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Exercise 2.4 · Q15

Q.A cubical gold block of dimensions of 9cm x 11cm x 12 cm is melted and recasted into spherical balls of radius 3 mm. Find the number of spherical balls formed.

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A 9×11×129\times11\times12 cm gold block melted and recast into spheres of radius 3 mm yields exactly 10,500 balls.

Volume is conserved on recasting: Volume of original solidVolume of one new solid=\dfrac{\text{Volume of original solid}}{\text{Volume of one new solid}}= number of new solids formed. Volume(cuboid)=l×b×h=l\times b\times h; Volume(sphere)=43πr3=\dfrac43\pi r^3.

  1. Volume of the gold block =9×11×12=1188 cm3=9\times11\times12=1188\text{ cm}^3.
  2. Radius of each ball =3=3 mm =0.3=0.3 cm (convert to cm to match units).
  3. Volume of one ball =43πr3=43π(0.3)3=43π(0.027)=0.036π cm3=\dfrac43\pi r^3=\dfrac43\pi(0.3)^3=\dfrac43\pi(0.027)=0.036\pi\text{ cm}^3.
  4. Using π=227\pi=\dfrac{22}{7}: Volume of one ball =0.036×227=0.7927≈0.11314 cm3=0.036\times\dfrac{22}{7}=\dfrac{0.792}{7}\approx0.11314\text{ cm}^3. …

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