Question-1 :-  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. (Take π = 22/7)

    
    Given that:-
    Height(h) of a frustum of a cone = 14 cm
    The diameters of its two circular ends are 4 cm and 2 cm.
    Radius(r1) of upper base of the glass = 4/2 = 2 cm
    Radius(r2) of lower base of the glass = 2/2 = 1 cm

    Therefore, Capacity of glass (Volume of frustum of cone) = 1/3 x πh x (r12 + r22 + r1r2)
    = 1/3 x 22/7 x 14 x [(2 x 2) + (1 x 1) + (2 x 1)]
    = 1/3 x 22/7 x 14 x [4 + 1 + 2]
    = 1/3 x 22/7 x 14 x 7
    = 308/3
    = 102.66 cm3
    Therefore, the capacity of the glass is 102.66 cm3.
    

Question-2 :-  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. (Take π = 22/7)

    
    Given that:-
    Slant height of a frustum of a cone = 4 cm
    Perimeter of upper circular end of frustum = 18 cm
    2πr1 = 18
    r1 = 18/2π
    r1 = 9/π cm
    Perimeter of lower end of frustum = 6 cm
    2πr2 = 6
    r2 = 6/2π
    r2 = 3/π cm

    Therefore, CSA of frustum = π(r1 + r2)l
    = π x (9/π + 3/π) x 4
    = π x 12/π x 4
    = 48 cm2
    Therefore, the curved surface area of the frustum is 48 cm2.
    

Question-3 :-  A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig.). 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. (Take π = 22/7)

    
    Given that:-
    Radius(r2) at upper circular end = 4 cm 
    Radius(r1) at lower circular end = 10 cm 
    Slant height (l) of frustum = 15 cm

    Therefore, Area of material used for making the fez = CSA of frustum + Area of upper circular end
    = π[r1 + r2]l + πr22
    = 22/7 x [10 + 4] x 15 + 22/7 x 4 x 4
    = 22/7 x 14 x 15 + 22/7 x 4 x 4
    = 22/7 x (210 + 16)
    = 22/7 x 226
    = 4972/7
    = 710.28 cm2
    Therefore, the area of material used for making it is 710.28 cm2.
    

Question-4 :- 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 Rs. 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm². (Take π = 3.14)

    
    Given that:-
    Radius(r1) of upper end of container = 20 cm 
    Radius(r2) of lower end of container = 8 cm 
    Height(h) of container = 16 cm
    Slant height(l) of frustum = √(r1 - r2)² + h2²
    = √(20 - 8)² + 16²
    = √144 + 256
    = √400
    = 20 cm

    Therefore, Capacity of container (Volume of frustum) = 1/3 x πh x (r12 + r22 + r1r2)
    = 1/3 x 3.14 x 16 x [400 + 64 + (20 x 8)]
    = 1/3 x 3.14 x 16 x [464 + 160]
    = 1/3 x 3.14 x 16 x 624
    = 10449.92 cm3
    = 10.45 Litres

    Cost of 1 litre milk = Rs 20
    Cost of 10.45 litre milk = 10.45 × 20 = Rs 209
    Area of metal sheet used to make the container = π[r1 + r2]l + πr22
    = π(20 + 8) x 20 + π(8)2
    = π[(28 x 20) + 64]
    = π[560 + 64]
    = 624π cm2
    Cost of 100 cm2 metal sheet = Rs 8
    Cost of 624π cm2 metal sheet = (624 x 3.14 x 8)/100 = Rs 156.75
    Therefore, the cost of the milk which can completely fill the container is Rs 209 
    The cost of metal sheet used to make the container is Rs 156.75.
    

Question-5 :-  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 1/16 cm, find the length of the wire. (Take π = 22/7)

    
    Given that:-
    Height of right circular cone = 20 cm

    In ΔAEG,
    EG/AG = tan 30o
    EG/AG = 1/√3
    EG = AG/√3
    EG = 10/√3 = 10√3/3 cm

    In ΔABD,
    BD/AD = tan 30o
    BD/AD = 1/√3
    BD = AD/√3
    BD = 20/√3 = 20√3/3 cm

    Radius(r1) of upper end of frustum = 10√3/3 cm
    Radius(r2) of lower end of container = 20√3/3 cm
    Height(h) of container = 10 cm
    Volume of frustum = 1/3 x πh x (r12 + r22 + r1r2)
    = 1/3 x 22/7 x 10 x [(10√3/3)2 + (20√3/3)2 + (10√3/3 x 20√3/3)]
    = 1/3 x 22/7 x 10 x [100/3 + 400/3 + 200/3]
    = 220/21 x 700/3
    = 22000/9 cm3

    Now, Radius(r) of wire = 1/16 x 1/2 = 1/32 cm
    Volume of wire = Area of cross-section × Length 
    = πr2 x l
    = π(1/32)2 x l

    Volume of frustum = Volume of wire
    22000/9 = 22/7 x (1/32)2 x l
    7000/9 x 1024 = l
    l = 796444.44 cm = 7964.44 m 
    Therefore, The length of the wire = 7964.44 m