Question

A copper sphere of radius 4.2 cm is melted and recast into a wire of length 1260 cm.find the thickness of the wire

Answer 1

Answer:

volume of sphere = volume of cylinder

4/3*$\pi$*r$x^{3}$ = $\pi$*r$x^{2}$*h

4/3*22/7*4.2*4.2*4.2 = 22/7*r$x^{2}$*1260

r$x^{2}$ = (4/3*4.2*4.2*4.2)/1260

r$x^{2}$ = 0.0784

r = 0.28

thickness = 2 * radius

              = 2 * 0.28

              = 0.56cm

Step-by-step explanation:

Answer 2

For sphere,

r1=4.2 cm

For cylindrical wire,

let radius of wire be x cm

r2= x cm

h= 1260 cm

since, sphere is converted into wire

volume of sphere = volume of cylinder

4/3 πr1³= πr2²h

4/3 × 4.2 × 4.2 = x² × 1260

x² = 4 × 4.2 × 4.2 × 4.2 / 3 × 1260

x² = 1.4 × 4.2 × 4.2 /315

x²=0.0784

x²=0.28 cm