A cylindrical jug of hight 24cm radius of base 7cm is full of milk to be served by the glasses in the shape of frustum of cone. Fi

Question

A cylindrical jug of hight 24cm radius of base 7cm is full of milk to be served by the glasses in the shape of frustum of cone. Find the number of glasses. That could be served if each glass is of hight 9cm and radii of its bases are 2cm and 1cm

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Kennedy 3 months 2021-07-17T13:54:59+00:00 1 Answers 0 views 0

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    2021-07-17T13:55:59+00:00

    \large\underline{\sf{Solution-}}

    Dimensions of cylindrical jug

    Radius of base of cylindrical jug, r = 7 cm

    Height of cylindrical jug, h = 24 cm

    So,

    Quantity of milk contained in cylindrical jug is equals to

    \rm :\longmapsto\:Volume_{(cylindrical \: jug)} = \pi \:  {r}^{2} \: h

    \rm :\longmapsto\:Volume_{(cylindrical \: jug)} = \dfrac{22}{7} \times 7 \times 7 \times 24

    \bf :\longmapsto\:Volume_{(cylindrical \: jug)} = 3696 \:  {cm}^{3}

    Dimensions of frustum shape glass

    Height of glass, H = 9 cm

    Radius of one end, x = 1 cm

    Radius of other end, y = 2 cm

    So,

    Quantity of milk contained in frustum shape glass is

    \rm :\longmapsto\:Volume_{(frustum \: glass)} = \dfrac{\pi \: H}{3}( {x}^{2}  +  {y}^{2}  + xy)

    \rm :\longmapsto\:Volume_{(frustum \: glass)} = \dfrac{22}{7} \times\dfrac{1}{3}  \times 9 \times ( {2}^{2}  +  {1}^{2}  + 2 \times 1)

    \rm :\longmapsto\:Volume_{(frustum \: glass)} = \dfrac{22}{7}  \times 3 \times 7

    \bf :\longmapsto\:Volume_{(frustum \: glass)} = 66 \:  {cm}^{3}

    Hence, number of glasses in which the whole quantity of milk in cylindrical jug be served in frustum shape glass is

    \rm :\longmapsto\:Number_{(glasses)} = \dfrac{Volume_{(cylindrical \: jug)}}{Volume_{(frustum \: glass)}}

    \rm :\longmapsto\:Number_{(glasses)} = \dfrac{3696}{66}

    \bf :\longmapsto\:Number_{(glasses)} = 56

    Additional Information :-

    Volume of cylinder = πr²h

    T.S.A of cylinder = 2πrh + 2πr²

    Volume of cone = ⅓ πr²h

    C.S.A of cone = πrl

    T.S.A of cone = πrl + πr²

    Volume of cuboid = l × b × h

    C.S.A of cuboid = 2(l + b)h

    T.S.A of cuboid = 2(lb + bh + lh)

    C.S.A of cube = 4a²

    T.S.A of cube = 6a²

    Volume of cube = a³

    Volume of sphere = 4/3πr³

    Surface area of sphere = 4πr²

    Volume of hemisphere = ⅔ πr³

    C.S.A of hemisphere = 2πr²

    T.S.A of hemisphere = 3πr²

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