2
S.A of
4. Rachel, an engineering student, was
asked to make a model shaped like a
cylinder with two cones at

1 thought on “2<br />S.A of<br />4. Rachel, an engineering student, was<br />asked to make a model shaped like a<br />cylinder with two cones at”

1. [tex]\large\underline{\sf{Solution-}}[/tex]

   Given :-

   • Diameter of conical part and cylindrical part, d = 3 cm.

   So,

   • Radius of conical part and cylindrical part, r = 1.5 cm

   Also,

   Total height of model = 12 cm

   Height of conical part, h = 2 cm

   So, Height of cylindrical part, H = 12 – 2 – 2 = 8 cm

   Now,

   We have to find the volume of air contained in the model.

   We know,

   [tex]\green{ \boxed{ \bf \: Volume_{(cylinder)} = \pi \: {r}^{2}h}} [/tex]

   and

   [tex]\green{ \boxed{ \bf \: Volume_{(cone)} = \frac{1}{3} \pi \: {r}^{2}h}} [/tex]

   ↝ Volume of air contained in model is

   [tex]\rm :\longmapsto\:Volume_{(air \: contained)} = Volume_{(cylindrical \: part)} + Volume_{(conical \: part)}[/tex]

   [tex]\rm \: = \: \:\pi \: {r}^{2}H + 2 \times \dfrac{1}{3}\pi \: {r}^{2}h[/tex]

   [tex]\rm \: = \: \:\pi \: {r}^{2}{\bigg(H + \dfrac{2}{3}h \bigg) }[/tex]

   [tex]\rm \: = \: \:\dfrac{22}{7} \times 1.5 \times 1.5 \times {\bigg(8 + \dfrac{2}{3} \times 2 \bigg) }[/tex]

   [tex]\rm \: = \: \:\dfrac{22}{7} \times 1.5 \times 1.5 \times {\bigg(8 + \dfrac{4}{3} \bigg) }[/tex]

   [tex]\rm \: = \: \:\dfrac{22}{7} \times 1.5 \times 1.5 \times {\bigg(\dfrac{24 + 4}{3} \bigg) }[/tex]

   [tex]\rm \: = \: \:\dfrac{22}{7} \times 1.5 \times 1.5 \times {\bigg(\dfrac{28}{3} \bigg) }[/tex]

   [tex]\rm \: = \: \:66 \: {cm}^{3} [/tex]

   [tex]\bf\implies \:Volume_{(air \: contained \: in \: model)} = \: \:66 \: {cm}^{3} [/tex]

   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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