A copper wire,3 ,mm in diameter,is wound abou | vedclass

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(B) It can be observed that $1$ round of wire will cover $3 \,mm$ $(0.3 \,cm)$ height of the cylinder.Number of rounds $= \frac{	ext{Height of cylind

Vedclass · Accepted Answer

$789.41$

Vedclass · Answer

(B) It can be observed that $1$ round of wire will cover $3 \,mm$ $(0.3 \,cm)$ height of the cylinder.Number of rounds $= \frac{	ext{Height of cylinder}}{	ext{Diameter of wire}} = \frac{12 \,cm}{0.3 \,cm} = 40$ rounds.Length of wire required in $1$ round $=$ Circumference of the base of the cylinder $= 2 \pi r = 2 \pi 	imes 5 = 10 \pi \,cm$.Total length of wire in $40$ rounds $= 40 	imes 10 \pi = 400 \pi \,cm = 400 	imes 3.14159 \approx 1256.64 \,cm$.Radius of the wire $= \frac{0.3 \,cm}{2} = 0.15 \,cm$.Volume of the wire $=$ Area of cross-section of wire $	imes$ Total length of wire $= \pi (0.15)^2 	imes 1256.64 \approx 88.826 \,cm^3$.Mass of the wire $=$ Volume $	imes$ Density $= 88.826 \,cm^3 	imes 8.88 \,g/cm^3 \approx 788.77 \,g$.Using $\pi \approx \frac{22}{7}$ for calculation: Length $= 400 	imes \frac{22}{7} = \frac{8800}{7} \approx 1257.14 \,cm$. Volume $= \pi 	imes (0.15)^2 	imes \frac{8800}{7} = \frac{22}{7} 	imes 0.0225 	imes \frac{8800}{7} \approx 88.93 \,cm^3$. Mass $= 88.93 	imes 8.88 \approx 789.7 \,g$. The closest option is $789.41 \,g$.

$A$ copper wire,$3 \,mm$ in diameter,is wound about a cylinder whose length is $12 \,cm$ and diameter $10 \,cm$,so as to cover the curved surface of the cylinder. Find the mass of the wire,assuming the density of copper to be $8.88 \,g/cm^3$. (in $g$)

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