A solid metallic sphere of diameter 28cm is melted and recast into a number of smaller cones, each of diameter $4\dfrac{2}{3}$ cm and height 3cm. Find the number of cones so formed.
Answer
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Hint: In this particular question use the concept that after melting the volume remain constant so the volume of the metallic sphere is equal to the total volume of the cones, where volume of sphere is $\dfrac{4}{3}\pi {R^3}$ cubic units and volume of cone is $\dfrac{1}{3}\pi {r^2}h$ where symbols have their usual meanings so use these concepts to reach the solution of the question.
Complete step-by-step answer:
Given data:
A solid metallic sphere of diameter (D) 28cm.
As we all know that radius is half of the diameter, so the radius of the sphere is, $R = \dfrac{D}{2} = \dfrac{{28}}{2} = 14$cm.
Now as we know that the volume (V) of the sphere is given as $\dfrac{4}{3}\pi {R^3}$ cubic cm.
$ \Rightarrow V = \dfrac{4}{3}\pi {R^3} = \dfrac{4}{3}\pi {\left( {14} \right)^3}{\text{ c}}{{\text{m}}^3}$
Now according to the question solid metallic sphere is melted and cast into small cones.
The diameter of the each cone is d = $4\dfrac{2}{3} = \dfrac{{14}}{3}$cm and height of the each cone is, h = 3cm
So the radius of the each cone is, $r = \dfrac{d}{2} = \dfrac{{\dfrac{{14}}{3}}}{2} = \dfrac{7}{3}$cm.
Now as we know that the volume (${V_1}$) of the cone is given as, $\dfrac{1}{3}\pi {r^2}h$ cubic units.
$ \Rightarrow {V_1} = \dfrac{1}{3}\pi {r^2}h = \dfrac{1}{3}\pi {\left( {\dfrac{7}{3}} \right)^2}\left( 3 \right){\text{ c}}{{\text{m}}^3}$
Now let x number of cones are made.
So the total volume of the cones is x times the volume of one cone.
So the total volume of x cones = $x.\dfrac{1}{3}\pi {\left( {\dfrac{7}{3}} \right)^2}\left( 3 \right){\text{ c}}{{\text{m}}^3}$
Now as we know that after melting the volume remains constant so the volume of the metallic sphere is equal to the total volume of the cones.
$ \Rightarrow x.\dfrac{1}{3}\pi {\left( {\dfrac{7}{3}} \right)^2}\left( 3 \right) = \dfrac{4}{3}\pi {\left( {14} \right)^3}$
Now simplify it we have,
$ \Rightarrow x{\left( {\dfrac{7}{3}} \right)^2} = \dfrac{4}{3}{\left( {14} \right)^3}$
$ \Rightarrow x = \dfrac{{4{{\left( {14} \right)}^3}\left( {{3^2}} \right)}}{{3\left( {{7^2}} \right)}} = 4 \times 3\left( {2 \times 2 \times 14} \right) = 672$ Cones.
So there are a total 672 such cones formed.
So this is the required answer.
Note: Whenever we face such types of questions the key concept we have to remember is the formula of volume of sphere as well as cone which is both stated above, and always remember that after melting volume remain constant, so simplify calculate the volume of sphere and volume of one cone where as necessary parameters are given in the question, so if there are x such cones so the x times the volume of one cone is equal to the volume of sphere so equate them and simplify as above we will get the required number of such cones.
Complete step-by-step answer:
Given data:
A solid metallic sphere of diameter (D) 28cm.
As we all know that radius is half of the diameter, so the radius of the sphere is, $R = \dfrac{D}{2} = \dfrac{{28}}{2} = 14$cm.
Now as we know that the volume (V) of the sphere is given as $\dfrac{4}{3}\pi {R^3}$ cubic cm.
$ \Rightarrow V = \dfrac{4}{3}\pi {R^3} = \dfrac{4}{3}\pi {\left( {14} \right)^3}{\text{ c}}{{\text{m}}^3}$
Now according to the question solid metallic sphere is melted and cast into small cones.
The diameter of the each cone is d = $4\dfrac{2}{3} = \dfrac{{14}}{3}$cm and height of the each cone is, h = 3cm
So the radius of the each cone is, $r = \dfrac{d}{2} = \dfrac{{\dfrac{{14}}{3}}}{2} = \dfrac{7}{3}$cm.
Now as we know that the volume (${V_1}$) of the cone is given as, $\dfrac{1}{3}\pi {r^2}h$ cubic units.
$ \Rightarrow {V_1} = \dfrac{1}{3}\pi {r^2}h = \dfrac{1}{3}\pi {\left( {\dfrac{7}{3}} \right)^2}\left( 3 \right){\text{ c}}{{\text{m}}^3}$
Now let x number of cones are made.
So the total volume of the cones is x times the volume of one cone.
So the total volume of x cones = $x.\dfrac{1}{3}\pi {\left( {\dfrac{7}{3}} \right)^2}\left( 3 \right){\text{ c}}{{\text{m}}^3}$
Now as we know that after melting the volume remains constant so the volume of the metallic sphere is equal to the total volume of the cones.
$ \Rightarrow x.\dfrac{1}{3}\pi {\left( {\dfrac{7}{3}} \right)^2}\left( 3 \right) = \dfrac{4}{3}\pi {\left( {14} \right)^3}$
Now simplify it we have,
$ \Rightarrow x{\left( {\dfrac{7}{3}} \right)^2} = \dfrac{4}{3}{\left( {14} \right)^3}$
$ \Rightarrow x = \dfrac{{4{{\left( {14} \right)}^3}\left( {{3^2}} \right)}}{{3\left( {{7^2}} \right)}} = 4 \times 3\left( {2 \times 2 \times 14} \right) = 672$ Cones.
So there are a total 672 such cones formed.
So this is the required answer.
Note: Whenever we face such types of questions the key concept we have to remember is the formula of volume of sphere as well as cone which is both stated above, and always remember that after melting volume remain constant, so simplify calculate the volume of sphere and volume of one cone where as necessary parameters are given in the question, so if there are x such cones so the x times the volume of one cone is equal to the volume of sphere so equate them and simplify as above we will get the required number of such cones.
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