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Hint: We are going to solve the given problem by using the formula of finding the volume of a spherical object.

It is known that the volume of a sphere with radius $r$ is given by $\frac{4}{3}\pi {r^3}$

Let the radius of the original ball be $R$.

So the volume of the original spherical ball with $$R = 3$$ is equal to $$\frac{4}{3}\pi {(3)^3}$$

Now the original ball is melted to form three balls of different sizes.

Let the radii of three resulting spherical balls be ${r_1},{r_2},{r_3}$ respectively.

Given ${r_1} = 1.5cm$ & ${r_2} = 2cm$.

$ \Rightarrow {V_{total}} = {V_1} + {V_2} + {V_3}$

$ \Rightarrow \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi ({r_1}^3 + {r_2}^3 + {r_3}^3)$

$ \Rightarrow {r_3}^3 = {3^3} - {1.5^3} - {2^3} = 15.625$

Hence ${r_3} = 2.5cm$

$\therefore $The radius of the third ball = 2.5cm

Note: In the given problem it is said that the spherical ball of radius 3cm is melted to form three spherical balls. So the volume of the original spherical ball is equal to the sum of the volumes of the three balls.

It is known that the volume of a sphere with radius $r$ is given by $\frac{4}{3}\pi {r^3}$

Let the radius of the original ball be $R$.

So the volume of the original spherical ball with $$R = 3$$ is equal to $$\frac{4}{3}\pi {(3)^3}$$

Now the original ball is melted to form three balls of different sizes.

Let the radii of three resulting spherical balls be ${r_1},{r_2},{r_3}$ respectively.

Given ${r_1} = 1.5cm$ & ${r_2} = 2cm$.

$ \Rightarrow {V_{total}} = {V_1} + {V_2} + {V_3}$

$ \Rightarrow \frac{4}{3}\pi {R^3} = \frac{4}{3}\pi ({r_1}^3 + {r_2}^3 + {r_3}^3)$

$ \Rightarrow {r_3}^3 = {3^3} - {1.5^3} - {2^3} = 15.625$

Hence ${r_3} = 2.5cm$

$\therefore $The radius of the third ball = 2.5cm

Note: In the given problem it is said that the spherical ball of radius 3cm is melted to form three spherical balls. So the volume of the original spherical ball is equal to the sum of the volumes of the three balls.

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