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.)
[tex]\large\underline{\sf{Solution-}}[/tex]
Given :-
So,
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