Question

A copper rod of diameter 1 cm and length 8 cm is drawn in to a wire of length 18 m of uniform thickness. find the thickness of wire​

Answer 1

❏ Formula Used:-

For a right circular cylinder of base radius r and height h,

$\sf\longrightarrow\boxed{ L.S.A=2\pi r h}$

$\sf\longrightarrow\boxed{ T.S.A.=2\pi r (r+h)}$

$\sf\longrightarrow \boxed{Volume=\pi r{}^{2}h}$

Where, •L.S.A.=Curved Surface area.

•T.S.A.=Total Surface Area.

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❏ Solution:-

Q) A copper rod of diameter 1 cm and length 8 cm is drawn in to a wire of length 18 m of uniform thickness. find the thickness of wire.

And)

•For the copper rod ➾

•diameter(D)=1 cm

•radius(R)= $\frac{1}{2}\:cm=0.5 \:cm$

•length(L)=8 cm

$\therefore Volume_{rod}=(\pi\times (0.5){}^{2}\times 8) \:cm{}^{3}$

•For the copper wire➾

•radius(r)=?

•length(l)=18 m=1800 cm

$\therefore Volume_{wire}=(\pi\times (r){}^{2}\times 1800) \:cm{}^{3}$

Now, as the rod is drawn to a wire ,so the volume of the Rod & wire is same .

$\therefore Volume_{wire}=Volume_{rod}$

$\therefore \pi\times (r){}^{2}\times 1800=\pi\times (0.5){}^{2}\times 8$

$\sf\longrightarrow r{}^{2}=\frac{0.5\times0.5\times8}{1800}$

$\sf\longrightarrow r{}^{2}=\frac{1}{900}$

$\sf\longrightarrow r=\sqrt{\frac{1}{900}}$

$\sf\longrightarrow r=\frac{1}{30}\:cm$

$\sf\longrightarrow d=2r=\cancel2\times\frac{1}{\cancel{30}}\:cm$

$\sf\longrightarrow d=\frac{1}{15} \:cm$

$\sf\longrightarrow\boxed{\red{ d=0.067 \:cm \:\:(almost)}}$

$\therefore$ Thinness of the wire= 0.067 cm.

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$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

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