Question

the radius of a solid iron sphere in 8 centime. 8 rings of iron plate of external radius 6 2/3 cm and thickness 3 cm are made by melting this sphere. find the internal diameter of each ring
|hint : diameter = radius/2|​

Answer 1

Given :

• Radius of iron sphere = 8 cm.
• External radii of iron ring (R) = 20/3 cm.
• Height (thickness) of iron ring (h) = 3 cm.
• Total number of rings made out of sphere = 8.

To Find :

• Internal diameter of the iron ring (2r).

Solution :

We know,

• Formula for finding the volume of a sphere = $\rm{\dfrac{4}{3}{\pi}r^3}$

Therefore,

Volume of solid iron sphere =

$\implies\rm{\dfrac{4}{3}\pi\times{}8^3\:cm^3}$

$\implies\rm{\dfrac{2048}{3}\pi\:cm^3}$

$\hrulefill$

Now, let the internal radius of each ring be r cm. Also, each ring forms a hollow cylinder of given dimensions of external radii, internal radii and height.

Then,

Volume of each ring = External volume - Internal volume

$\implies\rm{{\pi}R^2h-{\pi}r^2h}$

$\implies\rm{{\pi}h(R^2-r^2)}$

$\implies\rm{3{\pi}\left[\left(\dfrac{20}{3}\right)^2-r^2\right]\:cm^3}$

Then, volume of 8 such rings =

$\implies\rm{8\times{}3{\pi}\left(\dfrac{400}{9}-r^2\right)\:cm^3}$

$\implies\rm{24{\pi}\left(\dfrac{400}{9}-r^2\right)\:cm^3}$

Also, Volume of 8 rings = Volume of the sphere

$\implies\rm{24{\pi}\left(\dfrac{400}{9}-r^2\right)=\dfrac{2048}{3}\pi}$

$\implies\rm{\dfrac{400}{9}-r^2=\dfrac{2048}{3}\pi\times{}\dfrac{1}{24{\pi}}}$

$\implies\rm{r^2=\dfrac{400}{9}-\dfrac{256}{9}}$

$\implies\rm{r^2=\dfrac{144}{9}=16}$

$\implies\bold{r=4\:cm.}$

Hence, internal diameter of each ring = 2 × 4 = 8 cm.

Answer 2

Step-by-step explanation:

$x \sqrt[ | < \frac{ \sqrt{2} }{5} | ]{555555555555555}$

the radius of a solid iron sphere in 8 centime. 8 rings of iron plate of external radius 6 2/3 cm and thickness 3 cm are made by melting this sphere. find the internal diameter of each ring

|hint : diameter = radius/2|