A spherical ball of lead, 3 cm in diameter, is melted and recast into three spherical balls. The diameters of two of these balls are \(\frac{3}{2}\) cm and 2 cm, respectively. Find the diameter of the third ball.

Question

Question

This question was previously asked in

SSC CGL 2023 Tier-I Official Paper (Held On: 17 Jul 2023 Shift 4)

Answer 1

Given:

Diameter of spherical ball (D)= 3 cm

Diameter of 1st small ball (D₁)= 1.5 cm

Diameter of 2nd small ball (D₂)= 2 cm

Concept used:

Total volume of small spherical balls = Volume of large spherical balls

Formula used:

Volume of spherical ball = (4/3) × π × R³

Calculation:

Let, the diameter of 3rd small spherical ball = D₃

Volume of (1st small spherical ball + 2nd spherical ball + 3rd spherical ball) = volume of large spherical ball

⇒ 4/3 π × (D1/2)³ + 4/3 π × (D2/2)3 + 4/3 π × (D3/2)3 = 4/3 π (D/2)³

⇒ 4/3 π × [(1.5/2)3 + (2/2)3 + (D3/2)3 ]= 4/3 π (3/2)3

⇒ [(3.375/8) + 1 + (D3/2)3 ] = 3.375

⇒ (D3/2)3 = 2.375 - (3.375/8)

⇒ (D3/2)3 = (19 - 3.375)/8

⇒ D3 = ³√15.625 = 2.5

∴ The correct answer is 2.5.

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