Question
Download Solution PDFA 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.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
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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