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Hint: First of all find the volume of the sphere of radius 3 cm and volume of the spherical ball of diameter 0.6 cm. Then equate the volume of the sphere to the volume of the total number of balls that is,

Volume of sphere = n (Volume of one spherical ball)

Here, find the value of n which is the number of balls formed.

We are given a solid sphere of radius 3 cm which is melted and cast into small spherical balls each of diameter 0.6 cm. Here, we have to find the number of balls formed.

Before proceeding with the question, we must know that when one object of a particular shape is converted into other object/objects of the same or different shape, then volume always remains constant. That is, the volume of the object before and after the conversion would be the same, while the other quantities such as surface area, length, breadth, radius, etc. could change.

Now, we are given a solid sphere of radius 3 cm. We will first find the volume of this sphere.

We know that the volume of the sphere, \[V=\dfrac{4}{3}\pi {{r}^{3}}\].

By putting the volume of r = 3 cm, we get,

Volume of the sphere, \[V=\dfrac{4}{3}\pi {{\left( 3 \right)}^{3}}\text{ cubic cm}\]

Therefore, we get \[V=\dfrac{4}{3}\pi \left( 27 \right)\text{ cubic cm}\]

Or, \[V=36\pi \text{ cubic cm}\]

Now, we are given that this solid sphere is melted and cast into small spherical balls of diameter 0.6 cm.

As we know, \[\text{Radius}\left( r \right)=\dfrac{\text{Diameter}}{2}\]

Therefore, we get radius of balls \[=\dfrac{0.6}{2}=0.3\text{ }cm\]

Let us take the number of balls to be â€˜nâ€™.

Now, we will find the volume of each spherical ball.

As we know that volume of sphere \[=\dfrac{4}{3}\pi {{r}^{3}}\], therefore we get,

Volume of each spherical ball, \[{{V}_{1}}=\dfrac{4}{3}\pi {{r}^{3}}\]

We have found that r = 0.3 cm.

Therefore, we get, the volume of each spherical ball

\[{{V}_{1}}=\dfrac{4}{3}\pi {{\left( 0.3 \right)}^{3}}\text{ cubic cm}\]

\[{{V}_{1}}=\dfrac{4}{3}\pi \left( 0.027 \right)\text{ cubic cm}\]

Therefore, we get \[{{V}_{1}}=0.036\pi \text{ cubic cm}\]

Since one spherical ball has volume \[{{V}_{1}}=0.036\pi \text{ cubic cm}\], therefore, â€˜nâ€™ spherical balls would have volume \[n{{V}_{1}}=n\left( 0.036\pi \right)\text{cubic cm}\].

As we already know that when one object is melted and cast into another object, then volume remains constant, therefore, we get,

(Volume of sphere of radius 3 cm) = (Volume of n spherical balls)

That is, \[V=n{{V}_{1}}\]

By putting the values of \[V=36\pi \] and \[{{V}_{1}}=0.036\pi \], we get

\[36\pi =n\left( 0.036\pi \right)\]

By dividing both sides with \[0.036\pi \], we get

\[\Rightarrow \dfrac{36\pi }{0.036\pi }=n\]

By simplifying the above equation, we get

\[\Rightarrow n=\dfrac{36\pi }{36\pi }\times 1000\]

Therefore, we get, \[n=1000\text{ balls}\]

Hence, a solid sphere of radius 3 cm is melted and cast into 1000 spherical balls of diameter 0.6 cm.

Note: Students often make this mistake of using the formulas of volume, area, etc. without properly looking whether the radius or diameter is given in the question. They must properly read and if the diameter is given, first convert it into radius and then only put it into the formula to get the desired value. Also, students must remember that whenever any 3-dimensional object is converted into another, the volume always remains constant.

Volume of sphere = n (Volume of one spherical ball)

Here, find the value of n which is the number of balls formed.

We are given a solid sphere of radius 3 cm which is melted and cast into small spherical balls each of diameter 0.6 cm. Here, we have to find the number of balls formed.

Before proceeding with the question, we must know that when one object of a particular shape is converted into other object/objects of the same or different shape, then volume always remains constant. That is, the volume of the object before and after the conversion would be the same, while the other quantities such as surface area, length, breadth, radius, etc. could change.

Now, we are given a solid sphere of radius 3 cm. We will first find the volume of this sphere.

We know that the volume of the sphere, \[V=\dfrac{4}{3}\pi {{r}^{3}}\].

By putting the volume of r = 3 cm, we get,

Volume of the sphere, \[V=\dfrac{4}{3}\pi {{\left( 3 \right)}^{3}}\text{ cubic cm}\]

Therefore, we get \[V=\dfrac{4}{3}\pi \left( 27 \right)\text{ cubic cm}\]

Or, \[V=36\pi \text{ cubic cm}\]

Now, we are given that this solid sphere is melted and cast into small spherical balls of diameter 0.6 cm.

As we know, \[\text{Radius}\left( r \right)=\dfrac{\text{Diameter}}{2}\]

Therefore, we get radius of balls \[=\dfrac{0.6}{2}=0.3\text{ }cm\]

Let us take the number of balls to be â€˜nâ€™.

Now, we will find the volume of each spherical ball.

As we know that volume of sphere \[=\dfrac{4}{3}\pi {{r}^{3}}\], therefore we get,

Volume of each spherical ball, \[{{V}_{1}}=\dfrac{4}{3}\pi {{r}^{3}}\]

We have found that r = 0.3 cm.

Therefore, we get, the volume of each spherical ball

\[{{V}_{1}}=\dfrac{4}{3}\pi {{\left( 0.3 \right)}^{3}}\text{ cubic cm}\]

\[{{V}_{1}}=\dfrac{4}{3}\pi \left( 0.027 \right)\text{ cubic cm}\]

Therefore, we get \[{{V}_{1}}=0.036\pi \text{ cubic cm}\]

Since one spherical ball has volume \[{{V}_{1}}=0.036\pi \text{ cubic cm}\], therefore, â€˜nâ€™ spherical balls would have volume \[n{{V}_{1}}=n\left( 0.036\pi \right)\text{cubic cm}\].

As we already know that when one object is melted and cast into another object, then volume remains constant, therefore, we get,

(Volume of sphere of radius 3 cm) = (Volume of n spherical balls)

That is, \[V=n{{V}_{1}}\]

By putting the values of \[V=36\pi \] and \[{{V}_{1}}=0.036\pi \], we get

\[36\pi =n\left( 0.036\pi \right)\]

By dividing both sides with \[0.036\pi \], we get

\[\Rightarrow \dfrac{36\pi }{0.036\pi }=n\]

By simplifying the above equation, we get

\[\Rightarrow n=\dfrac{36\pi }{36\pi }\times 1000\]

Therefore, we get, \[n=1000\text{ balls}\]

Hence, a solid sphere of radius 3 cm is melted and cast into 1000 spherical balls of diameter 0.6 cm.

Note: Students often make this mistake of using the formulas of volume, area, etc. without properly looking whether the radius or diameter is given in the question. They must properly read and if the diameter is given, first convert it into radius and then only put it into the formula to get the desired value. Also, students must remember that whenever any 3-dimensional object is converted into another, the volume always remains constant.

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