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A solid cylindrical rod whose height is 9 times of its radius is melted and recast into spherical balls whose radius is half of the cylindrical radius. Then, number of spherical balls will be –

A solid cylindrical rod whose height is 9 times of its radius is melted and recast into spherical balls whose radius is half of the cylindrical radius. Then, number of spherical balls will be – Correct Answer 54

Given:

The radius of cylindrical = 1/9 times of the height of the cylindrical rod

The radius of the spherical ball = ½ times the radius of the cylindrical rod

Concept used:

The volume of the cylindrical rod = πr2h

The volume of the spherical ball = 4/3 πR3

Calculation:

Suppose, n number of spherical balls.

Suppose the radius of the cylindrical rod is x cm and the height of the rod is 9x cm.

The radius of spherical ball will be (x / 2) cm.

⇒ πr2h = n × 4 / 3 × πR3

⇒ π(x)2(9x) = n × 4 / 3 × π(x / 2)3

⇒ 9 = n × 4 / 3 × (1 / 8)

⇒ n = 54

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Simplify the value of $$\frac{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} \times {\text{0}}{\text{.3}} - {\text{3}} \times 0.9 \times {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}}}}{{{\text{0}}{\text{.9}} \times {\text{0}}{\text{.9 + 0}}{\text{.2}} \times {\text{0}}{\text{.2 + 0}}{\text{.3}} \times {\text{0}}{\text{.3}} - 0.9 \times {\text{0}}{\text{.2}} - {\text{0}}{\text{.2}} \times {\text{0}}{\text{.3}} - 0.3 \times 0.9}} = ?$$