Question

The inner and outer diameter of a hollow
cylindrical iron pipe 6 metre long are 3.5
cm and 4.2 cm. If the weight of one
cubic decimetre of iron is 5 kg, what is
the weight of the pipe ?​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Weight\:of\:pipe=50.82\:kg}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: {\implies Length \: of \: cylindrical \: pipe = 6 \: m} \\ \\ \tt: \implies Inner \:diameter (r_{1}) = 3.5 \: cm \\ \\ \tt: \implies Outer \:diameter (r_{2}) = 4.2 \: cm \\ \\ \tt: \implies Weight \: of \: 1 {dm}^{3} = 5 \: kg \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: {\implies Weight \: of \: cylindrical \: pipe = ?}$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies {Volume \: of \: pipe = \pi ( {( r_{2}) }^{2} - ( r_{1})^{2} )h }\\ \\ \tt: \implies Volume \: of \: pipe = \frac{22}{7} \times ( {4.2}^{2} - {3.5}^{2} ) \times 600\\ \\ \tt: \implies Volume \: of \: pipe = \frac{22}{7} \times (17.64 - 12.25) \times 600 \\ \\ \tt: \implies Volume \: of \: pipe = \frac{22}{7} \times 5.39 \times 600 \\ \\ \tt: \implies Volume \: of \: pipe =22 \times 0.77 \times 600 \\ \\ \tt: \implies Volume \: of \: pipe =10164 \: {cm}^{3} \\ \\ \green{\tt: \implies Volume \: of \: pipe =10.164 \: {dm}^{3} } \\ \\ \bold{For \: weight : } \\ \tt: \implies 1 {dm}^{3} = 5 \: kg \\ \\ \tt: \implies 10.164 \: {dm}^{3} = 5 \times 10.164 = 50.82 \: kg\\ \\ \green{\tt \therefore Weight \: of \: pipe = 50.82 \: kg}$

Answer 2

Given :-

Inner diameter(r) =3.5m

Outer diameter(R) =4.2 m

Length of the hollow cylinder(h) =600cm

And,

1 dm²= 5kg

To find :-

Weight of the pipe =?

Formula used :-

$Volume \: of \: hollow \: cylinder =\pi( {R \: }^{2} - {r}^{2} )h$

Solution :-

$Volume \: of \: pipe = \pi( {4.2}^{2} - {3.5}^{2} ) \times 600$

$Volume \: of \: pipe = 3.14 \times( 17.64 - 12.25) \times 600 \: {cm}^{3}$

$Volume \: of \: pipe = 3.14 \times 5.39 \times 600 \: {cm}^{3}$

$Volume \: of \: pipe = 10164 \: {cm}^{3}$

Now,

1 cm =1/10dm

1 cm³=1/1000 dm³

10164cm³=10164×1/1000=10,164dm³

Now According to the question, we get

1 dm ³ = 5kg

10,164 dm³=10,164×5= 50.82kg

Therefore the weight of the pipe is 50.82 kg.