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

⇝ 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 = $\dfrac{2\text x}{2} =\text x$

• Height of Cylinder = 2x

Hence,

❒ Using Formula of Volume of Cylinder :

$\text V_{(Cylinder)} = \pi \times {\text x}^{2} \times 2\text x$

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

❒ 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,

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

Answer 2

Answer:

Given :-

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

To Find :-

• What is the required fraction of the original log which is cut away.

Solution :-

$\bigstar$ From a cylindrical log whose height is equal to its diameter.

Let,

$\mapsto \bf Height =\: 2y$

$\mapsto \bf Diameter = 2y$

Now, we have to find the radius,

As we know that :

$\clubsuit$ Radius Formula :

$\mapsto \sf\boxed{\bold{\pink{Radius =\: \dfrac{Diameter}{2}}}}$

$\implies \sf Radius =\: \dfrac{\cancel{2}y}{\cancel{2}}$

$\implies \sf Radius =\: \dfrac{y}{1}$

$\implies \sf\bold{\green{Radius =\: y}}$

Now, we have to find the volume of cylinder :

As we know that :

$\clubsuit$ Volume Of Cylinder Formula :

$\mapsto \sf\boxed{\bold{\pink{Volume_{(Cylinder)} =\: {\pi}r^2h}}}$

where,

• π = Pie or 22/7
• r = Radius
• h = Height

Hence,

Given :

• Radius = y
• Height = 2y

According to the question by using the formula we get,

$\implies \sf Volume_{(Cylinder)} =\: \dfrac{22}{7} \times y^2 \times 2y$

$\implies \sf Volume_{(Cylinder)} =\: \dfrac{22}{7} \times 2y^3$

$\implies \sf\bold{\purple{Volume_{(Cylinder)} =\: \dfrac{44}{7}y^3}}$

Hence, the volume of cylinder is 44/7y³.

Again, we have to find the volume of Sphere :

As we know that :

$\clubsuit$ Volume Of Sphere Formula :

$\mapsto \sf\boxed{\bold{\pink{Volume_{(Sphere)} =\: \dfrac{4}{3}{\pi}r^3}}}$

where,

• π = Pie or 22/7
• r = Radius

Hence,

Given :

• Radius = y

According to the question by using the formula we get,

$\implies \sf Volume_{(Sphere)} =\: \dfrac{4}{3} \times \dfrac{22}{7} \times y^3$

$\implies \sf\bold{\purple{Volume_{(Sphere)} =\: \dfrac{88}{21}y^3}}$

Hence, the volume of sphere is 88/21y³.

Now, we have to find the required fraction :

$\longrightarrow \sf Required\: Fraction =\: \dfrac{Volume_{(Cylinder)}}{Volume_{(Sphere)}}$

$\longrightarrow \sf Required\: Fraction =\: \dfrac{\dfrac{44}{7}\cancel{y^3}}{\dfrac{88}{21}\cancel{y^3}}$

$\longrightarrow \sf Required\: Fraction =\: \dfrac{\dfrac{44}{7}}{\dfrac{88}{21}}$

$\longrightarrow \sf Required\: Fraction =\: \dfrac{44}{7} \times \dfrac{21}{88}$

$\longrightarrow \sf Required\: Fraction =\: \dfrac{44 \times 21}{7 \times 88}$

$\longrightarrow \sf Required\: Fraction =\: \dfrac{\cancel{924}}{\cancel{616}}$

$\longrightarrow \sf\bold{\red{Required\: Fraction =\: \dfrac{3}{2}}}$

${\small{\bold{\underline{\therefore\: The\: fraction\: of\: the\: original\: log\: which\: is\: cut\: away\: is\: \dfrac{3}{2}\: .}}}}$