Question
A hollow metallic cylinder whose external radius is 4.3 cm and internal radius is 1.1 cm and the whole length is 4 cm is melted and recast into a solid cylinder of 12 cm long. Find the diameter of a solid cylinder
✓
Answer
External radius of the hollow cylinder R = 4.3 cm
Internal radius of the hollow cylinder r = 1.1 cm
Length of the cylinder (h) = 4 cm
Length of the solid cylinder (H) = 12 cm
Let the radius of the solid cylinder be x
Volume of the solid cylinder = Volume of the hollow cylinder
$\begin{aligned} & \pi r^2 H=\pi h\left(R^2-r^2\right) \\ & x^2 \times 12=4\left(4.3^2-1.1^2\right) \\ & 12 x^2=4 \times 5.4 \times 3.2 \\ & 12 x^2=4(4.3+1.1)(4.3-1.1) \\ & x^2=\frac{4 \times 5.4 \times 3.2}{12} \\ & =\frac{4 \times 54 \times 32}{12 \times 100} \\ & =\frac{54 \times 32}{3 \times 100} \\ & =\frac{18 \times 32}{100} \\ & x=\sqrt{\frac{18 \times 32}{100}} \\ & =\sqrt{\frac{2 \times 9 \times 2 \times 16}{100}} \\ & =\frac{2 \times 3 \times 4}{10}\end{aligned}$
= 2.4 cm
Diameter of the solid cylinder
= 2 × 2.4
= 4.8 cm
Internal radius of the hollow cylinder r = 1.1 cm
Length of the cylinder (h) = 4 cm
Length of the solid cylinder (H) = 12 cm
Let the radius of the solid cylinder be x
Volume of the solid cylinder = Volume of the hollow cylinder
$\begin{aligned} & \pi r^2 H=\pi h\left(R^2-r^2\right) \\ & x^2 \times 12=4\left(4.3^2-1.1^2\right) \\ & 12 x^2=4 \times 5.4 \times 3.2 \\ & 12 x^2=4(4.3+1.1)(4.3-1.1) \\ & x^2=\frac{4 \times 5.4 \times 3.2}{12} \\ & =\frac{4 \times 54 \times 32}{12 \times 100} \\ & =\frac{54 \times 32}{3 \times 100} \\ & =\frac{18 \times 32}{100} \\ & x=\sqrt{\frac{18 \times 32}{100}} \\ & =\sqrt{\frac{2 \times 9 \times 2 \times 16}{100}} \\ & =\frac{2 \times 3 \times 4}{10}\end{aligned}$
= 2.4 cm
Diameter of the solid cylinder
= 2 × 2.4
= 4.8 cm
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