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A metallic sphere of radius 6 cm is melted and drawn into a wire, whose radius of cross section is 8 cm. What is the length of the wire?
Question

A metallic sphere of radius 6 cm is melted and drawn into a wire, whose radius of cross section is 8 cm. What is the length of the wire?

A.

3.5 cm

B.

5 cm

C.

4.5 cm

D.

4 cm

Correct option is C

Given:
Radius of the sphere = 6 cm
Radius of the cross section of the wire = 8 cm
Formula Used:
Volume of a sphere:
Vsphere=43πr3V_{\text{sphere}} = \frac{4}{3} \pi r^3​​
Where r is the radius of the sphere.
Volume of a cylinder (which represents the wire):
Vcylinder=πrwire2hV_{\text{cylinder}} = \pi r_{\text{wire}}^2 h​​
Where rwirer_{\text{wire}}​ is the radius of the cross section of the wire and h is the length of the wire. 
Solution:  
Volume of sphere=43×π×r3Volume of cylinder=π×r2×hHere the sphere is melted into a long wire. Assume the wire as a cylinder.Therefore,Sphere radius=6 cmCylinder cross-sectional radius=8 cmWe need to find h=?43×π×6×6×6=π×8×8×hSolving, we get h=4.5 cm\text{Volume of sphere} = \frac{4}{3} \times \pi \times r^3 \\\text{Volume of cylinder} = \pi \times r^2 \times h \\\text{Here the sphere is melted into a long wire. Assume the wire as a cylinder.} \\\text{Therefore,} \\\text{Sphere radius} = 6 \, \text{cm} \\\text{Cylinder cross-sectional radius} = 8 \, \text{cm} \\\text{We need to find } h = ? \\\frac{4}{3} \times \pi \times 6 \times 6 \times 6 = \pi \times 8 \times 8 \times h \\\text{Solving, we get } h = 4.5 \, \text{cm}

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