
A solid wax sphere of radius 5.6 cm is melted to make conical candles of diameters of base 5.6 cm and height 3.2 cm. Find the number of such candles prepared.
Answer
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Hint: In this question, we will use the concept that the volume of wax sphere will be equal to the combined volume of all the conical candles. We will use the following formulae for this: volume of sphere $=\dfrac{4}{3}\pi {{r}^{3}}$ and volume of a cone $\dfrac{1}{3}\pi {{r}^{2}}h$. Find the individual volumes of sphere and cone and then divide these to get the final answer.
Complete step-by-step answer:
We are given that a solid wax sphere of radius 5.6 cm is melted to make conical candles of diameters of base 5.6 cm and height 3.2 cm.
We need to find the number of such candles prepared.
We will first calculate the volume of the sphere and the candles separately.
Then we will divide the volume of the sphere by volume of candle to get the answer.
Given radius of the sphere = 5.6 cm
We know that the volume of sphere $=\dfrac{4}{3}\pi {{r}^{3}}$, where r is the radius of the sphere.
Using this we will calculate the volume of wax in the sphere.
Volume of wax in the sphere $=\dfrac{4}{3}\pi {{\left( 5.6 \right)}^{3}}$ …(1)
Now, we will find the volume of a conical candle.
We know that the volume of a cone $\dfrac{1}{3}\pi {{r}^{2}}h$,
Where r is the radius of the base of the cone and h is the height of the cone.
We are given that the diameter of the base of the cone is 5.6 cm.
So, its radius will be the half of the diameter which is 2.8 cm. Also the height is 3.2 cm.
Volume of the conical candle $\dfrac{1}{3}\pi {{\left( 2.8 \right)}^{2}}\times 3.2$ …(2)
Now, let n number of such candles be formed.
Now, we will divide (1) by (2) to get n.
$n=\dfrac{\dfrac{4}{3}\pi {{\left( 5.6 \right)}^{3}}}{\dfrac{1}{3}\pi {{\left( 2.8 \right)}^{2}}\times 3.2}$
Cancelling $\dfrac{\pi }{3}$, we get the following:
$n=\dfrac{4\times 5.6\times 5.6\times 5.6}{2.8\times 2.8\times 3.2}$
$n=\dfrac{16\times 5.6}{3.2}=28$
So, there will be 28 candles prepared from the wax sphere.
$n=\dfrac{16\times 5.6}{3.2}=28$
Note: It is important to know the formulae for volumes of various solid bodies. Like in this question, we will use the volume of sphere and cone. These are: volume of sphere $=\dfrac{4}{3}\pi {{r}^{3}}$and volume of a cone $\dfrac{1}{3}\pi {{r}^{2}}h$.
Complete step-by-step answer:
We are given that a solid wax sphere of radius 5.6 cm is melted to make conical candles of diameters of base 5.6 cm and height 3.2 cm.
We need to find the number of such candles prepared.
We will first calculate the volume of the sphere and the candles separately.
Then we will divide the volume of the sphere by volume of candle to get the answer.
Given radius of the sphere = 5.6 cm
We know that the volume of sphere $=\dfrac{4}{3}\pi {{r}^{3}}$, where r is the radius of the sphere.
Using this we will calculate the volume of wax in the sphere.
Volume of wax in the sphere $=\dfrac{4}{3}\pi {{\left( 5.6 \right)}^{3}}$ …(1)
Now, we will find the volume of a conical candle.
We know that the volume of a cone $\dfrac{1}{3}\pi {{r}^{2}}h$,
Where r is the radius of the base of the cone and h is the height of the cone.
We are given that the diameter of the base of the cone is 5.6 cm.
So, its radius will be the half of the diameter which is 2.8 cm. Also the height is 3.2 cm.
Volume of the conical candle $\dfrac{1}{3}\pi {{\left( 2.8 \right)}^{2}}\times 3.2$ …(2)
Now, let n number of such candles be formed.
Now, we will divide (1) by (2) to get n.
$n=\dfrac{\dfrac{4}{3}\pi {{\left( 5.6 \right)}^{3}}}{\dfrac{1}{3}\pi {{\left( 2.8 \right)}^{2}}\times 3.2}$
Cancelling $\dfrac{\pi }{3}$, we get the following:
$n=\dfrac{4\times 5.6\times 5.6\times 5.6}{2.8\times 2.8\times 3.2}$
$n=\dfrac{16\times 5.6}{3.2}=28$
So, there will be 28 candles prepared from the wax sphere.
$n=\dfrac{16\times 5.6}{3.2}=28$
Note: It is important to know the formulae for volumes of various solid bodies. Like in this question, we will use the volume of sphere and cone. These are: volume of sphere $=\dfrac{4}{3}\pi {{r}^{3}}$and volume of a cone $\dfrac{1}{3}\pi {{r}^{2}}h$.
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