Metallic spheres of radii 6 cm, 8 cm and 10 cm, respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.
Answer
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Hint: Volume of the 3 spheres when melted to form a single sphere will be equal. Equate the sum of volume of the 3 spheres with the volume of the new sphere to find the radius of the new sphere.
Three metallic spheres of radii 6 cm, 8 cm and 10 cm are to be melted and formed into a single sphere. Let the radius of the new sphere be ‘r’.
Let the radius of the three metallic spheres be \[r1,{\text{ }}r2\] and \[r3\].
So we have, $r1 = 6,r2 = 8,r3 = 10$
When the three spheres are melted and formed into a single sphere, their volumes will remain the same. So, we can write it as,
(Volume of sphere 1 + Volume of sphere 2 + Volume of sphere 3)=Volume of the new sphere ..(1)
Volume of sphere 1= $\dfrac{4}{3}\pi r{1^3}$
$
= \dfrac{4}{3} \times \pi \times {(6)^3} \\
= \dfrac{{864}}{3}\pi c{m^3} \\
$
Volume of sphere 2= $\dfrac{4}{3}\pi r{2^3}$
$
= \dfrac{4}{3} \times \pi \times {(8)^3} \\
= \dfrac{{2048}}{3}\pi c{m^3} \\
$
Volume of sphere 3= $\dfrac{4}{3}\pi r{3^3}$
$
= \dfrac{4}{3} \times \pi \times {(10)^3} \\
= \dfrac{{4000}}{3}\pi c{m^3} \\
$
So as per equation (1), we need to add all the three volumes and equate it to the volume of the new sphere with radius ‘r’
Volume of new sphere= \[\dfrac{{864}}{3}\pi + \dfrac{{2048}}{3}\pi + \dfrac{{4000}}{3}\pi \]
$
= \dfrac{\pi }{3}\left( {864 + 2048 + 4000} \right) \\
= \dfrac{{6912}}{3}\pi \\
= 2304\pi c{m^3} \\
$
Volume of new sphere= $\dfrac{4}{3}\pi {r^3}$
So, by equating both we get,
$
\dfrac{4}{3}\pi {r^3} = 2304\pi \\
{r^3} = \dfrac{{6912}}{4} \\
{r^3} = 1728 \\
$
Expressing in terms of exponents where the exponent needs to be 3, we get
${r^3} = {12^3}$
When exponents are the same, the bases are equal.
$r = 12$ cm
Hence the radius of the new sphere formed is 12 cm.
Note: In these types of problems, it is better to keep $\pi $ till the end because it will be easier to calculate and in the end it might get cancelled. This also avoids the possibility of calculation errors.
Three metallic spheres of radii 6 cm, 8 cm and 10 cm are to be melted and formed into a single sphere. Let the radius of the new sphere be ‘r’.
Let the radius of the three metallic spheres be \[r1,{\text{ }}r2\] and \[r3\].
So we have, $r1 = 6,r2 = 8,r3 = 10$
When the three spheres are melted and formed into a single sphere, their volumes will remain the same. So, we can write it as,
(Volume of sphere 1 + Volume of sphere 2 + Volume of sphere 3)=Volume of the new sphere ..(1)
Volume of sphere 1= $\dfrac{4}{3}\pi r{1^3}$
$
= \dfrac{4}{3} \times \pi \times {(6)^3} \\
= \dfrac{{864}}{3}\pi c{m^3} \\
$
Volume of sphere 2= $\dfrac{4}{3}\pi r{2^3}$
$
= \dfrac{4}{3} \times \pi \times {(8)^3} \\
= \dfrac{{2048}}{3}\pi c{m^3} \\
$
Volume of sphere 3= $\dfrac{4}{3}\pi r{3^3}$
$
= \dfrac{4}{3} \times \pi \times {(10)^3} \\
= \dfrac{{4000}}{3}\pi c{m^3} \\
$
So as per equation (1), we need to add all the three volumes and equate it to the volume of the new sphere with radius ‘r’
Volume of new sphere= \[\dfrac{{864}}{3}\pi + \dfrac{{2048}}{3}\pi + \dfrac{{4000}}{3}\pi \]
$
= \dfrac{\pi }{3}\left( {864 + 2048 + 4000} \right) \\
= \dfrac{{6912}}{3}\pi \\
= 2304\pi c{m^3} \\
$
Volume of new sphere= $\dfrac{4}{3}\pi {r^3}$
So, by equating both we get,
$
\dfrac{4}{3}\pi {r^3} = 2304\pi \\
{r^3} = \dfrac{{6912}}{4} \\
{r^3} = 1728 \\
$
Expressing in terms of exponents where the exponent needs to be 3, we get
${r^3} = {12^3}$
When exponents are the same, the bases are equal.
$r = 12$ cm
Hence the radius of the new sphere formed is 12 cm.
Note: In these types of problems, it is better to keep $\pi $ till the end because it will be easier to calculate and in the end it might get cancelled. This also avoids the possibility of calculation errors.
Last updated date: 24th Sep 2023
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Total views: 362.4k
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