Question

The capacity of a cylindrical tank is 6160cm^3. it's base diameter is 28cm find the depth of the tank

Answer 1

The capacity of the cylindrical tank is the same as its volume, i.e.,

$\longrightarrow\sf{V=6160\ cm^3}$

Since the tank is cylindrical in shape,

$\longrightarrow\sf{\pi r^2h=6160\quad\quad\dots(1)}$

where $\sf{r}$ is the base radius and $\sf{h}$ is the depth of the tank in cm.

Given that the base diameter of the tank is 28 cm. Hence base radius is,

$\longrightarrow\sf{r=\dfrac{28}{2}}$

$\longrightarrow\sf{r=14\ cm}$

Then (1) becomes,

$\longrightarrow\sf{\pi(14)^2h=6160}$

Then depth of the tank is,

$\longrightarrow\sf{196\pi h=6160}$

$\longrightarrow\sf{h=\dfrac{6160}{196\pi}}$

Taking $\sf{\pi=\dfrac{22}{7},}$

$\longrightarrow\sf{h=\dfrac{6160\times7}{196\times22}}$

$\longrightarrow\sf{\underline{\underline{h=10\ cm}}}$

Hence the depth of the tank is 10 cm.

Answer 2

Step-by-step explanation:

The capacity of the cylindrical tank is the same as its volume, i.e.,

\longrightarrow\sf{V=6160\ cm^3}⟶V=6160 cm

3

Since the tank is cylindrical in shape,

\longrightarrow\sf{\pi r^2h=6160\quad\quad\dots(1)}⟶πr

2

h=6160…(1)

where \sf{r}r is the base radius and \sf{h}h is the depth of the tank in cm.

Given that the base diameter of the tank is 28 cm. Hence base radius is,

\longrightarrow\sf{r=\dfrac{28}{2}}⟶r=

2

28

\longrightarrow\sf{r=14\ cm}⟶r=14 cm

Then (1) becomes,

\longrightarrow\sf{\pi(14)^2h=6160}⟶π(14)

2

h=6160

Then depth of the tank is,

\longrightarrow\sf{196\pi h=6160}⟶196πh=6160

\longrightarrow\sf{h=\dfrac{6160}{196\pi}}⟶h=

196π

6160

Taking \sf{\pi=\dfrac{22}{7},}π=

7

22

,

\longrightarrow\sf{h=\dfrac{6160\times7}{196\times22}}⟶h=

196×22

6160×7

\longrightarrow\sf{\underline{\underline{h=10\ cm}}}⟶

h=10 cm

Hence the depth of the tank is 10 cm.