Question

2.2dm^3 of lead is to be drawn into a cylindrical wire o.50cm in diameter. find the length of the wire?​

Answer 1

refer the attachment hope its help u

$\over \mid \ good \: night \mid \:$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm \: Volume_{(lead)} = 2.2 \: {dm}^{3}$

We know,

$\boxed{\sf{  \:\rm \: \: 1 \: dm \: = \: 10 \: cm \: \: }}$

So, using this

$\rm \: Volume_{(lead)} = 2.2 \times 10 \times 10 \times 10 \: {cm}^{3}$

$\rm\implies \:Volume_{(lead)} = 2200 \: {cm}^{3}$

Now, further given that,

2.2dm^3 of lead is to be drawn into a cylindrical wire 0.50cm in diameter.

Diameter of cylindrical wire = 0.50 cm

So, Radius of cylindrical wire = 0.25 cm

Let assume that the length of wire be h cm

We know,

Volume of cylinder of radius r and height h is given by

$\boxed{\sf{  \:\rm \: Volume_{(cylinder)} \: = \: \pi \: {r}^{2} \: h \: \: }}$

Now, as it is given that 2.2dm^3 of lead is to be drawn into a cylindrical wire 0.50cm in diameter.

$\rm\implies \:Volume_{(lead)} \: = \: Volume_{(cylinderical \: wire)}$

$\rm \: 2200 = \dfrac{22}{7} \times 0.25 \times 0.25 \times h$

$\rm \: 100 = \dfrac{1}{7} \times \dfrac{25}{100} \times \dfrac{25}{100} \times h$

$\rm \: 100 = \dfrac{1}{7} \times \dfrac{1}{4} \times \dfrac{1}{4} \times h$

$\rm \: h = 100 \times 7 \times 4 \times 4$

$\rm \: h \: = \: 11200 \: cm$

$\rm\implies \:\rm \: h \: = \: 112 \: m$

So, Length of wire = 112 m

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Additional Information :-

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²