Question

The radius of the cylinder is half its height and area of the inner part is 616 sq. cm. How many litres of milk approximately can it contain?
A.1.4 L
B.1.5 L
C.1.7 L
D.1.9 L

Answer 1

Hint: Here we are given that, \[r = \dfrac{h}{2}\] and area of inner part of cylinder is given as 616 sq. cm, we use the formula of surface area of cylinder, and substitute the value of r and first find the value of height of cylinder, then using that we find the volume of cylinder.

Complete step-by-step answer:

Here, we consider the cylinder as a right circular cylinder since the base of the given cylinder is a circle.
Let, the height and radius of the cylinder is h and r respectively.
Then, the base area \[ = \pi {r^2}\] and the whole surface \[ = \left( {2\pi rh + 2\pi {r^2}} \right)\]and the volume\[ = \pi {r^2}h\]
Here, given, radius of the cylinder is half its height, i.e. \[r = \dfrac{h}{2}\]
We know, the formula that the whole surface of a right cylinder
\[
   = \left( {area{\text{ }}of{\text{ }}two{\text{ }}base} \right) + \left( {area{\text{ }}of{\text{ }}the{\text{ }}lateral{\text{ }}surface} \right) \\
   = \left( {2\pi rh + 2\pi {r^2}} \right) \\
 \]
Here given, the area of the inner part is 616 sq. cm.
\[
  \therefore 2\pi rh + 2\pi {r^2} = 616 \\
   \Rightarrow 2\pi \left( {\dfrac{h}{2}} \right)h + 2\pi {\left( {\dfrac{h}{2}} \right)^2} = 616\left[ {\because r = \dfrac{h}{2}} \right] \\
   \Rightarrow \pi {h^2} + \dfrac{{\pi {h^2}}}{4} = 616 \\
   \Rightarrow \pi {h^2}\left( {1 + \dfrac{1}{4}} \right) = 616 \\
   \Rightarrow \pi {h^2} \cdot \dfrac{5}{4} = 616 \\
 \]
On simplifying and substituting the value of \[{{\pi }}\], and simplify we get,
\[ \Rightarrow {h^2} = \dfrac{{28 \times 28}}{5}\]
\[ \Rightarrow {{\text{h}}^2} = \dfrac{{784}}{5}\]
Taking square root on both the sides we get,
\[ \Rightarrow h = \dfrac{{28}}{{\sqrt 5 }}\]
Now, we find the volume of the cylinder,
\[
   = \pi {r^2}h \\
   = \pi {\left( {\dfrac{h}{2}} \right)^2}h \\
 \]
\[ = \dfrac{\pi }{4} \cdot {h^3}\]
On substituting the value of h we get,
\[
   = \dfrac{\pi }{4} \cdot {\left( {\dfrac{{28}}{{\sqrt 5 }}} \right)^3} \\
   = \dfrac{{22}}{7} \times \dfrac{{28 \times 28 \times 28}}{{5\sqrt 5 }} \times \dfrac{1}{4} \\
   = 1542.7 \\
 \]
\[\therefore \]The volume of the cylinder\[{\text{ = 1542}}{\text{.7 }}c{m^3}\]\[ = \dfrac{{1542.7}}{{100}}{\text{ }}{m^3}\]\[ = 1.5{\text{ }}{m^3} = 1.5{\text{ }}L\left[ {\because 1{\text{ }}L = 1{\text{ }}{m^3}} \right]\]
So, the cylinder contains 1.5 L milk approximately.
Hence, option (B) is correct.

Note: Here we apply the formula of cylinder, which are, the whole surface area of a right cylinder \[ = \left( {2\pi rh + 2\pi {r^2}} \right)\]units and volume \[ = \pi {r^2}h\]units, where r and h are the radius and height of the cylinder.
Note that we have to consider the entire surface area of the cylinder and not just the lateral surface area.