A metallic cylinder has radius 3 cm and height 5 cm. To reduce

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Q) A metallic cylinder has radius 3 cm and height 5 cm. To reduce its weights, a conical hole is drilled in the cylinder. The conical hole has a radius of $\frac{3}{2}$ cm and its depth $\frac{8}{9}$ cm. Calculate the ratio of the volume of metal left in the cylinder to the volume of metal taken out in conical shape.

Ans:

Step 1: ∵ Volume of the cylinder = π r ² h

For cylinder, radius = 3 cm and height = 5 cm (given)

∴ Cylinder’s volume = π × (3)² × 5 = π × 9 × 5 = 45 π cm³

Step 2: ∵ Volume of the cone = $\frac{1}{3}$ π r ² h

For conical hole, radius = $\frac{3}{2}$ cm and height = $\frac{8}{9}$ cm (given)

∴ Conical hole’s volume = $\frac{1}{3} \pi (\frac{3}{2})^2 (\frac{8}{9})$

= $\frac{1}{3} \pi (\frac{9}{4}) (\frac{8}{9})$

= $\frac{2}{3}$ π cm³

Step 3: ∵ Volume of remaining body = Cylinder volume – Conical hole’s volume

= 45 π – $\frac{2}{3}$ π

= $\frac{133}{3}$ π cm³

Step 4: ∵ Ratio of volumes of metal left to metal taken out:

= Volume of remaining body : Volume of conical hole

= $\frac{133}{3}\pi : \frac{2}{3}\pi$

= 133 : 2 = 66.5 : 1

The required ratio of metal’s volume is 66.5 : 1

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