Question

Find the height of the cylinder whose volume is 1.54 m3 and diameter of base is 140 cm.

Answer 1

$\begin{gathered}\begin{gathered}\bf \:Given - \begin{cases} &\sf{Volume_{(Cylinder)} = 1.54 \: {m}^{3} } \\ &\sf{Diameter_{(Cylinder)} = 140 \: cm = 1.4 \: m} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\sf{Height_{(Cylinder)}}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}$

We know,

$\bf :\longmapsto\: \boxed{ \bf{Volume_{(Cylinder)} = \pi \: {r}^{2} \: h}}$

where,

• r = radius of cylinder

• h = height of cylinder

$\large\underline{\bold{Solution-}}$

Given that,

• Volume of cylinder = 1.54 m³

• Diameter of cylinder = 1.4 m

So,

• Radius of cylinder, r = 0.7 m

• Let Height of cylinder be 'h' m.

Using the formula of Volume of cylinder, we have

$\rm :\longmapsto\:Volume_{(Cylinder)} \: = \: \pi \: {r}^{2} \: h$

On Substituting the values, we get

$\rm :\longmapsto\:1.54 = \dfrac{22}{7} \times 0.7 \times 0.7 \times h$

$\rm :\longmapsto\:\dfrac{154}{100} = \dfrac{22}{7} \times \dfrac{7}{10} \times \dfrac{7}{10} \times h$

$\rm :\longmapsto\:h \: = \: 1 \: m$

$\boxed{ \bf{Hence, \: height \: of \: cylinder, \: h =1 m \: or \: 100 cm}}$

Additional Information :-

$\boxed{ \bf{ \:CSA_{(Cylinder)} = 2\pi \: rh }}$

$\boxed{ \bf{ TSA_{(Cylinder)} = 2\pi \: r(h + r)}}$

where,

• r = radius of cylinder

• h = height of cylinder

Answer 2

Given: Volume of cylinder = 1.54 m and Diameter of cylinder = 140 cm

$radius (r) = \frac{d}{2} = \frac{140}{2} = 70cm$

$volume \: \: of \: \: cylinder \: \: = {\pi \: r}^{2} h$

$= > 1.54 = \frac{22}{7} \times 0.7 \times 0.7 \times h \\ = > h = \frac{1.54 \times 7}{22 \times 0.7 \times 0.7} \\ \\ = > \frac{154 \times 7 \times 10 \times 10 }{22 \times 7 \times 7 \times 100} = 1m$

Hence, the height of the cylinder is 1 m.