the density p of earth depends upon the radius r , the acceleration due to gravity g on the surface of the earth and the gravitational constant G . Derive a relation between them using dimensional method.
Answers
Explanation:
Answer:
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 }Internalradius
\rm{\implies \: 5 \: cm \: - 1cm \: }⟹5cm−1cm
\bf{\implies \: r = \: 4 \: cm}⟹r=4cm
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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}ExternalVolume
\rm{ \: \frac{22}{7} \times( {5)}^{2} \times 35 }
7
22
×(5)
2
×35
\rm{ \: \frac{22}{7} \times 25\times 35 }
7
22
×25×35
\rm{ \: \frac{22 \times 25 \times 35}{7} }
7
22×25×35
\rm{ \frac{19250}{7} }
7
19250
\bf{2750}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}ExternalVolume
\rm{ \: \frac{22}{7} \times( {4)}^{2} \times 35 }
7
22
×(4)
2
×35
\rm{ \: \frac{22}{7} \times16 \times 35 }
7
22
×16×35
\rm{ \: \frac{22 \times 16 \times 35}{7} }
7
22×16×35
\rm{ \: \frac{12320}{7} }
7
12320
\bf{1760}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}2750−1760
\bf{990 \: cm³}990cm³
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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}➠7.722kg