Question

the length of a metallic pipe is 70 cm in diameter is 10 cm and thickness is 5 M if density of metal used in 8 gram per centimetre cube then find mass of the pipe​

Answer 1

Answer:

The mass of the pipe is 7539824 grams

Step-by-step explanation:

Diameter of outer circle = 70 cm

=> Radius of outer circle, R = 35 cm

Given thickness = 10 cm

=> Radius of inner circle, r = 35 - 10 = 25 cm

Length of the pipe, h = 5 m = 500 cm

Hence volume of the pipe,

V = πR²h - πr²h

= πh(R²-r²)

= πh(R+r)(R-r)

= 3.14 x 500 x (35+25)(35-25)

= 3.14 x 500 x 60 x 10

=> V = 942478 cm³

Given density of metal = 8 g/cm³

We know that mass is defined as the product of volume and density.

Mass = volume x density

= 942478 x 8

= 7539824 grams

Hence the mass of the pipe is 7539824 grams

Answer 2

$step \: by \: step \: eplanatation \\ \: \: \\ Diameter \: of \: the \: outer \: \: circle \: \ is \: 70 \: cm </p><p> \\ </p><p>Radius \: of \: the \: outer \: circle \: \: \:is \: 70/2 = 35 \\ </p><p></p><p>Given \: Thickness \: is = 10 cm</p><p></p><p> \\ Radius \: of \: the \: inner \: circle = </p><p></p><p>r = - 10 = 25 \: \\ length \: of \: the \: pipe = 5m = 500cm \\ hence \: volume \: of \: the \: pipe \: \pi \: r ^{2}h \: - \pi \: r ^{2}h \\ \pi \: h( {r}^{2} - r ^{2} ) \\ \pi h(r + r)(r - r) \\ 3.14 \times 500(35 + 25)(35 - 25) \\ 3.14 \times 500 \times 60 \times 10 \\ v = 942478 \: cm ^{3} \\ given \: density \: of \: metal \: 8g |cm^{3} \\ we \: know \: that \: mass \: is \: defined as \: the \: product \: of \: volume \: and \: density \: \\ mass \: = volume \times density \\ 942478 \times 8 = 7539824 \: grams$

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