Video: CBSE Class X • Pack 4 • 2015 • Question 20

CBSE Class X • Pack 4 • 2015 • Question 20

11:37

Video Transcript

504 cones each of diameter 3.5 centimetres and height three centimetres are melted and recast into a metallic sphere. Find the diameter of the sphere and then find its surface area. Use 𝜋 is equal to 22 over seven.

Now for a problem like this, it’s always good to summarise the information given by the question. Here we’ll be working with cones. And the question tells us that the number of these cones, which we’ll call uppercase 𝑁, is equal to 504. The question also tells us that the diameter of each cone is 3.5 centimetres. That is to say, the diameter of the circle which forms the cone’s base is 3.5 centimetres. We’ll call this lowercase 𝑑.

Now we also know that the radius of any circle is half of its diameter. From the information, we can therefore conclude that the radius of the circle which forms the base of the cone is equal to 3.5 over two centimetres.

In order to avoid having to work with decimal numbers, let’s make sure we have an integer on the top and bottom of this fraction by multiplying the top and bottom by two. We can then see that 3.5 over two is the same as seven over four, and we can substitute this value in for the radius.

We now have that the radius of the circle which forms the cone’s base is seven over four centimetres, and we’ll call this lowercase 𝑟. Finally, we’re told that the height of the cone is three centimetres, and we’ll call this height ℎ.

Now the question goes on to say that all 504 cones are melted and recast into a single metallic sphere. Now for the purpose of this question, we’re gonna assume that no material is lost during the melting and recasting process. And from this, we can therefore conclude that our single sphere will have the same volume as our 504 original cones.

Now the question asked us to find the diameter of this sphere. And we can use the volume relationship that we’ve just identified to do so. Let’s start by finding the volume of our cones. The formula for the volume of a cone is given here, as one-third times 𝜋 times the radius of the cone squared times the height.

Now since we know the radius and the height of our cones, we can use this information to work out the volume. To find the volume of one cone, the first thing we can do is to substitute in the known values for the radius and height of the cone. And here we’ve shown our two substitutions. Looking at this first line of working, we see that we have a factor of a third and a factor of three, which can be cancelled out, since three times one-third is equal to one. We now have that the volume of one cone is equal to 𝜋 times seven over four squared.

Next, we recall that when squaring a fraction, we can individually square the top and bottom half. And so the volume of one cone is equal to 𝜋 times seven squared over four squared. We’re gonna choose to leave the expression in this form for now without evaluating the squares.

The next thing we can do is to recognise that we have an expression for the volume of one cone. But we want to find an expression for the volume of 504 cones. In order to do this, we can multiply both sides of our equation by 504. Doing so, we find that the volume of 504 cones is equal to 504 times 𝜋 times seven squared over four squared.

Now that we have this equation, we can look back at our question. And we can recall that the volume of 504 cones is equal to the volume of our one sphere. Here we have drawn our sphere, which has a radius of uppercase 𝑅.

Now the formula for the volume of a sphere is four over three times 𝜋 times the radius cubed. Since the radius of the sphere is half the diameter, we can use this formula to find the radius and then answer our question. In order to do this, we can equate the expression for the volume of our 504 cones with the expression for the volume of a sphere. We can then solve this new equation to find uppercase 𝑅.

Let’s tidy things up to make some room for our calculations. Here we have equated the volume of one sphere with the volume of our 504 cones. We can immediately see a factor of 𝜋 on both sides of the equation, and this can be cancelled out.

We now want to isolate the uppercase 𝑅 term on the left-hand side of the equation. And to do so, we multiply both sides of our equation by three over four, since three over four times four over three is equal to one. We now have that 𝑅 cubed is equal to 504 times seven squared over four squared times three over four. Let’s combine all of our terms into one fraction to make things easier to work with.

Now you may be able to see that, in order to find 𝑅, we’ll eventually need to take the cube root of both sides of the equation. Without a calculator, this might actually be a tricky process. And so in order to help us, we recall the following fact. Taking the cube root of 𝑎 times 𝑏 over 𝑐 or any product can be found by taking the cube root of the individual terms. This means that, for the right-hand side of our equation, if we can work toward a fraction where the individual terms are identifiable cubes, then this will make taking the cube root of them far easier.

Let’s now illustrate this with an example. On the bottom of our fraction, we can see that we have a four squared multiplied by a four. And we should be able to understand that this is equal to four cubed. Relating this to the rule that we’ve just mentioned, we can individually take the cube root of the four cubed on the bottom half of our fraction. Taking the cube root of four cubed is equal to four, which is a nice integer number.

Now although we can do this for the bottom half of our fraction, currently, the top half is far more complicated. This is mainly due to the factor of 504 that we have. One technique we might want to use to simplify this is to decompose 504 into its prime factors. Let’s do this now.

504 divided by two is 252, and two is of course a prime number. 252 can be written as the product of two and 126. And again, two is a prime factor. 126 can be written as two times 63, which gives us another prime factor of two. Two is no longer a factor, and we now divide 63 into three parts. So 63 is three times 21. Finally, 21 is three times seven, and we’re left with all of our prime factors.

From this, we can conclude that 504 equals two times two times two, which is two cubed, times three times three again, which is three squared, and finally times seven. And writing this in, we have that 504 is equal to two cubed times three squared times seven.

Let’s now perform a substitution of these values and see how it can help us. Here we see that we’ve performed the substitution of 504 as a product of its prime factors. For the top half of our fraction, we now see that we can combine like terms. We have two cubed times three squared times three which is equal to three cubed times seven times seven squared which is equal to seven cubed. And this is all divided by four cubed.

This is great. Our fraction is now expressed as the cube of many numbers, and all of these numbers are the cube of some integer. Using the rule that we mentioned earlier, we now understand that we can take the cube root of these individual terms. And this is the same as taking the cube root of the entire fraction. This allows us to take the cube root of both sides of our equation in a far more simple way, since the cube root of a cubed number is just the number itself.

We’re therefore left with 𝑅 is equal to two times three times seven divided by four. Two times three times seven is equal to six times seven, which is 42. 𝑅 is therefore equal to 42 divided by four. Dividing both the top and bottom half of this fraction by two to simplify gives us that 𝑅 is equal to 21 over two. Remembering our units, we can say that 𝑅 is equal to 21 over two centimetres. You could also write this as 10.5 centimetres, but we’re gonna choose to leave it as a fraction for now.

Now here we must remember that we haven’t actually answered our question yet, since it asked us for the diameter of the sphere and not the radius. We know that the diameter of any sphere is two times the radius. If we call the diameter of our sphere uppercase 𝐷, we therefore have that uppercase 𝐷 is equal to two times uppercase 𝑅, which is two times 21 over two. Of course, we can cancel the factor of two and of half. And we find that the diameter of the sphere is 21 centimetres, which satisfies the first part of our question.

Now that we have this information, we’re ready to move on to the final part of our question, which is finding the surface area of the newly formed sphere. The formula for a surface area of a sphere is equal to four times 𝜋 times the radius squared. However, since we have a nice round number for our diameter, we can choose to work with this instead.

We already know that the diameter of the sphere is equal to two times its radius. We can therefore say that the diameter squared is equal to two times the radius all squared. Squaring the individual terms in these brackets, we find that the diameter squared is equal to four times the radius squared. If we compare this to the formula for the surface area of a sphere, we can see that, on the right-hand side of both equations, we have a four times the radius squared. We can therefore perform a substitution of this with the diameter squared. And we find that the surface area of a sphere is equal to 𝜋 times the diameter squared.

We can now use this newly found formula in our calculations. Here in the formula for the surface area of the sphere, we substitute in the value that we have for the diameter and the approximate value of 𝜋 which the question gives us, which is 22 over seven. Again, this might be a little tricky to evaluate without a calculator.

To help us, we can decompose 21 into seven times three. We can then take the square of the individual terms inside the brackets, to find that 21 squared becomes seven squared times three squared. Combining this all into one fraction, we find it is 22 times seven times seven times three times three all divided by seven. At the very least, this lets us get rid of one factor of seven on the top and bottom half of the fraction.

We’re now able to write seven times three times three as seven times nine, which is 63. Our surface area is hence given by 22 times 63. To work this out, we can use a familiar column method. Two times 63 is 126. We then add in a zero for our next column, since we’re doing 20 times 63. And we can also write this in, since 20 times 63 is 1260.

Adding these two numbers together, we get that 22 times 63 is equal to 1386. The surface area of our sphere is therefore 1386 centimetres squared, since, remember, this is an area, not a length. Now that we’ve completed this step, we have fully answered the question. And we have found that the sphere which is formed by the melting and recasting above 504 original cones has a surface area of 1386 centimetres squared and a diameter of 21 centimetres.

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