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How many balls each of radius $1cm$ can be made by melting a bigger ball whose diameter is $8cm$?

seo-qna
Last updated date: 19th Jul 2024
Total views: 348k
Views today: 7.48k
Answer
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Hint: First we have to define what the terms we need to solve the problem are. Since the diameter of the bigger ball is given as $8cm$(centimeter) and radius is the half of the diameter and thus radius of the bigger ball is $\dfrac{8}{2} = 4$cm. and we need to find total how many balls with the radius of minimum $1cm$.

Complete step-by-step solution:
Since assume that the radius of the smallest ball is $1cm$(from given), and since the bigger ball’s diameter is eight centimeters also the radius is four centimeters;
Now we need to calculate how many bigger balls can be melt into exactly one centimeter.
Hence, we use the formula for the volume which is Volume=$\dfrac{4}{3}\pi {r^3}$
First we find the bigger balls radius of volume which is $r = 4$and hence substitute that radius in the volume formula we get; Volume of the radius $r = 4$ $ \Rightarrow $$\dfrac{4}{3}\pi {r^3}$$ \Rightarrow \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {(4)^3}$(the cubic root of four is sixty four) thus we get Volume = $\dfrac{{256}}{3}\pi $(also the centimeter cube )
Now we find the radius of the smallest balls $1cm$volume = $\dfrac{4}{3}\pi {r^3}$$ \Rightarrow \dfrac{4}{3}\pi {(1)^3}$=$\dfrac{4}{3}\pi $(\[c{m^3}\])
Hence, we have the volume of the radius four and also the radius one centimeter
Now we will divide the both volumes to find the total number of balls melted at one radius.
Thus, Number of balls with each of radius $1cm$= $\dfrac{{\dfrac{{256\pi }}{3}}}{{\dfrac{{4\pi }}{3}}} \Rightarrow \dfrac{{256\pi }}{3} \times \dfrac{3}{{4\pi }}$(cross multiplying divided into multiplication) hence further solving we get Number of balls = $64$(melted at one centimeter)

Note: Since volume formula is $\dfrac{4}{3}\pi {r^3}$; r is the radius and $\pi $is the pie or $\dfrac{{22}}{7}$and also half of the diameter is the radius, because diameter is the overall length to the given problem but we need only half to find use of the volume. We can write the above expression as $n\times \text{Volume of small balls} = \text{Volume of bigger ball}$ where n= number of small balls. Here we are melting the bigger ball into smaller balls so the volume of the bigger ball will be distributed in small balls.