Example 10 - Chapter 13 Class 10 Surface Areas and Volumes (Term 2)
Last updated at May 29, 2018 by Teachoo
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Example 12 Important Deleted for CBSE Board 2022 Exams
Example 13 Deleted for CBSE Board 2022 Exams
Example 14 Deleted for CBSE Board 2022 Exams
Surface Area and Volume Formulas Important
Chapter 13 Class 10 Surface Areas and Volumes (Term 2)
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Transcript
Example 10 A copper rod of diameter 1 cm and length 8 cm is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire. Volume of copper rod = Volume of wire Volume of copper rod Copper rod is in form of cylinder with Diameter of rod = 1 cm So, radius = r = 1/2 cm And, length = height = h = 8 cm Volume of copper rod = 2 = (1/2)^2 8 = 1/4 8 = 2 cm3 Volume of wire Wire is form of cylinder with radius = r length = height = h = 18 m = 18 100 = 1800 cm Volume of wire = 2 = 2 1800 = 1800 2 cm2 Now, Volume of copper rod = Volume of wire 1800 2=2 1800 2=2 r2 = 2/1800 r2 = 2/1800 r2 = 1/900 r = (1/900) r = (1/(30 30)) r = (1/ 30 ^2 ) r = 1/30 So, r = 1/30 cm Thickness of wire = Diameter of wire = 2r = 2 1/30 = 1/15 cm
