Question

find the cube root of (1.5)^3 send in process request for all plz​

Answer 1

Answer:

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}}}}

V=

π

πr

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

2

2x

=x

Height of Cylinder = 2x

Hence,

❒ Using Formula of Volume of Cylinder :

\begin{gathered} \text V_{(Cylinder)} = \pi \times {\text x}^{2} \times 2\text x \\ \end{gathered}

V

(Cylinder)

=π×x

2

×2x

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

:⟼

V

(Cylinder)

=2

π

πx

3

❒ When the greatest possible sphere has been taken out :

\begin{gathered} \small\red{\text{Radius of Sphere = Radius of Cylinder}} \\ \end{gathered}

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}}}}}

V

(Sphere)

=

3

4

π

πr

3

❒ Using Formula of Volume of Sphere :

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

V

(Sphere)

=

3

4

×π×x

3

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

:⟼

V

(Sphere)

=

3

4

π

πx

3

Now,

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

Required Fraction=

Vol. of log taken out

Volume of log

Hence,

\begin{gathered} \text{Required Fraction} = \frac{ \text V_{(\text Cylinder)} }{ \text V_{(\text Sphere)}} \\ \end{gathered}

Required Fraction=

V

(Sphere)

V

(Cylinder)

\begin{gathered} = \dfrac{2\pi {\text x}^{3} }{ \dfrac{4}{3}\pi {\text x}^{3} } \\ \end{gathered}

=

3

4

πx

3

2πx

3

\begin{gathered} = \dfrac{6 \: \cancel{\pi\text{x}^{3}} }{4 \: \cancel{\pi\text {x}^{3} }} \\ \end{gathered}

=

4

πx

3

6

πx

3

\begin{gathered} = \dfrac{6}{4} \\ \end{gathered}

=

4

6

\begin{gathered} = \bf \dfrac{3}{2} \\ \end{gathered}

=

2

3

So,

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

RequiredFraction=

2

3

RequiredFraction=

2

3

Answer 2

Answer: This is your answer