Question

the diameter of a sphere is 9 cm It is melted and drawn into a wire of diameter 3 cm what will be the length of the wire​

Answer 1

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Q) The diameter of a sphere is 9 cm . It is melted and drawn into a wire of diameter 3 cm . What will be the length of the wire ?

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☆ Concept :

• In these type of Questions , where , one object/shape is melted and recasted into a new object/shape , the volume remains constant .

• In this question , the volume of the given sphere must be equal to the volume of the new object formed , that is a wire , which is cylindrical in shape .

• You must know the formulas required to solve the Question .

☆ Given :

• Diameter of Sphere = 9cm
• Radius of wire = 3cm

☆ To Find :

• The length (h) of the new wire formed .

☆ Solution :

$\rm \: radius_{sphere} = \frac{diameter}{2}$

$\implies \boxed{ \rm \: radius _{sphere} = \frac{9}{2} \: cm}$

Now ,

Since , the volume of both the sphere and the wire would be same so ,

$\rm \: volume_{sphere} = volume_{wire}$

$\mapsto \rm \: \frac{4}{3} \cancel{ \pi} \: {(r _{sphere})}^{3 } = \cancel{\pi} \: {(r _{wire} )}^{2} \times h$

$\rm \: cancel \: \pi \: from \: both \: sides$

$\mapsto \rm \: \frac{4}{3} \times {( \frac{9}{2}) }^{3} = {3}^{2} \times h$

$\mapsto \rm \: \frac{ \cancel4}{ \cancel3} \times \frac{ \cancel9 \times \cancel9 \times 9}{ \cancel2 \times \cancel2 \times 2 } = \cancel3 \times \cancel3 \times h$

$\mapsto \rm \: \frac{3 \times 9}{2} \: cm = h$

$\large\mapsto \boxed{ \rm \: h = \frac{27}{2} \: cm = 13.5 \: cm}$

So , The length of the wire would be 13.5 cm

☆ Know More :

$\rm \: 1. \: volume _{sphere} = \frac{4}{3} \pi {r}^{3}$

$\rm \: 2. \: volume _{cylinder} = \pi {r}^{2} h$

$\rm \: 3. \: csa _{sphere} = tsa _{sphere} = 4 \pi {r}^{2}$

$\rm \: 4. \: csa _{cylinder} = 2 \pi \: rh$

$\rm \:5. \: tsa_{cylinder} = 2 \pi \: r(r + h)$