Question

From a cylindrical log whose height is equal to its diameter, the greatest possible sphere has been taken out. What is the fraction of the original log which is cut away?​

Answer 1

Answer:

The answer is ⅔

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Answer 2

⇝ Given :-

From a cylindrical log whose height is equal to its diameter, the greatest possible sphere has been taken out.

⇝ To Find :-

The fraction of the original log which is cut away.

⇝ Solution :-

As Height and Diameter of cylindrical Log are equal.

Let,

Height and Diameter be = 2 x

★ We Know Volume of Cylinder is :

\large \boxed{ \red{{ \boxed{ \bf V = \pmb{\pi}{r} {}^{2}h}}}}

Where,

r = Radius of Cylinder

h = Height of Cylinder

We Have,

Diameter of Cylinder = 2x

So,

Radius of Cylinder =

$[tex] \dfrac{2\text x}{2} =\text x$[/tex]

Height of Cylinder = 2x

Hence,

❒ Using Formula of Volume of Cylinder :

$[tex] \text V_{(Cylinder)} = \pi \times {\text x}^{2} \times 2\text x$[/tex]

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf V_{(Cylinder)} = 2 \pmb\pi{x}^{3} } } }}$

❒ When the greatest possible sphere has been taken out :

$\small\red{\text{Radius of Sphere = Radius of Cylinder}}$

Hence,

Radius of Sphere = x

★ We Know Volume of a Sphere is :

$\large \boxed{\red{ \boxed{\bf{V_{(Sphere)} = \dfrac{4}{3} \pmb\pi {r}^{3}}}}}$

❒ Using Formula of Volume of Sphere :

$\text V_{(Sphere)} = \frac{4}{3} \times\pi \times {\text x}^{3}$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf V_{(Sphere)} = \dfrac{4 \pmb\pi {x}^{3} }{3} }} }}$

Now,

$\small\text{Required Fraction} = \frac{ \text{Volume of log}}{ \text{Vol. of log taken out}}$

Hence,

$\text{Required Fraction} = \frac{ \text V_{(\text Cylinder)} }{ \text V_{(\text Sphere)}}$

$\dfrac{2\pi {\text x}^{3} }{ \dfrac{4}{3}\pi {\text x}^{3} }$

$= \dfrac{6 \: \cancel{\pi\text{x}^{3}} }{4 \: \cancel{\pi\text {x}^{3} }}$

$= \dfrac{6}{4}$

$= \bf \dfrac{3}{2}$

So,

$\large\underline{\pink{\underline{\frak{\pmb{\text Required \: Fraction =\dfrac{3}{2} }}}}}$