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  1. Circumference of the base of cylindrical vessel = 132 cm Height of vessel, h = 25 cm Let r be the radius of the cylindrical vessel. Step 1: Find the radius of vessel We know that, circumference of base = 2πr, so 2πr = 132 (given) r = (132/(2 π)) r = 66×7/22 = 21 Radius is 21 cm Step 2: Find the voluRead more

    Circumference of the base of cylindrical vessel = 132 cm

    Height of vessel, h = 25 cm

    Let r be the radius of the cylindrical vessel.

    Step 1: Find the radius of vessel

    We know that, circumference of base = 2πr, so

    2πr = 132 (given)

    r = (132/(2 π))

    r = 66×7/22 = 21

    Radius is 21 cm

    Step 2: Find the volume of vessel

    Formula: Volume of cylindrical vessel = πr2h

    = (22/7)×212×25

    = 34650

    Therefore, volume is 34650 cm3

    Since, 1000 cm3 = 1L

    So, Volume = 34650/1000 L= 34.65L

    Therefore, vessel can hold 34.65 litres of water.

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  2. Inner radius of cylindrical pipe, say r1 = diameter1/ 2 = 24/2 cm = 12cm Outer radius of cylindrical pipe, say r2 = diameter2/ 2 = 28/2 cm = 14 cm Height of pipe, h = Length of pipe = 35cm Now, the Volume of pipe = π(r22-r12)h cm3 Substitute the values. Volume of pipe = 110×52 cm3 = 5720 cm3 Since,Read more

    Inner radius of cylindrical pipe, say r1 = diameter1/ 2 = 24/2 cm = 12cm

    Outer radius of cylindrical pipe, say r2 = diameter2/ 2 = 28/2 cm = 14 cm

    Height of pipe, h = Length of pipe = 35cm

    Now, the Volume of pipe = π(r22-r12)h cm3

    Substitute the values.

    Volume of pipe = 110×52 cm3 = 5720 cm3

    Since, Mass of 1 cm3 wood = 0.6 g

    Mass of 5720 cm3 wood = (5720×0.6) g = 3432 g or 3.432 kg. Answer!

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  3. tin can will be cuboidal in shape Dimensions of tin can are Length, l = 5 cm Breadth, b = 4 cm Height, h = 15 cm Capacity of tin can = l×b×h= (5×4×15) cm3 = 300 cm3 Plastic cylinder will be cylindrical in shape. Dimensions of plastic can are: Radius of circular end of plastic cylinder, r = 3.5cm HeiRead more

    1. tin can will be cuboidal in shape

    Ncert solutions class 9 chapter 13-8

    Dimensions of tin can are

    Length, l = 5 cm

    Breadth, b = 4 cm

    Height, h = 15 cm

    Capacity of tin can = l×b×h= (5×4×15) cm3 = 300 cm3

    1. Plastic cylinder will be cylindrical in shape.

    Ncert solutions class 9 chapter 13-9

    Dimensions of plastic can are:

    Radius of circular end of plastic cylinder, r = 3.5cm

    Height , H = 10 cm

    Capacity of plastic cylinder = πr2H

    Capacity of plastic cylinder = (22/7)×(3.5)2×10 = 385

    Capacity of plastic cylinder is 385 cm3

    From results of (i) and (ii), plastic cylinder has more capacity.

    Difference in capacity = (385-300) cm3 = 85cm3

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  4. CSA of cylinder = 94.2 cm2 Height of cylinder, h = 5cm (i) Let radius of cylinder be r. Using CSA of cylinder, we get 2πrh = 94.2 2×3.14×r×5 = 94.2 r = 3 radius is 3 cm (ii) Volume of cylinder Formula for volume of cylinder = πr2h Now, πr2h = (3.14×(3)2×5) (using value of r from (i)) = 141.3 VolumeRead more

    CSA of cylinder = 94.2 cm2

    Height of cylinder, h = 5cm

    (i) Let radius of cylinder be r.

    Using CSA of cylinder, we get

    2πrh = 94.2

    2×3.14×r×5 = 94.2

    r = 3

    radius is 3 cm

    (ii) Volume of cylinder

    Formula for volume of cylinder = πr2h

    Now, πr2h = (3.14×(3)2×5) (using value of r from (i))

    = 141.3

    Volume is 141.3 cm3

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  5. (i) Rs 20 is the cost of painting 1 m2 area. Rs 1 is the cost to paint 1/20 m2 area So, Rs 2200 is the cost of painting = (1/20×2200) m2 = 110 m2 area The inner surface area of the vessel is 110m2. (ii) Radius of the base of the vessel, let us say r. Height (h) = 10 m and Surface area formula = 2πrhRead more

    (i) Rs 20 is the cost of painting 1 m2 area.

    Rs 1 is the cost to paint 1/20 m2 area

    So, Rs 2200 is the cost of painting = (1/20×2200) m2

    = 110 m2 area

    The inner surface area of the vessel is 110m2.

    (ii) Radius of the base of the vessel, let us say r.

    Height (h) = 10 m and

    Surface area formula = 2πrh

    Using result of (i)

    2πrh = 110 m2

    2×22/7×r×10 = 110

    r = 1.75

    Radius is 1.75 m.

    (iii) Volume of vessel formula = πr2h

    Here r = 1.75 and h = 10

    Volume = (22/7)×(1.75)2× 10 = 96.25

    Volume of vessel is 96.25 m3

    Therefore, the capacity of the vessel is 96.25 m3 or 96250 litres.

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  6. Height of cylindrical vessel, h = 1 m Capacity of cylindrical vessel = 15.4 litres = 0.0154 m3 Let r be the radius of the circular end. Now, Capacity of cylindrical vessel = (22/7)×r2×1 = 0.0154 After simplifying, we get, r = 0.07 m Again, total surface area of vessel = 2πr(r+h) = 2×22/7×0.07×(0.07+Read more

    Height of cylindrical vessel, h = 1 m

    Capacity of cylindrical vessel = 15.4 litres = 0.0154 m3

    Let r be the radius of the circular end.

    Now,

    Capacity of cylindrical vessel = (22/7)×r2×1 = 0.0154

    After simplifying, we get, r = 0.07 m

    Again, total surface area of vessel = 2πr(r+h)

    = 2×22/7×0.07×(0.07+1)

    = 0.44×1.07

    = 0.4708

    Total surface area of vessel is 0.4708 m2

    Therefore, 0.4708 m2 of the metal sheet would be required to make the cylindrical vessel.

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  7. Radius of pencil, r1 = 7/2 mm = 0.7/2 cm = 0.35 cm Radius of graphite, r2 = 1/2 mm = 0.1/2 cm = 0.05 cm Height of pencil, h = 14 cm Formula to find, volume of wood in pencil = (r12-r22)h cubic units Substitute the values, we have = [(22/7)×(0.352-0.052)×14] = 44×0.12 = 5.28 This implies, volume of wRead more

    Ncert solutions class 9 chapter 13-10

    Radius of pencil, r1 = 7/2 mm = 0.7/2 cm = 0.35 cm

    Radius of graphite, r2 = 1/2 mm = 0.1/2 cm = 0.05 cm

    Height of pencil, h = 14 cm

    Formula to find, volume of wood in pencil = (r12-r22)h cubic units

    Substitute the values, we have

    = [(22/7)×(0.352-0.052)×14]

    = 44×0.12

    = 5.28

    This implies, volume of wood in pencil = 5.28 cm3

    Again,

    Volume of graphite = r22h cubic units

    Substitute the values, we have

    = (22/7)×0.052×14

    = 44×0.0025

    = 0.11

    So, the volume of graphite is 0.11 cm3.

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  8. Diameter of cylindrical bowl = 7 cm Radius of cylindrical bowl, r = 7/2 cm = 3.5 cm Bowl is filled with soup to a height of4cm, so h = 4 cm Volume of soup in one bowl= πr2h (22/7)×3.52×4 = 154 Volume of soup in one bowl is 154 cm3 Therefore, Volume of soup given to 250 patients = (250×154) cm3= 3850Read more

    Diameter of cylindrical bowl = 7 cm

    Radius of cylindrical bowl, r = 7/2 cm = 3.5 cm

    Bowl is filled with soup to a height of4cm, so h = 4 cm

    Ncert solutions class 9 chapter 13-11

    Volume of soup in one bowl= πr2h

    (22/7)×3.52×4 = 154

    Volume of soup in one bowl is 154 cm3

    Therefore,

    Volume of soup given to 250 patients = (250×154) cm3= 38500 cm3

    = 38.5litres. Answer!

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  9. Let ABCD be a rhombus with AC and BD as its diagonals. We know that the diagonals of a rhombus bisect each other at right angles. Let O be the intersecting point of both the diagonals. Let AC=24cm and BD=32cm OA=AC/2 OA= 24/2=12cm OB=BD/2 OB=32/2=16cm In rt.ΔAOB by Pythagoras theorem we have AB²=OA²Read more

    Let ABCD be a rhombus with AC and BD as its diagonals.
    We know that the diagonals of a rhombus bisect each other at right angles.
    Let O be the intersecting point of both the diagonals.
    Let AC=24cm and BD=32cm
    OA=AC/2
    OA= 24/2=12cm
    OB=BD/2
    OB=32/2=16cm
    In rt.ΔAOB by Pythagoras theorem we have
    AB²=OA²+OB²
    =(12)²+(16)²
    =144+256
    =400
    AB=20cm
    Hence, each side of the rhombus is of length 20cm
    Area of rhombus=1/2*AC*BD
    =1/2*24*32
    =12*32
    Area of rhombus=base*altitude
    12*32=20*h
    19.2cm=h

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