24 cm and 6 cm respectively, then
    r = 6 cm, h = 24 cm
    Let radius of the sphere be r1 cm,
    Since, the volume of clay in the form of a cone and the sphere remains the same. Therefore
    1 third πr squared straight h space space equals space 4 over 3 πr subscript 1 cubed rightwards double arrow 6 space straight x space 6 space straight x space 24 space equals space 4 straight r subscript 1 cubed rightwards double arrow space straight r subscript 1 cubed equals 6 space straight x space 6 space straight x space 6 rightwards double arrow space straight r subscript 1 equals space 6 space cm.

    Question 67
    CBSEENMA10008871
    Wired Faculty App

    A right circular cone is 8 cm heigh and the radius of its base is 2 cm. The cone is melted and re-cast into a sphere. Determine the diameter of the sphere.

    Solution

    Let r cm be the radius and h cm be the height of the cone, then
    r = 2 cm, h = 8 cm
    Let r1 cm be the radius of the sphere.
    Since, sphere is made out of the cone, so their volumes are equal.
    Therefore,
    1 third πr squared straight h space equals 4 over 3 πr subscript 1 cubed rightwards double arrow space straight r squared straight h space space equals space 4 straight r subscript 1 cubed rightwards double arrow space straight r subscript 1 cubed equals 1 fourth straight r squared straight h rightwards double arrow space straight r subscript 1 cubed equals 1 fourth straight x space 2 space straight x space 2 space straight x space 8 rightwards double arrow space straight r subscript 1 cubed equals space 2 space straight x space 2 space straight x space 2 rightwards double arrow space straight r subscript 1 equals space 2 space cm
    Hence, diameter of sphere = 4 cm.

    Question 68
    CBSEENMA10008872
    Wired Faculty App

    A solid cylinder of metal 8 m high and 4 m diameter is melted and recast into a cone of diameter 3 m. Find the height of the cone.

    Solution
    Let h cm be the height and r cm be the radius of a cylinder, then
                            h = 8 m
    and                   r = 4 over 2 space equals space 2 space m 
    Let r1 m be the radius and m be the height of the cone, then
    straight r subscript 1 equals 3 over 2 equals 1.5 space straight m comma straight h subscript 1 equals ?            

    Since a cone is made out of the solid cylinder, so their volumes are equal.
    Therefore,
    rightwards double arrow space space space πr squared straight h space equals space 1 third πr subscript 1 squared straight h subscript 1 rightwards double arrow space space space 2 space straight x space 2 space straight x space 8 space equals space 1 third straight x space 1.5 space straight x space 1.5 space straight x space straight h subscript 1 rightwards double arrow space space space straight h subscript 1 equals fraction numerator 2 space straight x space 2 space straight x space 8 space straight x space 3 over denominator 1.5 space straight x space 1.5 end fraction space space space space space space space space space space space equals space fraction numerator 96 over denominator 2.25 end fraction equals 42.66 space straight m

    Question 69
    CBSEENMA10008873
    Wired Faculty App

    A cone is 8.4 cm high and the radius of its base is 2.1 cm. it is melted and recast into a sphere. Find the radius of the sphere.

    Solution

    Let r cm be the radius and h cm be the height of a cylinder, then
    r = 2.1. cm, h = 8.4 cm
    Let r1 cm be the radius of a sphere.
    Since a sphere is made out of a cone, so their volumes are equal.
    Therefore,
    rightwards double arrow space space space space space space space 1 third πr squared straight h space space space equals space 4 over 3 πr subscript 1 cubed rightwards double arrow space space space space space space space space straight r squared straight h space equals space 4 straight r subscript 1 cubed rightwards double arrow space space space 2.1 space straight x space 2.1 space straight x space 8.4 space equals space 4 space straight x space straight r subscript 1 cubed rightwards double arrow space space space straight r subscript 1 cubed space equals space fraction numerator 2.1 space straight x space 2.1 space straight x space 8.4 over denominator 4 end fraction rightwards double arrow space space straight r subscript 1 cubed equals space 2.1 space straight x space 2.1 space straight x space 2.1 rightwards double arrow space space straight r subscript 1 space equals space 2.1 space cm

    Question 70
    CBSEENMA10008874
    Wired Faculty App

    A well of diameter 14 m is dug 15 m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 7 m to form an embankment. Find the height of the embankment.

    Solution
    Let h m be the height and r m be the radius of the well, then
        h = 15 m,  r = 14 over 2 equals space 7 space m.
    Now,  
    Volume of earth dug out = πr squared straight h
                                
                              equals space open parentheses 22 over 7 straight x space 7 space straight x space 7 space straight x space 15 close parentheses space straight m cubed
          [ because shape of well is similar to cylinder]
                              = ( 22 x 7 x 15) m3
                               
                              = 2310 m3

    Let H m be the required height of the embankment.
    Since the shape of the embankment will be like the shape of cylinder of internal radius (r2) 7 m and external radius (R1) = 7 + 7 = 14 m. Now,
    the volume of embankment
    = volume of the earth dug out
    ⇒ π(R12 –r12) H = 2310
    ⇒ π(142 – 72) H = 2310
    ⇒ π (196 – 49) H = 2310
    ⇒ π(147) H = 2310
    rightwards double arrow space space space space space space space space straight H space equals space fraction numerator 2310 space straight x space 7 over denominator 147 space straight x space 22 end fraction space space space space space space space space space space space space space space space equals space 16170 over 3234 space equals space 5 space straight m

    Hence, the height of the embakment = 5m
    Alternative Method :
    Height of the embankment
    equals fraction numerator volume space of space the space earth space dug space out over denominator Area space of space the space embankment end fraction

    Note :
    Area of the embankment
    = Area of ring
    = π R2 – π r2
    = π (R2 – π r2)
    = π (R + r) (R – r)


    Question 71
    CBSEENMA10008875
    Wired Faculty App

    An agriculture field is in the form of a rectangle of length 20 m, and width 14 m. A 10 m deep well of diameter 7 m is dug in a comer of the field and the earth taken out of the well is spread evenly over the remaining part of the field. Find the rise in the level of the field.

    Solution

    Since shape of the well is just like the shape of cylinder, so we consider well as cylinder.
    Let r m be the radius and h m be the height of the well, then 
    straight r space equals fraction numerator 7 degree over denominator 2 end fraction straight m comma space straight h space left parenthesis depth right parenthesis space equals space 10 straight m
    Now,
    Volume of earth dug out = πr squared straight h
                              equals space open parentheses 22 over 7 x 7 over 2 x 7 over 2 x 10 close parentheses m cubed equals 385 space m cubed
    Also, we have, 
    Length of the rectangular field = 20 m
    Breadth of the field = 14 m
    Therefore, are of the field = (20 x 14) m2
                                          = 280 m2
    And
       Area of the remaining part of the field
                   = Area of field - Area of the base
                     
                         equals space open parentheses 280 minus 77 over 2 close parentheses straight m cubed
    Since,
    Volume of the raised field
             = Volume of the earth dug out
    rightwards double arrow space space open parentheses 483 over 2 close parentheses straight h space equals space 358 rightwards double arrow space space space straight h space equals space fraction numerator 358 space straight x space 2 over denominator 483 end fraction rightwards double arrow space space space straight h space equals space 1.6 space left parenthesis approx right parenthesis
    Hence, rise in the level of the field = 1.6 m (apporx.)

    Question 72
    CBSEENMA10008876
    Wired Faculty App

    A metallic sphere of radius 10.5 cm is melted and recast into 5 small cones each of radius 3.5 cm and height 3 cm. Find how many cones are obtained.

    Solution

    Let R cm be the radius of the sphere, then
    R = 10.5 cm
    Now, Volume of sphere
    equals 4 over 3 πR cubed equals space open parentheses 4 over 3 straight x space straight pi space straight x space 10.5 space straight x space 10.5 space straight x space 10.5 space close parentheses space cm cubed equals 1543.5 space straight pi space cm cubed

    Let r cm be the radius and h cm be the height of the cone, then
    r = 3.5 cm and h = 3 cm
    Now, Volume of each cone
    Therefore
    Total number of cones
     
      equals fraction numerator Volume space of space sphere over denominator Volume space of space one space cone end fraction equals space fraction numerator 1543.5 straight pi over denominator 12.25 straight pi end fraction equals 126 space cones


    Question 73
    CBSEENMA10008877
    Wired Faculty App

    Find the number of coins 1.5 cm in diameter and 0.2 cm thick to be melted to form a right circular cylinder whose height is 8 cm and diameter 6 cm.

    Solution

    Since shape of the coin having thickness will be like a cylinder. So, we consider coin as a cylinder.
    Let r cm be the radius and h cm be the height (thickness) of a coin, then
    straight r space equals space fraction numerator 1.5 over denominator 2 end fraction cm comma space space space and space straight h space equals space 0.2 space cm
    Now, Volume of each coin
                    equals space πr squared straight h equals space open parentheses straight pi space straight x space fraction numerator 1.5 over denominator 2 end fraction space straight x space fraction numerator 1.5 over denominator 2 end fraction straight x space 0.2 close parentheses space cm cubed equals space 0.225 space straight pi space space space cm cubed.
    Let R cm be the radius and H cm be the height of a cylinder, then
    straight R space equals space 6 over 2 equals 3 space cm space and space straight H space equals space 8 space cm
    Now, Volume of cylinder
             equals space πR squared straight H equals space left parenthesis straight pi space straight x space 3 space straight x space 3 space straight x space 8 right parenthesis space cm cubed equals space 72 space straight pi space space cm cubed
    Therefore,
    Total number of coins = fraction numerator Volume space of space cylinder over denominator Volume space of space one space coin end fraction
                                    
                                    = fraction numerator 72 straight pi over denominator 0.225 straight pi end fraction equals 320

    Question 74
    CBSEENMA10008878
    Wired Faculty App

    Spherical marbles of diameter 1.4 cm each are dropped in a cylindrical beaker of radius 3.5 cm containing some water. Find the number of marbles that should be dropped into the beaker so that the water level in the beaker rises by 5.6 cm.

    Solution

    Let ‘R’ cm be the radius of the sphere, then
    R = 10.5 cm
    Now, volume of sphere
    equals 4 over 3 πr cubed equals open parentheses 4 over 3 straight x space straight n space straight x space 10.5 space straight x space 10.5 space straight x space 10.5 close parentheses space cm cubed

    Let, ‘h’ cm be the rise in water level, and ‘r’ cm be the radius of the beaker. Then
    r = 3.5 cm., h = 5.6 cm
    Now, volume of cylinder = πr2h
    = (π x 3.5 x 3.5 x 5.6) cm2
    Therefore
    Required no. of marbles
    equals space fraction numerator volume space of space cylinder space left parenthesis beaker right parenthesis over denominator volume space of space one space sphereical space marbles end fraction equals fraction numerator straight pi space straight x space 3.5 space straight x space 3.5 space straight x space 5.6 over denominator begin display style 4 over 3 end style xπ space straight x space 10.5 space straight x space 10.5 space straight x space 10.5 space straight x space 10.5 end fraction equals space fraction numerator 3.5 space straight x space 3.5 space straight x space 5.6 space straight x space 3 over denominator 10.5 space straight x space 10.5 space straight x space 10.5 end fraction
    Hence, the required no. of marbles = 150

    Question 75
    CBSEENMA10008879
    Wired Faculty App

    The radius of a conical solid is melted and recaste into a number of cylindrical solids if the radius of the cone is three times the radius of the cylinder and the height of the cone is four times the height of the cylinder, find the number of cylindrical solids.

    Solution

    Let R cm be the radius and H cm be the height of cone.
    And r cm be the radius and h cm be the height of a cylinder, then
    R = 3r
    and    H = 4h
    Now,
    Volume of cone = 1 third space πR squared straight H
    and Volume of cylinder = π r2h Therefore,
    The number of cylindrical solids
    equals space fraction numerator Volume space of space cone over denominator Volume space of space one space cylinder end fraction equals space fraction numerator begin display style 1 third end style πR squared straight H over denominator πr squared straight h end fraction equals space fraction numerator begin display style 1 third end style straight pi left parenthesis 3 straight r right parenthesis squared left parenthesis 4 straight h right parenthesis over denominator πr squared straight h end fraction equals space fraction numerator begin display style 1 third end style straight pi space straight x space 9 straight r squared straight x 4 straight h over denominator πr squared straight h end fraction equals space 12

    Question 76
    CBSEENMA10008880
    Wired Faculty App

    Water flows at the rate of 10 m per minute through a cylindrical pipe having its diameter as 5 mm. How much time will it take to fill a conical vessel whose diameter of the base is 40 cm and depth 24 cm.

    Solution
    Let r cm be the radius and h cm be the height of a cylindrical pipe, then

    straight r space equals space fraction numerator 5 mm over denominator 2 space straight x space 10 end fraction space equals space 1 fourth space cm and space space space straight h space equals space 10 space straight m space equals space 1000 space cm

    [length of the water column in cylindrical pipe will become the height of cylindrical pipe]

     
    Question 77
    CBSEENMA10008881
    Wired Faculty App

    Water is flowing at the rate of 0.70 m/sec through a circular pipe, whose internal diameter is 2 cm into a cylindrical tank the radius of whose base is 40 cm. Determine the increase in the water level in 1/2 hour.

    Solution
    Let R m be the radius and H m be the height of a circular pipe, then
                straight R space equals space 2 over 2 equals 1 over 100 equals 0.01 space straight m and space space space space straight H space equals space 0.70 space straight m

    [length of water column = Height of circular pipe]
    Volume of water that flow's out of the circular pipe in 1 sec
    = πR2H
    = π x 0.01 x 0.01 x 0.70
    = 0.00007 π m3.
    Therefore, Volume of water that flows out of
    the circular pipe in 1 half hr (30 x 60) sec
                        = (0.00007 straight pi x 1800) cm3 
                        = 0.126 straight pi space straight m cubed
    Let r m be the radius and h m be the height of cylindrical tank, then
    r = 40 cm = 40 over 100 = 0.4 m, h = ?
    Now, Volume of water in the cylindrical lank up to a height of h m
    = πr2h =  π x 0.4 x 0.4 x h = 0.1 6π h m2.
    Since, Volume of the water flown into the tank
    = Volume of the water that flows through the pipe in half an hour.
    ⇒ 0.16 π h = 0.126 π
    rightwards double arrow space space space straight h space equals space fraction numerator 0.126 over denominator 0.16 end fraction space space space space space space space space space equals space 0.7875 space straight m
    Hence, increase in water level in 1/2 hr = 7875 m.


    Question 78
    CBSEENMA10008882
    Wired Faculty App

    Water in a canal 30 dm wide and 12 dm deep, is flowing velocity of 10 km/hr. How much area will it irrigate if 8 cm of standing water is required for irrigation?

    Solution

    We have,
    Width of the canal = 30 dm = 300 cm = 3 m
    Depth of the canal = 12 dm = 120 cm = 1.2 m
    It is given that the water is flowing with velocity 10 km/hr. Therefore length of the water column formed in bold 1 over bold 2 hour = 5 km = 5000 m
    ∵ Volume of the water flowing in  bold 1 over bold 2 hour

    = Volume of the cuboid of length = 5000 m,
    Width = 3 m and depth = 1.2 m
    = 5000 x 3 x 1.2 m3 = 18000 m3.
    Suppose xm2 area is irrigated in 1 half hour. Then,
    straight x cross times 8 over 100 equals 18000 rightwards double arrow space space straight x space equals space 1800000 over 8 straight m squared rightwards double arrow space space straight x space equals space 225000 space straight m squared
    Hence, the canal irrigates 22500 m2 area in bold 1 over bold 2 hour.

     

    Question 79
    CBSEENMA10008883
    Wired Faculty App

    A hemispherical tank full of water is emptied by a pipe at the rate of   litres per second. How much time will it take to half empty the tank, if the tank is 3 m in diameter.

    Solution

    Let r m be the radius of the hemispherical tank, then
                                 straight r space equals space 3 over 2 straight m.
    Now, volume of hemispherical tank 
                                 equals 2 over 3 πr cubed equals space open parentheses 2 over 3 straight x 22 over 7 straight x 3 over 2 straight x 3 over 2 straight x 3 over 2 close parentheses space straight m cubed equals space 99 over 14 straight m cubed equals 99000000 over 14 space cm cubed
    And, Volume of water to be emptied
                            equals 1 half (Volume of hemispherical tabk)
                               
                            equals open parentheses 1 half straight x 99000000 over 14 close parentheses space cm cubed equals space 99000000 over 28 straight x 1 over 1000 space litres equals space 99000 over 28 space litres
    Hence,
    Time taken to half empty the tank
                       
                             equals space 99000 over 28 straight x 7 over 25 space seconds
                              = 16.5 minutes.

    Sponsor Area

    Question 80
    CBSEENMA10008884
    Wired Faculty App

    Water is flowing at the rate of 5 km per hour through a pipe of diameter 14 cm into a rectangular tank which is 50 m long and 44 m wide. Determine the time in which the level of water in the tank will rise by 7 cm.

    Solution
    Let r cm be the radius and h cm be the height of a pipe (cylinder), then
               
                    r = 7 cm = 7 over 10 space straight m
    And, h = 5 km = 5000 m

    [distance covered in 1 hr. = height of pipe]
    Now, Volume of water that flows out of the circular pipe in 1 hour
    equals space πr squared straight h equals space open parentheses 22 over 7 straight x 7 over 100 straight x 7 over 100 straight x 5000 close parentheses straight m cubed equals space 77 space straight m cubed
    Let l, b and h are respectively the length, breath and height of the tank (cuboid), then
                             l = 50 m
                            b = 44 m
    and                   h =  7 cm = 7 over 100 m         
    Now, Volume of water that flows in the rectangular tank       
              = l x b x h
              equals space open parentheses 50 space straight x space 44 space straight x space 7 over 100 close parentheses space straight m cubed space equals space 154 space straight m cubed
    because time taken to flow 77 m3 of water in to the tank K = 1 hr
    therefore time taken to flow 154 m3 of water in to the tank = 154 over 77 equals space 2 space hrs   
    Hence, the level of water in the tank rise by 7 cm in 2 hrs.    

    Question 81
    CBSEENMA10008885
    Wired Faculty App

    Water is flowing at the rate of 3 km/ hr through a circular pipe of 20 cm external diameter into a circular cistern of diameter 10m and depth 2 m. In how much time will the cistern be filled.

    Solution
    Let R cm be the radius and H cm be the height of a circular pipe, then
    straight R space equals space 20 over 2 space equals space 10 space cm space equals space 1 over 10 space straight m

    H (depth) = 3 km = 3000 m.
    Now, Volume of water that flows out of the circular pipe in 1 hour = πR2H
    equals space open parentheses straight pi space straight x space 1 over 10 straight x 1 over 10 straight x 3000 close parentheses space straight m cubed equals space 30 straight pi space straight m cubed
    Let r m be the radius and h m be the depth (height) of a circular cistern, then
    straight r space equals space 10 over 2 equals 5 space straight m space and space straight h space equals space 2 space straight m

    Now, Volume of water that flows in the circular cistern
    = π r2h
    = π x 5 x 5 x 2 = 50 π m3.
    Hence,
    Time taken to fill up the cistern
    equals space space space space space space space space space space space space space Volume space of space water space that space flows space space space space space fraction numerator out space the space circular space pipe space in space 1 space hr over denominator Volume space of space water space that space flows space in space the space cistem end fraction equals fraction numerator 50 straight pi over denominator 30 straight pi end fraction equals fraction numerator 5 straight pi over denominator 3 straight pi end fraction equals space 5 over 2 equals 1 space hour space 40 space minutes

    Question 82
    CBSEENMA10008886
    Wired Faculty App

    An iron solid sphere of radius 3 cm is melted and recast into small spherical balls of radius 1 cm each. Assuming that there is no wastage in the process, find the number of small spherical balls made from the given sphere.

    Solution
    Let R be the radius of solid sphere, then R = 3cm
    So,   Volume (V1) = 4 over 3 πR cubed
                 
                       equals space open parentheses 4 over 3 straight x 22 over 7 straight x 3 space straight x space 3 space straight x space 3 close parentheses space cm cubed
    Let r be the radius of small spherical ball, then
                            r = 1 cm
    So, Volume (V2) = 4 over 3 space πr cubed
                             equals space open parentheses 4 over 3 straight x 22 over 7 straight x space 1 space straight x space 1 straight x space 1 close parentheses space cm cubed
    Now, Required number of small spherical balls
    equals space fraction numerator Volume space of space solid space sphere space left parenthesis straight V subscript 1 right parenthesis over denominator Volume space of space one space small space spherical space ball space left parenthesis straight V subscript 2 right parenthesis end fraction equals space fraction numerator begin display style 4 over 3 end style space straight x space begin display style 22 over 7 end style straight x space 3 space straight x space 3 space straight x space 3 over denominator begin display style 4 over 3 end style straight x space begin display style 22 over 7 end style straight x space 1 space straight x space 1 space straight x space 1 space end fraction
    = 3 x 3 x 3 = 27
    Question 83
    CBSEENMA10008887
    Wired Faculty App

    A spherical copper shell, of external diameter 18 cm, is melted and recast into a solid cone of base radius 14 cm and height 4 3 over 7 space cm. Find the inner diameter of the shell.

    Solution

    Let R be the external radius of the copper shell r be the internal radius, h be the height of the conical part and r1 be its radius.
    Now, External d of copper shell = 18 cm.

    Question 84
    CBSEENMA10008888
    Wired Faculty App

    A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank in his field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at the rate of 6km/h, in how much time will the tank be filled?  

    Solution

    Speed of the water
    = 6 k.m./hr. = 6000 m/hr.
    i.e. length of the water column (h) = 6000 m
    And, internal radius of the pipe (r)
    equals space 10 space cm. space space space equals space 1 over 10
    therefore Volume if water that flows in 1 hour = πr squared straight h
    equals space open parentheses straight pi space cross times space 1 over 10 cross times 1 over 10 cross times 6000 close parentheses straight m cubed equals 60 space straight pi space straight m cubed
    Now, radius of the base of the tank (R)
    equals 10 over 2 space equals space 5 space straight m
    And, Depth of the tank (h = 2m)
    equals space straight pi space straight R squared space straight x space straight H equals space straight pi space straight x space 5 space straight x space 5 space straight x space 2 space equals space 50 space πm cubed
    Hence, Required time to fill the tank by pipe
    fraction numerator Volume space of space the space tank over denominator Volume space of space water space that space flows space in space 1 space hour end fraction equals space space space fraction numerator 50 straight pi over denominator 60 straight pi end fraction equals 5 over 6 hr.

    Question 86
    CBSEENMA10008890
    Wired Faculty App

    Three cubes each of side 5 cm are joined end to end. Find the surface area of the resulting cuboid.

    Solution

    Solution not provided.
    Ans. 350 cm2 

    Question 88
    CBSEENMA10008892
    Wired Faculty App

    Three cubes of side 6 cm are joined end to end. Find the surface area of the resulting cuboid.

    Solution

    Solution not provided.
    Ans. 504 cm2 

    Question 105
    CBSEENMA10008909
    Wired Faculty App

    How many balls each of radius 1 cm can be made from a solid sphere of lead of radius 8 cm.

    Solution

    Solution not provided.
    Ans. 512

    Question 109
    CBSEENMA10008913
    Wired Faculty App

    The diameter of a sphere is 42 cm. It is melted and drawn into a cylindrical wire of 28 cm diameter. Find the length of the wire.

    Solution

    Solution not provided.
    Ans.  63 cm           

    Question 111
    CBSEENMA10008915
    Wired Faculty App

    How many spherical lead shots each 0.3 cm in diameter can be made from a rectangular solid (9 x 11 x 12) cm3.

    Solution

    Solution not provided.
    Ans.  84000               

    Sponsor Area

    Question 125
    CBSEENMA10008929
    Wired Faculty App

    A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the height of the cylinder is 20 cm and radius of the base is 3.5 cm, find the total surface area of the article.

    Solution

    Let r cm be the radius, h cm be the height of the cylinder then
    r = 3.5 cm
    and    h = 20 cm.
    Let r1 be the radius of the hemisphere, then
    r1 = 3.5 cm
    Now,
    Total surface area of the area of the article
    = curved surface area of the cylinder
    + 2 (curved surface area of hemisphere)
    = 2πrh + 2 (2 πr12)
    = 2 πrh + 4 πr2    [∵ r = r1]
    = 2 π r(h + 2r)
    equals space 2 space straight x space 22 over 7 straight x space 3.5 space left parenthesis 20 space plus 2 space straight x space 3.5 right parenthesis equals space 2 space straight x space 22 over 7 space straight x space 3.5 space straight x space 27 space equals space 594 space cm squared

    Question 126
    CBSEENMA10008930
    Wired Faculty App

    The decorative block shown in fig. is made of two solids - a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block.
    Solution

    Let the side (edge) of the cube be a cm, then a
    = 5 cm.
    Let radius of the hemisphere be r, then
    r = 2.1 cm
    Now,
    The total surface area of the block
    = T.S.A of cube - base area of hemisphere + C.S.A of hemisphere
    = 6 (side)2 – straight pir2 + 2rstraight pi2
    = 150 – straight pir2 + 2 straight pir2
    = 150 + straight pi2
    equals space open parentheses 150 plus 22 over 7 straight X space 2.1 space straight X space 2.1 close parentheses space cm squared equals space left parenthesis 150 space plus space 13.86 right parenthesis space cm squared equals space 163.86 space cm squared.

    Tips: -

    open parentheses bold Use bold space bold pi bold space bold equals bold space bold 22 over bold 7 close parentheses
    Question 127
    CBSEENMA10008931
    Wired Faculty App

    Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the birth - bath.   

    open parentheses bold pi bold space bold equals bold space bold 22 over bold 7 close parentheses bold space bold space

    Solution

    Let r cm be the radius and h cm be the height of the cylinder then,
    r = 30 cm,
    h = 1.45 m = 145 cm.
    Lei r1 cm be the radius of the hemisphere, then
    r1 = 30 cm
    Now,
    The Total Surface Area of the bird bath
    = C.S.A of cylinder + C.S.A of Hemisphere
    = 2 straight pi rh + 2 straight pi r1 2
    = 2straight pirh + 2straight pir2    [∵ r = r1]
    = [2 ∵ r (h + r)] cm2

    equals space open square brackets open parentheses 2 straight x 22 over 7 straight x 30 close parentheses left parenthesis 145 space plus space 30 right parenthesis close square brackets space cm squared equals space open parentheses 1320 over 7 straight x 175 close parentheses space cm squared
    = 33000 cm2
    = 3.3 m2

    Question 128
    CBSEENMA10008932
    Wired Faculty App

    A rocket is in the form of a circular cylinder closed at the lower end and a cone of the same radius is attached to the top The radius of the cylinder is 2.5 m, its height is 21 m and the slant height of the cone is 8 m. Calculate the total surface area of the rocket.

    Solution

    Let r m be the radius, and h m be the height of the cylinder, then
    r = 2.5 m and h = 21 m
    Let r m be the radius and l m be the slant height of the cone, then
    r1 = 2.5 m and l = 8 m
    Now,
    The Total Surface Area of the rocket
    = C.S.A of cone + C.S.A of cylinder + Area of base
    = straight pir1l + 2 straight pirh + straight pir2
    = straight pirl + 2 straight pirh + straight pir2 [r1 = r]
    = straight pir (l + 2h + r)

    equals space open square brackets 22 over 7 straight x space 2.5 space left parenthesis 8 space plus 2 space straight x space 21 space plus space 2.5 close square brackets space straight m squared equals space open square brackets 22 over 7 straight x space 2.5 space straight x space 52.5 close square brackets space straight m squared equals space open parentheses fraction numerator 2887.5 over denominator 7 end fraction close parentheses straight m squared space equals space 412.5 space straight m squared

    Question 129
    CBSEENMA10008933
    Wired Faculty App

    A toy is in the form of a cone mounted on a hemisphere of common base radius 7cm. The total height of the toy is 31 cm. Find the total surface area of the toy. 

    Solution
    Let r cm be the radius and h cm the height of the cone, then r = 7 cm, and h = 24 cm. Let l cm be the slant height of am cone, then
    l italic space italic equals italic space square root of straight r squared plus straight h squared end root space space equals space square root of left parenthesis 7 right parenthesis squared plus left parenthesis 24 right parenthesis squared end root space space equals space square root of 49 plus 576 end root space equals space square root of 625 equals 25 space cm

    Let r1 cm be the radius of hemisphere then
    r1 = 7 cm
    Now, Total surface area of toy
    = C.S.A of cone + C.S.A. of hemisphere
    equals space πrl space plus space 2 πr subscript 1 squared equals πrl plus 2 πl squared space left square bracket because space straight r subscript 1 equals space rl space right square bracket equals space 22 over 7 straight x 7 space left parenthesis 25 space plus space 2 space straight x space 7 right parenthesis equals space 22 space left parenthesis 25 space plus space 14 right parenthesis equals space left parenthesis 22 space straight x space 39 right parenthesis space cm squared space equals space 858 space cm squared


    Tips: -

    open parentheses bold Use bold space bold pi bold space bold equals bold space bold 22 over bold 7 close parentheses
    Question 130
    CBSEENMA10008934
    Wired Faculty App

    The interior of building is in the form of a right circular cylinder of radius 7 m and height 6 m, surmounted by a right circular cone of same radius and of vertical angle 60°. Find the cost of painting the building from inside at the rate of Rs. 30/m2. 

    Solution

    We have, r1 = Radius of the base of the cylinder = 7m
    r2 = Radius of the base of the cone = 7 m
    h1 = Height of the cylinder = 6m

    In right triangle EOC, we have
    sin space 30 degree space equals space OC over CE rightwards double arrow space space space 1 half equals 7 over CE rightwards double arrow space space CE space equals space 14 space straight m

    Thus, slant height (CE) = 14 m
    Now,Surface area of building = Curved surface area of cylinder + curved surface area of cone
    equals space 2 πr subscript 1 straight h subscript 1 plus πr subscript 1 straight l equals πr subscript 1 left parenthesis 2 straight h subscript 1 plus l right parenthesis equals space 22 over 7 space straight x space 7 space left parenthesis 2 space straight x space 6 space plus 14 right parenthesis equals open parentheses 22 over 7 space straight x space 7 space straight x space 26 close parentheses space straight m squared equals 572 space straight m squared

    Cost of painting the building from inside at the rate of Rs. 30/m2
    = Rs. (572 x 30) = Rs. 17160.  


    Question 131
    CBSEENMA10008935
    Wired Faculty App

    A solid is composed of a cylinder with hemispherical ends. If the whole length of the solid is 100 cm and the diameter of the hemispherical ends is 28 cm, find the cost of polishing the surface of the solid at the rate of 5 paise per sq. cm.

    Solution

    We have,
    r = radius of the cylinder
    = radius of hemispherical ends = 14 cm
    h = height of cylinder = 72 cm
    ∴ Total surface area = curved surface area of the cylinder + surface areas of hemispherical ends
    = (2straight pirh +2 x 2straight pir2) cm2
    = 2straight pir (h + 2r) cm2


    equals space 2 space straight x space 22 over 7 straight x space 14 space left parenthesis 72 space plus space 2 space straight x space 14 right parenthesis space cm squared equals space open parentheses 2 space straight x space 22 over 7 straight x space 14 space straight x space 100 close parentheses space cm squared equals space 8800 space cm squared
    Now, cost of polishing the surface of the solid at the rate of 5 paise per sq. m

    equals open parentheses 8800 space straight x space 5 over 100 close parentheses space equals Rs. space 440

    Question 132
    CBSEENMA10008936
    Wired Faculty App

    A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy.
    Find the difference of the volumes of the cylinder and the toy.

    (Use π = 3.14).
    Solution

    Let r cm be the radius and h cm be the height of a cone then.
    straight r space equals space 4 over 2 equals 2 space cm space and space straight h space equals space 2 space cm

    Let r1 cm be the radius and h1 cm be the height of hemisphere then,
    r1 = 2 cm, h1 = 2 cm (∵ h1= r1]

    Let r2 cm be the radius and h2 cm be the height of cylinder, then
    r2 = 2 cm
    and    h2 = (2 + 2) cm = 4 cm
    Now,
    Volume of solid toy,
    Volume of cone + Volume of hemisphere
    equals space 1 third space πr squared straight h space plus space 2 over 3 space πr subscript 1 cubed equals space 1 third space πr squared straight h space plus space 2 over 3 space πr cubed space space space space space left parenthesis because space straight r space equals space straight r subscript 1 right parenthesis equals space 1 third space πr squared space left parenthesis straight h space plus space 2 straight r right parenthesis equals space 1 third straight x space 3.14 space straight x space 2 space straight x space 2 space left parenthesis 2 plus 2 plus 2 right parenthesis equals space 1 third straight x space 3.14 space straight x space 4 space straight x space 6 space equals space 25.12 space cm cubed
    Volume of cylinder, 
    equals space πr subscript 2 straight h subscript 2 equals space left parenthesis 3.14 space straight x space 2 space straight x space 2 space straight x space 4 right parenthesis space cm cubed equals space left parenthesis 3.14 space straight x space 6 right parenthesis space cm cubed equals space 50.24 space cm cubed
    Hence,
    (i) Volume of toy = 25.12 cm
    (ii) Difference of the volume of cylinder and the toy
    = (50.24 – 25.12) cm3
    = 25.12 cm3.


    Question 133
    CBSEENMA10008937
    Wired Faculty App

    An iron pole consisting of a cylindrical portion 110 cm high and base diameter 12cm is surmounted by a cone 9cm high. Find the mass of the pole it is given that 1cm3of iron has 8 gm mass.

    open parentheses bold Use bold space bold pi bold space bold equals bold space bold 355 over bold 113 close parentheses

    Solution

    Let r cm be the radius and h cm be the height of a cone then
    r = 6 cm, h = 9 cm

    Required mass of pole
    = (8 x 12780) gm = 102.240 k.g.

    Question 134
    CBSEENMA10008938
    Wired Faculty App

    A tent is in the form of a cylinder surmounted by a conical top. If the height and diameter of the cylinderical part are 2.1 m and 4m respectively, and the height of the top is 2.8m, find the area of the canvas used for making the tent. Find the cost of the canvas used for making the tent. Find the cost of the canvas of the tent at the rate of Rs. 500 per m2. Also, find the volume of air enclosed in the tent.

    Solution
    Let ‘r’ met. be the radius, l be the slant height and ‘h’ m be the height of the conical part. Then, r 2m, l = 2.8 m.
    h equals space square root of l to the power of italic 2 minus straight r squared end root equals square root of left parenthesis 2.8 right parenthesis squared minus left parenthesis 2 right parenthesis squared end root equals square root of left parenthesis 2.8 plus 2 right parenthesis left parenthesis 2.8 minus 2 right parenthesis end root equals space square root of 4.8 space straight x space 0.8 end root equals square root of 48 over 10 straight x 8 over 10 end root equals square root of 384 over 100 end root equals space square root of 3.84 space end root equals space 1.96 space straight m

    Let ‘R’ met. be the radius and ‘H’ met. be the height of the cylindrical part.
    Thus,
    R = 2m and H = 2.1m

    Now,
    Area of the canvas used
    = curved surface area of conical part + curved surface area of cylindrical part = straight pirl + 2straight piRH = straight pirl + 2straight pirH
    [∵ r = R] = straight pir (l + 2H)
    equals 22 over 7 straight x space 2 space left parenthesis 2.8 space plus space 1 space straight x space 2.1 right parenthesis space equals space 22 over 7 straight x space left parenthesis 2.8 space plus 4.2 right parenthesis equals space 22 over 7 straight x space 2 space straight x space 7 space equals space 44 straight m squared
    Cost of the canvas of the tent = Rs. 500 x 44 = Rs. 22,000
    And.
    Volume of air enclosed in the tent = Volume of the conical part + Volume of cylindrical part
    equals 1 third πr squared straight h space plus space πr squared straight h space equals space 1 third πr squared straight h space plus space space space space space space left square bracket because space straight r space equals space straight R right square bracket equals space πr squared space open square brackets 1 third straight h space plus space straight H close square brackets space equals 22 over 7 space straight x space 2 space straight x space 2 space open square brackets 1 third straight x space 1.96 space plus space 2.1 close square brackets equals space 22 over 7 space straight x space 4 space straight x space open square brackets fraction numerator 1.96 plus 6.3 over denominator 3 end fraction close square brackets equals 22 over 7 straight x space 14 space straight x space fraction numerator 8.26 over denominator 3 end fraction equals space 34.61 space straight m cubed space


    Question 135
    CBSEENMA10008939
    Wired Faculty App

    A toy is in the form of a cone mounted on a hemisphere with same radius. The diameter of the base of the conical portion is 7.0 cm and the total height of the toy is 14.5 cm. Find

    the volume of the toy open square brackets Use space straight pi space equals space 22 over 7 close square brackets.

    Solution

    Let r cm be the radius and h cm be the height of the cone, then
    r = 3.5 cm and h = 14.5 cm - 3.5 cm = 11 cm Let r1 cm be the radius of the hemisphere, then r1 = 3.5 cm
    Now,
    Volume of Toy
    = Volume of hemisphere + Volume of cone
    equals space 2 over 3 πr subscript 1 to the power of 1 plus 1 third πr squared straight h equals space 2 over 3 πr to the power of 1 plus 1 third πr squared straight h space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space straight r subscript 1 equals straight r right square bracket equals space 1 third πr squared left square bracket 2 straight r space plus straight h right square bracket equals space 1 third straight x 22 over 7 straight x space 3.5 space straight x space 3.5 space straight x space left square bracket space 2 space straight x space 3.5 space plus space 11 right square bracket equals space open parentheses 1 third straight x 22 over 7 straight x space 3.5 space straight x space 3.5 space straight x space 18 close parentheses space cm cubed equals space 231 space cm cubed space

    Question 136
    CBSEENMA10008940
    Wired Faculty App

    An iron pillar has lower part in the form of a right circular cylinder and the upper part in the form of a right circular cone. The radius of the base of each of the cone and cylinder is 8 cm. The cylindrical part is 240 cm high and the conical part is 36 cm high. Find the weight of the pillar if 1 cm3 of iron weighs 7.5 grams.

    open parentheses Take space straight pi space equals space 22 over 7 close parentheses

    Solution

    Let r1 and r2 cm be the radii of the base of the cylinder and cone respectively. Then, r1 = r2 = 8 cm Let h1 and h2 cm be the heights of the cylinder and the cone respectively. Then,
    h1 = 240 cm and h2 = 36 cm

    Now, Volume of the cylinder = straight pir12h1 cm3 = (straight pi x 8 x 8 x 240) cm3 = (straight pi x 64 x 240) cm3
    Volume of the cone = 1 third πr subscript 2 squared straight h subscript 2 cm cubed equals space open parentheses 1 third straight pi space straight x space 8 space straight x space 8 space straight x space 36 close parentheses space cm cubed equals space open parentheses 1 third straight pi space straight x space 64 space straight x space 36 close parentheses space cm cubed
    Total volume of the iron = Volume of cylinder + Volume of the cone
    equals space open parentheses straight pi space straight x space 64 space straight x space 240 plus 1 third straight pi space straight x space 64 space straight x 36 close parentheses space cm cubed equals space straight pi space straight x space 64 space left parenthesis 240 space plus space 12 right parenthesis space cm cubed equals space open parentheses 22 over 7 straight x space 64 space straight x space 252 close parentheses space cm cubed equals space left parenthesis 22 space straight x space 64 space straight x space 36 right parenthesis space cm cubed
    Hence, total weight of the pillar = Volume x Weight per cm3
    = (22 x 64 x 36 x 7.5) = 380.16 kg

    Question 137
    CBSEENMA10008941
    Wired Faculty App

    A bucket is made up of a metal sheet is in the form of a frustum of a cone of height 16cm with radii of its lower and upper ends at 8cm and 20cm respectively. Find the cost of the bucket, if the cost of metal sheet used is Rs. 15 per 100 cm2.

    Solution

    Let R and r be respectively the radii of bigger and smaller ends of the frustum
    Then, R = 20 cm and r = 8 cm
    Let h and l be the height and slant height of frustum, then
                   h = 16
    and space space space space space space space l italic space equals space square root of straight h squared plus left parenthesis straight R minus straight r right parenthesis squared end root space space space space space space space space space space space space space space equals space square root of left parenthesis 16 right parenthesis squared plus left parenthesis 20 minus 8 right parenthesis squared end root space space space space space space space space space space space space space space equals space square root of 256 plus 144 end root space space space space space space space space space space space space space space equals space square root of 400 equals 20 space cm
    Now,
    Curved surface area of the frustum
              equals space πl space left parenthesis straight R plus straight r right parenthesis equals space 22 over 7 straight x 20 space left parenthesis 20 plus 8 right parenthesis equals space 22 over 7 space straight x space 20 space straight x space 28 equals space 1760 space cm squared
    Total metal sheet required
             = C. S. A. + Area of base
           equals space 1760 space plus space 22 over 7 space straight x space 8 space straight x space 8
            = 1760 + 201.14
            = 1961.14 cm2
    Cost of metal sheet used
           equals space 1961.14 space straight x space 15 over 100

           = Rs 294.17

     

    Question 138
    CBSEENMA10008942
    Wired Faculty App

    The height of a cone is 30 cm. A small cone is cut off at the top by a plane parallel to the base. If its volume be 1 over 27 of the volume of the given cone, at which height above the base is the section made?

    Solution

    Let r and R be respectively the radii of smaller and bigger cone and h and H be their respective heights.
    Now,   Volume of smaller cone = 1 third πr squared straight h
    and     Volume of given cone = 1 third πR squared straight h
    It is given that :
    Volume of smaller cone
    equals 1 over 27 th space of the volume of the given cone
    rightwards double arrow space fraction numerator Volume space of space smaller space cone space over denominator Volume space of space the space given space cone end fraction equals 1 over 27 rightwards double arrow space fraction numerator begin display style 1 third end style πr squared straight h over denominator begin display style 1 third end style πR squared straight H end fraction equals 1 over 27 rightwards double arrow space space straight r squared over straight R squared space straight x space straight h over 30 equals 1 over 27 space space space space space space space space space open square brackets table row cell table row cell because space increment ABC end cell cell negative increment ADE end cell end table end cell row cell rightwards double arrow space straight r over straight R equals straight h over 30 end cell end table close square brackets rightwards double arrow space open parentheses straight h over 30 close parentheses cubed straight x straight h over 30 equals 1 over 27 rightwards double arrow space space open parentheses straight h over 30 close parentheses cubed straight x straight h over 30 equals 1 over 27 rightwards double arrow space space straight h over 30 equals 1 third
    rightwards double arrow     h = 10 cm
    Hence,  height of the frustum
    = 30 cm - 10 cm
    = 20 cm

     

    Question 139
    CBSEENMA10008943
    Wired Faculty App

    A hollow cone is cut by a plane parallel to the base and the upper portul is removed. If the curved surface of the remainder is  8 over 9 of the curved surface of the whole cone, find the ratio of the line-segment into which the cones altitude is divided by the plane.

    Solution

    Let r and R be respectively the radii of upper portion of cone and whole cone, and h and H be their respective heights.
    And let l and L be respectively the slant heights of upper portion of cone and whole cone.
    It is given that :
    fraction numerator Curved space surface space of space remaining space portion over denominator Curved space surface space of space the space whole space cone end fraction equals 8 over 9 therefore space space fraction numerator table row cell table row cell table row cell Curved space surface space of space the space upper end cell row cell portion space of space the space cone end cell end table end cell end table end cell end table over denominator Curved space portion space of space the space whole space cone end fraction equals 1 over 9 rightwards double arrow space space space space space space πrl over πRL equals 1 over 9 rightwards double arrow space space space space space straight r over straight R straight x space straight l over straight L equals 1 over 9 rightwards double arrow space space straight h over straight H straight x straight h over straight G equals 1 over 9 space space open square brackets table row cell because space increment ABC space minus space increment ADE end cell row cell rightwards double arrow space straight r over straight R equals straight l over straight L equals straight h over 30 end cell end table close square brackets rightwards double arrow space space space open parentheses straight h over straight H close parentheses squared equals open parentheses 1 third close parentheses squared rightwards double arrow space space straight h over straight H equals 1 third rightwards double arrow space space space straight H space equals space 3 straight h therefore space Required space height space equals space fraction numerator Height space of space upper space portion over denominator Height space of space the space frustum end fraction rightwards double arrow space space space space fraction numerator straight h over denominator 3 straight h minus straight h end fraction equals fraction numerator straight h over denominator 2 straight h end fraction equals 1 half.

    Question 140
    CBSEENMA10008944
    Wired Faculty App

    If the radii of the circular ends of a conical bucket, which is 16 cm high, are 20 cm and 8 cm, find the capacity and total surface area of the bucket.

    Solution

    Let R and r be the radii of top of bottom circular ends of a conical bucket, respectively and ‘h’ be the height of the bucket.
    ∴ h =16 cm, R = 20 cm,
    r = 8 cm
    Capacity of the conical bucket = Volume of the bucket
    equals space 1 third πrh space left parenthesis straight R squared plus straight r squared plus Rr right parenthesis equals space 1 third straight x 22 over 7 straight x space 16 space straight x space left parenthesis 20 squared plus 8 squared plus 20 space straight x space 8 right parenthesis space cm cubed space space space space space space space space space space space space space left square bracket because space straight h space equals space 16 comma space straight R space equals space 20 space cm comma space straight r space equals space 8 space cm right square bracket equals space 1 third space straight x space 22 over 7 space straight x space 16 space left parenthesis 400 space plus space 64 space plus space 160 right parenthesis space cm cubed equals space 1 third space straight x space 22 over 7 space straight x space 16 space left parenthesis 624 right parenthesis space cm cubed equals space fraction numerator 352 space straight x space 208 over denominator 7 end fraction space cm cubed minus 73216 over 7 space cm cubed
    = 10459.428 cm3
    = 10459.43 cm3
    Slant height of the bucket is given by
    l italic space equals space square root of straight h squared left parenthesis straight R minus straight r right parenthesis squared end root equals space square root of left parenthesis 16 right parenthesis squared plus left parenthesis 20 minus 8 right parenthesis squared end root equals space square root of 256 plus 144 end root space cm space equals space square root of 400 space cm space equals space 20 space cm
    Total surface area of the conical bucket
    = Curved surface area of the conical bucket
    + Area of the bottom
    equals space left parenthesis straight R space plus straight r right parenthesis space straight l space plus space πr squared equals space open square brackets 22 over 7 straight x left parenthesis 20 plus 8 right parenthesis space straight x space 20 plus 22 over 7 space straight x space left parenthesis 8 right parenthesis squared close square brackets space cm squared equals space 22 over 7 space left square bracket 28 space straight x space 20 space plus space 64 right square bracket space cm squared equals space 22 over 7 space straight x space left square bracket 560 space plus space 64 right square bracket space cm squared equals space 22 over 7 space straight x space left square bracket 624 right square bracket space cm squared equals space 13728 over 7 space cm squared space equals space 1961.1428 space cm squared equals space 1961.14 space cm squared

     

    Question 141
    CBSEENMA10008945
    Wired Faculty App

    A bucket is in the form of a frustum of a cone and holds 28.49 litres of milk. The radii of the top and bottom are 28 cm and 21 cm. Find the height of the bucket.

    Solution

    Let ‘R’ be the radius ‘r’ be the bottom of the frustum of cone. Then,
    R = 28 cm, r = 21 cm.
    let ‘h’ cm be the height of the frustum of a cone.
    Now,
    Volume of frustum of cone = (28.49 x 1000) cm3
    equals space 1 third πrh space left parenthesis straight R squared plus straight r squared plus Rr right parenthesis space equals space 28490 equals space 1 third space straight x space 22 over 7 space straight x space straight h space left square bracket left parenthesis 28 right parenthesis squared plus left parenthesis 21 right parenthesis squared space plus space 28 space straight x space 21 right parenthesis right square bracket space equals space 28490 equals space 1 third space straight x space 22 over 7 space straight x space straight h space left square bracket 784 plus 441 plus 588 right square bracket space equals space 28490 equals space 1 third space straight x space 22 over 7 space straight x space straight h space left square bracket space 1813 space right square bracket space equals space 28490 equals space straight h space equals space fraction numerator 28490 space straight x space 3 space straight x space 7 over denominator 22 space straight x space 1813 end fraction space equals space 15 space cm space
    Hence, Height of bucket = 15 cm.

    Question 142
    CBSEENMA10008946
    Wired Faculty App

    A bucket of height 8 cm made up of copper sheets is in the form of frustum of a right circular cone with radii of its lower ends as 3 cm and 9 cm respectively. Calculate
    (i) the height of the cone of which the bucket is a part.
    (ii) the volume of water which can be filled in the bucket.
    (iii) the area of copper sheet required to make the bucket.
    Solution
    Let h be the height, l the slant height and r and R the radii of the smaller and bigger ends of frustum of a cone, then
    h = 8 cm, r = 3 cm and R = 9 cm
    (i) Let h1 be the height of the cone of which the bucket is a part. Then,
                              space space space space space space space straight h subscript 1 equals fraction numerator hR over denominator straight R minus straight r end fraction rightwards double arrow space space straight h subscript 1 equals open parentheses fraction numerator 8 space straight x space 9 over denominator 9 minus 3 end fraction close parentheses space cm space equals space 12 space cm
    (ii)    Volume of the water which can be filled in the bucket
    = Volume of the frustum
    equals space straight pi over 3 space left parenthesis straight R squared plus straight R space. space straight r space plus space straight r squared right parenthesis space straight h equals space straight pi over 3 left parenthesis 9 squared plus 9 space straight x space 3 space plus space 3 squared right parenthesis space straight x space 8 equals space 312 space cm cubed
    (iii)    Area of the copper sheet required to make the bucket
    equals space straight pi space equals space left parenthesis straight R space plus straight r right parenthesis space 1 plus πr squared equals space straight pi space left parenthesis 9 plus 3 right parenthesis space straight x space 10 space plus straight pi space straight x space 3 squared equals space straight pi space left parenthesis 120 plus 9 right parenthesis space cm squared equals space 129 straight pi space cm squared left square bracket because space straight l space equals space square root of left parenthesis straight R minus straight r right parenthesis squared plus straight h squared end root space equals square root of left parenthesis 9 minus 3 right parenthesis squared plus 64 end root space equals space square root of 36 plus 64 end root space equals space square root of 100 space equals space 10 space cm space right square bracket
    Question 143
    CBSEENMA10008947
    Wired Faculty App

    A container (open at the top) made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find
    (i)    the cost of milk when it is completely filled with milk at the rate of Rs. 15 per litre.
    (ii)    the cost of metal sheet used, if it costs Rs. 5 per 100 cm2.    (Take π = 3.14)

    Solution

    Let l be the slant height of the frustum. We have r = 8 cm,
    R = 20 cm and h = 16 cm

    therefore space space space l italic space space equals space space square root of left parenthesis straight R minus straight r right parenthesis squared plus straight h squared end root space space space space space space space space equals space space square root of left parenthesis 20 minus 8 right parenthesis squared plus left parenthesis 16 right parenthesis squared end root space space space space space space space space equals space space space square root of left parenthesis 12 right parenthesis squared plus left parenthesis 16 right parenthesis squared end root equals square root of 144 plus 256 end root space space space space space space space space equals square root of 400 space equals space 20 space cm. space space
    Now, volume of the container
              equals straight pi over 3 left parenthesis straight r squared plus straight R squared plus straight R. straight r right parenthesis space straight x space straight h equals space straight pi over 3 space left curly bracket 8 squared plus 20 squared plus 20 space straight x space 8 right curly bracket space straight x space 16 space cm cubed equals space fraction numerator 3.14 space straight x space 624 space straight x space 16 over denominator 3 end fraction space cm cubed
    rightwards double arrow Volume of container = 10449.92 cm3
    rightwards double arrow Volume of container = fraction numerator 10449.92 over denominator 1000 end fraction space litres

    rightwards double arrow Volume of container = 10.44992 litres
       = 10.45 litres (Approx)

    ∴ Cost of milk at the rate of Rs. 15 per litre = Rs.
    (10.45 x 15) = Rs. 156.75
    Now, Total surface area of the frustum = π (R + r) l + straight pir2
    = {3.14 (20 + 8) x 20 + 3.14 x 82} cm2
    = 3.14 x (560 + 64) cm3
    = (3.14 x 624) cm3 = 1959.36 cm2
    therefore Cost of metal used = Rs. open parentheses fraction numerator 1959.36 space straight x space 5 over denominator 100 end fraction close parentheses
    = Rs. 97.96 (Approx)



    Question 144
    CBSEENMA10008948
    Wired Faculty App

    An open container made up of a metal sheet is in the form of a frustum of a cone of height 8 cm with radii of its lower and upper ends as 4 cm and 10 cm respectively.
    Find the cost of oil which can completely fill the container at the rate of Rs. 50 per litre. Also, find the cost of metal used, if it costs Rs. 5 per 100 cm2 (Use π = 3.14).

    Solution

    Let ‘l’ be the slant height of the frustum.
    We have, R = 10 cm, r = 4 cm and h = 8 cm
    therefore space space space l space space equals space square root of left parenthesis straight R minus straight r right parenthesis squared plus straight h squared end root space space space space space space space space space equals space square root of left parenthesis 10 minus 4 right parenthesis squared plus 8 squared end root space space space space space space space space space equals space square root of 36 plus 64 end root equals square root of 100 equals 10 space cm.
    Now, volume of container
    equals straight pi over 3 left curly bracket straight R squared plus straight r squared straight R. straight r right curly bracket space straight h equals space straight pi over 3 left curly bracket left parenthesis 10 right parenthesis squared plus left parenthesis 4 right parenthesis squared plus 10 space straight x space 4 right square bracket space straight x space 8 equals space fraction numerator 3.14 over denominator 3 end fraction left curly bracket 100 plus 16 plus 40 right curly bracket space straight x space 8 equals space open parentheses 3.14 space straight x space 52 space straight x space 8 close parentheses space cm cubed equals 1.30624 space litres space left parenthesis approx right parenthesis

    ∴ Cost of oil at the rate of Rs. 50 per litre
    = Rs. (1.30624 x 50) = Rs. 65.312.

    Now, Total surface area of frustum
    = straight pi (R + r) l + straight pir2 [∵ Top is open]
    = 3.14 (10 + 4) (10) + 3.14 (4) (4)
    = 3.14 x 140 + 3.14 x 16
    = 3.14 (140 + 16) = 3.14 x 156 = 489.84 cm2
    Thus the cost of metal used

    equals space Rs. space open parentheses 489.84 space straight x space 5 over 100 close parentheses = Rs. 24.49 (Approx).

    Question 145
    CBSEENMA10008949
    Wired Faculty App

    A bucket made up of a metal sheet is in the form of a frustum of a cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the volume of the bucket. Also find the cost of the bucket if the cost of metal sheet used is Rs. 20 per 100 cm2. (Use straight pi = 3.14)
    Solution

    Let R and r be respectively the radii of bigger and smaller ends of the frustum
    Then, R = 20 cm and r = 8 cm
    Let h and l be the height and slant height of frustum, then

    h = 16 cm
    and space space space space space straight l space equals space square root of straight h squared plus left parenthesis straight R minus straight r right parenthesis squared end root space space space space space space space space space space space space equals space square root of left parenthesis 16 right parenthesis squared plus left parenthesis 20 minus 8 right parenthesis squared end root space space space space space space space space space space space space equals space square root of 256 space plus space 144 end root space equals space square root of 400 space equals space 20 space cm
    Now, Curved surface area of thefrustum
    equals space πl space left parenthesis straight R plus straight r right parenthesis equals space 22 over 7 space straight x space 20 space left parenthesis 20 plus 8 right parenthesis equals space 22 over 7 space straight x space 20 space straight x space 28 space equals space 1760 space cm squared

    Total metal sheet required
    = C.S.A. + Are of base
    = 1760 + 22 over 7 x 8 x 8
    = 1760 + 201.14 = 1961.14 cm2 
    Cost of metal sheet used
    = 1961.14 x 20 over 100

    =  Rs. 383.23 (Approx)

    Question 146
    CBSEENMA10008950
    Wired Faculty App

    From a solid cylinder whose height is 8 cm and radius 6 cm, a conical cavity of height 8 cm and of base radius 6 cm, is hollowed out. Find the volume of the remaining solid correct to two places of decimals. Also find the total surface area of the remaining solid.
    (Take π = 3.1416).

    Solution
    Volume of the remaining solid Volume of the cylinder Volume of the cone
    equals open square brackets straight pi space straight x space 6 squared space straight x space 8 minus 1 third straight x space straight pi space straight x space 6 squared space straight x 8 close square brackets space cm cubed equals space 2 over 3 space straight x space 3.1416 space straight x space 36 space straight x space 8 space cm cubed equals space 192 space straight x space 3.1416 space cm cubed equals space 603.1872 space cm cubed equals 603.19 space cm cubed
    Slant height of the cone
    AB thin space equals space square root of BC squared plus AC squared end root equals square root of left parenthesis 8 right parenthesis squared plus left parenthesis 6 right parenthesis squared end root equals square root of 64 plus 36 end root equals space 10 space cm

    Now, total surface area of the remaining solid
    = Curved surface area of the cylinder +
    area of the base of the cylinder + curve
    surface area of the cone
    = (2straight pi x 6 x 8 + straight pi x 62 + straight pi x 6 x 10) cm2
    = (96 straight pi + 36 straight pi + 60straight pi) cm2
    = 192straight pi 192 x 3.1416
    = 603.1872 = 603.19 cm3

    Question 205
    CBSEENMA10009009
    Wired Faculty App

    In fig. AB is a diameter of the circle with centre O and OA = 7 cm. Find the area of the shaded region  open parentheses Use space straight pi space equals space 22 over 7 close parentheses.

    Solution

    Solution is not provided.
    Ans. 66.5 cm2

    Question 262
    CBSEENMA10009066
    Wired Faculty App

    Find the area of the shaded region in the given figure, where ABCD is a square of side 14 cm.

    Solution

    Solutionn not provided.
    Ans. 42 cm2 

    Question 277
    CBSEENMA10009081
    Wired Faculty App

     If the diameter of a semicircular protector is 14 cm. Find its perimeter.
    Solution

    Solutionn not provided.
    Ans. 36 cm.

    Question 294
    CBSEENMA10009594
    Wired Faculty App

    In Fig., a tent is in the shape of a cylinder surmounted by a conical top of the same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs. 500/sq. metre. (use π = 22/7)
    Solution

    In the given figure,

    Given diameter 3m
    For conical portion, 
    radius = 3/2 = 1.5m and length is 2.8m
    ∴ S1 = curved surface area of conical portion
    ∴ S1 = πrl
       = π x 1.5 x 2.1
    = 6.3π m2
    Area of canvas used for making tent = S1+S2
     =4.2π +6.3π
    = 10.5π
    =10.5 x (22/7)
    =33m2
    Thus, total cost of the canvas at the rate of Rs. 500 per m2 = 500 x 33 =16500
    Rs. 16500 is total cost.
    Question 295
    CBSEENMA10009599
    Wired Faculty App

    A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. (π = 22/7)
    Solution

    Let the radius of the conical vessel = r1 = 5 cm
    Height of the conical vessel = h1 = 24 cm
    Radius of the cylindrical vessel = r2
    Let the water rise upto the height of h2 cm in the cylindrical vessel.
    Now, volume of water in conical vessel = volume of water in cylindrical vessel
    therefore space 1 third straight pi subscript 1 straight r squared straight h subscript 1 space equals space straight pi subscript 2 straight r squared straight h subscript 2 therefore space straight r subscript 1 superscript 2 straight h space equals 3 straight r subscript 2 superscript 2 straight h subscript 2 superscript space therefore space 5 space straight x space 5 space straight x space 24 space equals space 3 space straight x space 10 space straight x space 10 space straight x space straight h subscript 2 therefore straight h subscript 2 space equals space fraction numerator 5 straight x space 5 straight x 24 over denominator 3 straight x 10 space straight x 10 end fraction space equals 2 space cm Thus comma space the space water space will space rise space upto space the space height space of space 2 space cm space in space the space cylinderical space vessel.

    Question 299
    CBSEENMA10009630
    Wired Faculty App

    Due to sudden floods, some welfare associations jointly requested the government to get 100 tents fixed immediately and offered to contribute 50% of the cost. If the lower part of each tent is of the form of a cylinder of diameter 4.2 m and height 4 m with the conical upper part of same diameter but of height 2.8 m, and the canvas to be used costs Rs. 100 per sq. m, find the amount, the associations will have to pay. What values are shown by these associations? [ use π = 22/7]
    Solution

    Height of the cylindrical part, H = 4 m

    Height of the conical part, h = 2.8 m
    Radius of the base, r = 2.1 m
    So comma space straight l equals space square root of straight r squared space plus space straight h squared end root rightwards double arrow space straight l space equals space square root of left parenthesis 2.1 right parenthesis squared space plus left parenthesis 2.8 right parenthesis squared end root rightwards double arrow space straight l space equals space square root of 12.25 end root rightwards double arrow space 1 space equals space 3.5 space straight m surface space area space of space canvas space to space be space used space space equals space CSA space of space conical space part space plus space CSA space of space cylindrical space part space equals space πrl space plus space 2 πrH space equals space πr space left parenthesis straight l plus 2 straight H right parenthesis space equals space 22 over 7 space straight x space 2.1 space straight x space left parenthesis 3.5 space plus 8 space right parenthesis equals space 75. space 9 space straight m squared
    Cost of the canvas = Rs 100/ m2

    So, total cost of the canvas to be used = 75.9 x 100 = Rs 7,590
    therefore,
    Amount that the association will have to pay for each tent
    =50% of Rs 7590
    = Rs (50/100) x 7590
    = Rs 3,795
    Therefore, Amount that the associations will have for 100 tent = Rs  3795 x 100
    = Rs 379500
    Hence, the amount that the associations will have to pay for 100 tents is Rs 3,79,500
    The associations are showing humanity and their helpful nature.
    Question 300
    CBSEENMA10009631
    Wired Faculty App

    A hemispherical bowl of internal diameter 36 cm contains liquid. This liquid is filled into 72 cylindrical bottles of diameter 6 cm. Find the height of the each bottle, if 10% liquid is wasted in this transfer.

    Solution
    Radius space of space the space hemispherical space bowl comma space straight r space equals space 36 over 2 space equals space 18 space cm Volume space of space the space hemispherical space bowl space equals space 2 over 3 πr cubed space equals space 2 over 3 space straight x space straight pi space straight x space 18 space straight x space 18 space straight x space 18 space equals space 3888 space straight pi space cm cubed Volume space of space the space liquid space transferred space space equals space 3888 space straight pi space minus 10 percent sign space of space 3888 space straight pi space equals space 3888 space straight pi space minus space 388.8 space straight pi space equals space 3499.2 space straight pi space cm cubed
    Radius of each cylindrical bottle, R = 6/2 = 3 cm
    Let the height cylindrical bottle be h
    volume of each cylindrical bottle = πR2h
     = π x 3 x3 x h
    Therefore, Total volume of 72 such cylindrical bottles  
    = π x 3 x3 x h x 72
     = 648 πh2 cm3
    According to the question,
    Total volume of 72 such cylindrical bottles = volume of the liquid transferred 
    Therefore, 648 πh = 3499.2π
    ⇒ h = 3499.2/648
    ⇒  h = 5.4  cm
    Hence, the height of each cylindrical bottle is 5.4 cm
    Question 301
    CBSEENMA10009632
    Wired Faculty App

    A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have ? Find the cost of painting the total surface area of the solid so formed, at the rate of Rs 5 per 100 sq. cm. [Use π =3.14]

    Solution

    Now,
    Total surface area of the solid  = Total surface area of the cube + curved surface area of the hemisphere - Area of the base of the hemisphere
     = 6a2 + 2πr2 -πr2
    = [ 6 x (10)2 + 2 x 3.14 x (5)2 - 3.14 x (5)2] cm2
     = (600 +157 -78.5 ) cm2
    = 678.5 cm2
    Cost of painting  = Rs 5 per 100 cm2
    Therefore, 
    Cost of painting the solid = 678.5 x (5/100) = Rs. 33.90
    Hence, the approximate cost of painting the solid so formed is Rs. 33.90

    Question 302
    CBSEENMA10009633
    Wired Faculty App

    504 cones, each of diameter 3.5 cm and height 3 cm, are melted and recast into a metallic sphere. Find the diameter of the sphere and hence find its surface area. [Use π =22/7] 

    Solution

    Given,
    Diameter of the cone = 3.5 cm
    ∴ Radius of the cone, r= 3.5 /2 cm
    Height of the cone, h = 3 cm
    Now, 
      volume space of space each space cone space equals space fraction numerator 1 over denominator 3 space end fraction πr squared straight h space equals space 1 third space straight x space space 22 over 7 space straight x space open parentheses fraction numerator 3.5 over denominator 2 end fraction close parentheses squared space straight x space 3 9.625 space cm squared Therefore comma space volume space of space 504 space space cone space equals space 504 space straight x space 9.625 space equals space 4851 space cm cubed
    Let the radius of metallic sphere be R
    It is given that cones are melted and recast into a metallic sphere.
    therefore the volume of the sphere = Volume of 504 cones
    rightwards double arrow space 4 over 3 straight pi space straight R cubed space equals space 4851 rightwards double arrow space 4 over 3 space straight x space 22 over 7 space straight x space straight R cubed space equals space 4851 rightwards double arrow space straight R cubed space equals space fraction numerator 4851 space straight x space 7 space straight x space 3 over denominator 4 space straight x space 22 end fraction rightwards double arrow space straight R cubed space equals space 1157.625 straight R space equals space 10.5 space cm Diameter space of space the space sphere space equals space 2 space straight R space equals space 2 space straight x space 10.5 space cm space space equals space 21 space cm Therefore comma surface space space area space of space the space sphere space space equals space space 4 straight pi space straight R squared space space equals space 4 space straight x space 22 over 7 space straight x space left parenthesis 10.5 right parenthesis squared space equals space 1386 space cm squared

    Question 303
    CBSEENMA10009643
    Wired Faculty App

    In Figure, PQRS is a square lawn with side PQ = 42 metres. Two circular flower beds are there on the sides PS and QR with centre at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts).

    Solution

    Area of the square lawn PQRS = 42 m x 42 m
    Let OP = OS = xm
    So, x2 + x2  = (42)2
    ⇒ 2x2 = 42 x 42
    ⇒ x2 = 21 x 42
    Now,
    area of sector POS=
      90 over 360 space x space πx squared space equals space 1 fourth space straight x space left parenthesis πx right parenthesis squared space.... space left parenthesis space straight i space right parenthesis equals 1 fourth space space straight x space 22 over 7 space straight x space 21 space straight x space 42 space straight m squared space.... space left parenthesis ii right parenthesis Also comma space Area space of space increment POS space equals space 1 fourth space straight x space Area space of space square space lawn space PQRS space equals space 1 fourth space straight x space left parenthesis space 42 space straight x space 42 right parenthesis squared space straight m squared space space space space space left parenthesis angle POQ space equals space 90 degree right parenthesis space.. space left parenthesis iii right parenthesis So comma space Area space of space flower space bed space PSP space equals space Area space of space Sector space POS space minus space Area space of space increment POS space equals space 1 fourth space straight x space 22 over 7 space straight x space 21 space straight x 42 space minus space 1 fourth space straight x space 42 space straight x space 42 space space left square bracket frome space left parenthesis ii right parenthesis space and space left parenthesis iii right parenthesis right square bracket space equals space 1 fourth space straight x space 21 space straight x space 42 space straight x open parentheses 22 over 7 space minus space 2 close parentheses equals space 1 fourth space straight x space 21 space straight x space 42 space straight x space open parentheses 8 over 7 close parentheses straight m squared Therefore space area space of space the space two space flower space beds space equals space 2 space straight x space 1 fourth space straight x space 21 space straight x space 42 space straight x open parentheses space 8 over 7 close parentheses space space equals space 504 space straight m squared Hence comma space the space total space area space of space the space two space flower space beds space 504 space straight m squared

    Question 304
    CBSEENMA10009644
    Wired Faculty App

    From each end of a solid metal cylinder, metal was scooped out in the hemispherical form of the same diameter. The height of the cylinder is 10 cm and its base is of radius 4.2 cm. The rest of the cylinder is melted and converted into a cylindrical wire of 1.4 cm thickness. Find the length of the wire. ( π = 22/7)

    Solution

    Height of the solid metal cylinder, h = 10 cm
    Radius of the solid metal cylinder, r = 4.2 cm
    therefore,
    Radius of each hemisphere = radius of the solid metal cylinder, r = 4.2 cm
    Now, 
    Volume of the rest of the cylinder = Volume of cylinder - 2 x volume of each hemisphere
      space equals space πr squared straight h space minus space 2 space straight x space 2 over 3 space πr cubed space equals space πr squared space open parentheses straight h space minus 4 over 3 straight r close parentheses equals straight pi space straight x space left parenthesis 4.2 space right parenthesis squared space open parentheses 10 space minus space 4 over 3 space straight x space 4.2 close parentheses space equals straight pi space straight x space left parenthesis 4.2 right parenthesis squared space straight x space left parenthesis 4.4 right parenthesis space cm cubed
    Thickness of the cylindrical wire = 1.4 cm
    Therefore, 
    Radius of the cylindrical wire, R = 1.4/2 = 0.7 cm
    Let the length of the wire be H cm.
    It is given that the rest of the cylinder is melted and converted into a cylindrical wire.
    therefore,
    Volume of the cylindrical wire = Volume of the rest of the cylinder
    ⇒ π x 0.7 x 0.7 x H = π x (4.2)2 x (4.4)
    rightwards double arrow space straight H space equals space fraction numerator 4.2 space straight x space 4.2 space straight x space 4.4 over denominator 0.7 space straight x space 0.7 end fraction space equals space 158.4 space cm
    Hence, the length of the wire is 158.4 cm

    Question 305
    CBSEENMA10009697
    Wired Faculty App

    The table below shown the salaries of 280 persons:

    Salary (In thousand) No. of Person
    5-10 49
    10-15 133
    15-20 63
    20-25 15
    25-30 6
    30-35 7
    35-40 4
    40-45 2
    45-50 1

    Calculate the median salary of the data.
    Solution

    Let N  = total frequency

    ∴ we have N = 280∴ N2 = 2802 = 140

    The cumulative frequency just greater than N/2 is 182 and the corresponding class is 10-15

    Thus 10-15 is the median class such that

    l = 10
    f =133
    F = 49
    h = 5
    Median  = l + N2-Ff x h  = 10 + 140-49133 x 5 = 13.42

    Class Frequency (F) Cumulative Frequency (f)
    5 -10 49 49
    10-15 133 182
    15-20 63 245
    20-25 15 260
    25-30 6 266
    30-35 7 273
    35-40 4 277
    40-45 2 279
    45-50 1 280
    Question 306
    CBSEENMA10009706
    Wired Faculty App

    The mean of the following distribution is 18. Find the frequency f of the class 19-21.

    Class 11-13 13-15 15-17 17-19 19-21 21-23 23-25
    Frequency 3 6 9 13 f 5 4

    Solution

    Σfiui = f-8we haveh = 2; A = 18,N = 40 +fΣfiui = f- 8X  = 18∴ Mean  = A  +h1NΣfiui18 = 18 + 2 140+ f(f-8)2 (f-8)40 +f = 0f- 8 = 0f = 8

    Class Mid values xi Frequency fi di = xi -18 ui = xi -182 fiui
    11-13 12 3 -6 -3 9
    13-15 14 6 -4 -2 -12
    15-17 16 9 -2 -1 -9
    17-19 18 13 0 0 0
    19-21 20 f 2 1 f
    21-23 22 5 4 2 10
    23-25 24 4 6 3 12
        Σ fi = 40 +f      
    Question 307
    CBSEENMA10009707
    Wired Faculty App

    The following distribution gives the daily income of 50 workers of a factory:

    Daily Income (In) 100-120 120-140 140-160 160-180 180-200
    Number of workers 12 14 8 6 10

    Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
    Solution

    Other than the given class intervals, we assum a class interval 80-100 with zero frequency

    Daily income Frequency Income less than  cumulative frequency
    100-120 12 120 12
    120-140 14 140 26
    140-160 8 160 34
    160-180 6 180 40
    180-200 10 200 50

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