A cylindrical rod of iron whose height is twelve times its radius, is melted and cast into spherical balls, each of one-third the radius of the cylinder. Find the number of spherical balls.

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This question was previously asked in

SSC CHSL Exam 2024 Tier-I Official Paper (Held On: 03 Jul, 2024 Shift 4)

Answer 1

Given:

Height of the cylindrical rod (h) = 12 × radius (r)

Radius of the spherical balls = r / 3

Formula Used:

Volume of cylinder = πr²h

Volume of sphere = \(\frac{4}{3}\pi R^3\)

Calculation:

Volume of cylinder = πr²(12r)

⇒ Volume of cylinder = 12πr³

Volume of one spherical ball = \(\frac{4}{3}\pi \left(\frac{r}{3}\right)^3\)

⇒ Volume of one spherical ball = \(\frac{4}{3}\pi \left(\frac{r^3}{27}\right)\)

⇒ Volume of one spherical ball = \(\frac{4}{81}\pi r^3\)

Number of spherical balls = Volume of cylinder / Volume of one spherical ball

⇒ Number of spherical balls = \(\frac{12\pi r^3}{\frac{4}{81}\pi r^3}\)

⇒ Number of spherical balls = 12 / \(\frac{4}{81}\)

⇒ Number of spherical balls = 12 × 81 / 4

⇒ Number of spherical balls = 243

The number of spherical balls is 243.

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