Question

The external diameter of a 35-cm-long metal pipe is 10 cm. If the pipe is 1 cm thick and the density of the metal is 7.8 g/cm^3, find the volume of the metal used in the pipe and the mass of the pipe.​

Answer 1

Given :

• Density of Metal = 7.8 g/cm³
• Inner diameter = 8 cm (10-2 cm)
• outer diameter = 10 cm

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To Find :

The mass of the pipe given

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Formula to be applied :

Volume of the pipe :

• Area of ring × Hieght of pipe

Density = Mass/volume

so, Mass = volume × Density

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Solution :

given, Inner diameter = 8 cm

Radius = diameter/2

so, inner radius = 8/2 = 4 cm

Outer diameter = 10 cm

so, radius = 5 cm

We know that ,

$\tt{ area \: of \: ring = \pi \: o {r}^{2} - \pi \: i {r}^{2} }$

$\implies \tt{area = \pi( {or}^{2} - {ir}^{2}) }$

Here, Ir = Inner Radius

Or = outer radius

so, area of ring:

$\tt{ a = \pi( {5}^{2} - {4}^{2}) }$

$\implies \tt{ a = \pi(25 - 16)}$

$\implies \tt{a = 3.14 \times 9}$

$\implies \tt{a = 28.26}$

so, area of ring = 28.26 cm²

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So, volume of pipe = Area of ring × length

$\tt{ \implies \: volume = 35 \times 28.26}$

$\implies \tt{volume = 989.1}$

Hence, Volume = 989.1 cm³

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Mass = Volume × Density

$\implies \tt{mass = 989.1 \times 7.8}$

$\red{ \underline{ \boxed{ \tt{ \therefore \: mass = 6154.98 \: g}}}}$

Hence , mass = 6154.98 g

Above is A rough figure to get an idea :)

Answer 2

Given :

➠ Density of Metal

➱ 7.8 g/cm³

➠ External diameter of metal pipe

➱ 10 cm

▶ External radius

➻ diameter /2

➻ 10 cm / 2

➠ External radius = 5 cm

▶ Height of pipe ➠ 35 cm

➠ Thickness = 1 cm

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To Find :

➠ The mass of the pipe given

➠ Volume of metal used in pipe

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Formula to be applied :

▶Internal radius

➠ External Radius - Thickness

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Solution :

➠ Place given value in formula

➠ r = R - 1 cm

$\tt{Internal \: radius }$

$\rm{\implies \: 5 \: cm \: - 1cm \: }$

$\bf{\implies \: r = \: 4 \: cm}$

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Formula to be applied :

▶ External volume of pipe

➠ π R² h

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➠ Place given value in formula

$\tt{External \: Volume}$

$\rm{ \: \frac{22}{7} \times( {5)}^{2} \times 35 }$

$\rm{ \: \frac{22}{7} \times 25\times 35 }$

$\rm{ \: \frac{22 \times 25 \times 35}{7} }$

$\rm{ \frac{19250}{7} }$

$\bf{2750}$

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Formula to be applied :

▶ External volume of pipe

➠ π r² h

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➠ Place given value in formula

$\tt{External \: Volume}$

$\rm{ \: \frac{22}{7} \times( {4)}^{2} \times 35 }$

$\rm{ \: \frac{22}{7} \times16 \times 35 }$

$\rm{ \: \frac{22 \times 16 \times 35}{7} }$

$\rm{ \: \frac{12320}{7} }$

$\bf{1760}$

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Formula to be applied :

▶ Volume of metal used in pipe

• ➠( External volume of pipe - Internal volume of pipe)

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➠ Place given value in formula

$\tt{2750 -1760}$

$\bf{990 \: cm³}$

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Given :

➠ Mass of 1 cm³ of metal is 7.8 gram

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Final Answer :

➠Therefore mass of pipe

➠ mass of 990 cm ³ metal

• = 990 × 7.8 gram

$\bf{➠7.722 \: kg}$