Questions & Answers

Question

Answers

A. 12

B. 6

C. $\sqrt[3]{{16}}$

D. $3\sqrt[3]{8}$

E. $2\sqrt[3]{{36}}$

Answer
Verified

Hint: In this question, we are supposed to combine three cheese balls with diameter 2 inches, 4 inches and 6 inches, when we combine the volume of the bigger cheese ball will be the sum of the volume of the three smaller cheese balls.

Volume of the bigger cheese ball = Sum of volume of the three smaller cheese balls.

Volume of new cheese ball = $\dfrac{4}{3}\pi \left( {{r_1}^3 + {r_2}^3 + {r_3}^3} \right)$

Putting the values of r in the above equation,

Volume of bigger cheese ball = $\dfrac{4}{3}\pi \left( {{2^3} + {4^3} + {6^3}} \right)$

Since the cheese ball is in spherical shape the volume can be taken as

Volume of bigger cheese ball =$\dfrac{4}{3}\pi {R^3}$

On equating both the equations,

$\dfrac{4}{3}\pi {R^3} = \dfrac{4}{3}\pi \left( {8 + 64 + 216} \right)$

Cancelling the common terms, we get,

${R^3} = \left( {8 + 64 + 216} \right)$

${R^3} = 288$

On solving it further, we get,

$R = \sqrt[3]{{288}}$

On factoring, we get,

$R = \sqrt[3]{{8 \times 36}}$

$R = 2\sqrt[3]{{36}}$ inches

Answer = Option E

Note: Do not make the mistake of equating the sum of areas or perimeter because with the change of structure, the value of these entities will also change, but the value of volume will remain the same.

Volume of the bigger cheese ball = Sum of volume of the three smaller cheese balls.

Volume of new cheese ball = $\dfrac{4}{3}\pi \left( {{r_1}^3 + {r_2}^3 + {r_3}^3} \right)$

Putting the values of r in the above equation,

Volume of bigger cheese ball = $\dfrac{4}{3}\pi \left( {{2^3} + {4^3} + {6^3}} \right)$

Since the cheese ball is in spherical shape the volume can be taken as

Volume of bigger cheese ball =$\dfrac{4}{3}\pi {R^3}$

On equating both the equations,

$\dfrac{4}{3}\pi {R^3} = \dfrac{4}{3}\pi \left( {8 + 64 + 216} \right)$

Cancelling the common terms, we get,

${R^3} = \left( {8 + 64 + 216} \right)$

${R^3} = 288$

On solving it further, we get,

$R = \sqrt[3]{{288}}$

On factoring, we get,

$R = \sqrt[3]{{8 \times 36}}$

$R = 2\sqrt[3]{{36}}$ inches

Answer = Option E

Note: Do not make the mistake of equating the sum of areas or perimeter because with the change of structure, the value of these entities will also change, but the value of volume will remain the same.

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