Question

2
S.A of
4. 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 3cm
and its length is 12cm. If each cone has a
height of 2cm, 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.)​

Answer 1

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

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,

$\green{ \boxed{ \bf \: Volume_{(cylinder)} = \pi \: {r}^{2}h}}$

and

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

↝ Volume of air contained in model is

$\rm :\longmapsto\:Volume_{(air \: contained)} = Volume_{(cylindrical \: part)} + Volume_{(conical \: part)}$

$\rm  \:  =  \: \:\pi \: {r}^{2}H + 2 \times \dfrac{1}{3}\pi \: {r}^{2}h$

$\rm  \:  =  \: \:\pi \: {r}^{2}{\bigg(H + \dfrac{2}{3}h \bigg) }$

$\rm  \:  =  \: \:\dfrac{22}{7} \times 1.5 \times 1.5 \times {\bigg(8 + \dfrac{2}{3} \times 2 \bigg) }$

$\rm  \:  =  \: \:\dfrac{22}{7} \times 1.5 \times 1.5 \times {\bigg(8 + \dfrac{4}{3} \bigg) }$

$\rm  \:  =  \: \:\dfrac{22}{7} \times 1.5 \times 1.5 \times {\bigg(\dfrac{24 + 4}{3} \bigg) }$

$\rm  \:  =  \: \:\dfrac{22}{7} \times 1.5 \times 1.5 \times {\bigg(\dfrac{28}{3} \bigg) }$

$\rm  \:  =  \: \:66 \: {cm}^{3}$

$\bf\implies \:Volume_{(air \: contained \: in \: model)}  =  \: \:66 \: {cm}^{3}$

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