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✧ A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radius of two of these balls are 1.5 cm and 2 cm. Find the radius of the third ball.

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Answered by SarcasticL0ve
10

A spherical ball of radius 3 cm is melted and recast into three spherical balls. The radius of two of these balls are 1.5 cm and 2 cm.

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Here we have,

  • Radius of large ball, r = 3 cm
  • Radius of two small ball, r and r = 1.5 cm and 2 cm

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We know that,

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\star\;{\boxed{\sf{\purple{Volume_{\;(sphere)} = \dfrac{4}{3} \pi r^3}}}}\\ \\

Now, Finding Volume of large ball,

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:\implies\sf \dfrac{4}{3} \pi r^3\\ \\

:\implies\sf \dfrac{4}{3} \times \dfrac{22}{7} \times 3 \times 3 \times 3\\ \\

:\implies\sf \dfrac{4}{ \cancel{3}} \times \dfrac{22}{7} \times \cancel{27}\\ \\

:\implies\sf 4 \times \dfrac{22}{7} \times 9\\ \\

:\implies{\boxed{\frak{\pink{{792}{7}\;cm^3}}}}\;\bigstar\\ \\

Volume of two smaller balls,

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:\implies\sf \dfrac{4}{3} \pi {(r_1)}^3+ \dfrac{4}{3} \pi {(r_2)}^3\\ \\

:\implies\sf \dfrac{4}{3} \times \dfrac{22}{7} \times 1.5 \times 1.5 \times 1.5 + \dfrac{4}{3} \times \dfrac{22}{7} \times 2 \times 2 \times 2\\ \\

:\implies\sf \dfrac{4}{3} \times \dfrac{22}{7} \bigg( \dfrac{27}{8} + 8 \bigg)\\ \\

:\implies\sf \dfrac{4}{3} \times \dfrac{22}{7} \bigg( \dfrac{27 + 64}{8} \bigg)\\ \\

:\implies\sf \dfrac{4}{3} \times \dfrac{22}{7} \times \dfrac{91}{8}\\ \\

:\implies{\boxed{\frak{\pink{{143}{3}\;cm^3}}}}\;\bigstar\\ \\

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Volume of third ball = Volume of large ball - Volume of 2 small balls

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:\implies\sf \dfrac{792}{7} - \dfrac{143}{3}\\ \\

:\implies\sf \dfrac{2376 - 1001}{21}\\ \\

:\implies{\boxed{\frak{\pink{{1375}{21}\;cm^3}}}}\;\bigstar\\ \\

Therefore,

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:\implies\sf \dfrac{4}{3} \pi {(r_3)}^3 = \dfrac{1375}{21}\\ \\

:\implies\sf \dfrac{4}{3} \times \dfrac{22}{7} \times {(r_3)}^3 = \dfrac{1375}{21}\\ \\

:\implies\sf {(r_3)}^3 = \dfrac{1375}{21} \times \dfrac{3}{4} \times \dfrac{7}{22}\\ \\

:\implies\sf {(r_3)}^3 = \dfrac{125}{8}\\ \\

:\implies\sf r = \sqrt[3]{\dfrac{125}{8}}\\ \\

:\implies{\boxed{\frak{\purple{r = \dfrac{5}{2}\;cm}}}}\;\bigstar\\ \\

\therefore Hence, radius of third ball is 5/2 or 2.5 cm.


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Answered by kcsshweta
8

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