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The lateral surface area of a cylinder of length 14m is 220${m^2}$. Find the volume of the cylinder.

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Last updated date: 22nd Mar 2024
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MVSAT 2024
Answer
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Hint: In this question use the direct formula for lateral surface area of the cylinder which is $\left( {2\pi rh} \right)$, the length of cylinder given refers to the height of the cylinder. The volume of the cylinder is directly given by the formula $\pi {r^2}h$.

Complete step-by-step answer:
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As we know that the lateral surface area (L.S.A) of the cylinder is $\left( {2\pi rh} \right)$ square units, where (r) and (h) is the radius and the length of the cylinder as shown in figure.

$ \Rightarrow L.S.A = 2\pi rh$ Sq. units.

Now it is given that the length of the cylinder is = 14 m (see figure).

And the L.S.A = 220 sq. meter.

$ \Rightarrow 220 = 2\pi r \times 14$

$ \Rightarrow r = \dfrac{{220}}{{28\pi }}$ Meter.

Now as we know that the volume (V) of the cylinder is = $\pi {r^2}h$ cubic units.

$ \Rightarrow V = \pi {r^2}h$ Cubic meter.

So substitute the value in this equation we have,

 $ \Rightarrow V = \pi {\left( {\dfrac{{220}}{{28\pi }}} \right)^2}\left( {14} \right)$ $m^3$.

Now simplify the above equation we have,

$ \Rightarrow V = \pi \times \dfrac{{{{\left( {220} \right)}^2}}}{{{{\left( {28} \right)}^2}{\pi
^2}}}\left( {14} \right)$

$ \Rightarrow V = \dfrac{{{{\left( {220} \right)}^2}}}{{{{\left( {28} \right)}^2}\pi }}\left( {14}
\right) = \dfrac{{{{\left( {220} \right)}^2}}}{{{{\left( {28} \right)}^2} \times
\dfrac{{22}}{7}}}\left( {14} \right) = \dfrac{{2200}}{{2 \times 4}} = 275{\text{ }}{{\text{m}}^3}$
$\left[ {\because \pi = \dfrac{{22}}{7}} \right]$

So this is the required volume of the cylinder.

Note: The lateral surface area is the area of the side surfaces of the cylinder. It is always advised to remember the basic formula like lateral surface area, curved surface area, total surface area, volume of basic shapes like cuboid, cube, cone etc. It helps save a lot of time while saving such problems as they are mostly formula based kinds of problems.
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