Question

The radius of the cylinder is half its height and area of the inner part is 616 sq. cm. How many liters 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 have to find the amount of liters of milk that the cylinder can contain. First we will find the radius and height of the cylinder using the area of the inner part of the cylinder given. Then we will use the formula of volume of the cylinder to find the amount of milk that the cylinder can contain in it. Volume of a cylinder is the amount of any material which can be contained in the cylinder.

Formula used:
We will use the following formula:
The area of the inner part of a cylinder is given by \[A = 2\pi rh + \pi {r^2}\] .
Volume of a cylinder is given by \[V = \pi {r^2}h\] where \[h,r\] are the height and radius respectively.

Complete step by step solution:
Let \[h\] and \[r\] be the height and radius of the cylinder respectively.
We are given that the radius of the cylinder is half its height.
So, \[r = \dfrac{h}{2}\]
We are also given that the area of the inner part is 616 sq. cm.
Area of the inner part\[ = 616{\rm{sq}}{\rm{.cm}}\]
We know that Area of the inner part of a cylinder is given by the sum of the area of the cylinder and the area of the bottom which is in the form of a circle. That means the area of the inner part of a cylinder is \[A = 2\pi rh + \pi {r^2}\] .
Thus, equating the area of the inner part of a cylinder to \[2\pi rh + \pi {r^2}\], we get
\[ \Rightarrow 2\pi rh + \pi {r^2} = 616\]
Substituting \[h = h\] and \[r = \dfrac{h}{2}\], we get
\[ \Rightarrow 2\pi \times \dfrac{h}{2} \times h + \pi {\left( {\dfrac{h}{2}} \right)^2} = 616\]
Simplifying the expression, we get
\[ \Rightarrow \pi \times h \times h + \pi \left( {\dfrac{{{h^2}}}{4}} \right) = 616\]
\[ \Rightarrow \pi \times {h^2} + \pi \left( {\dfrac{{{h^2}}}{4}} \right) = 616\]
Taking LCM on left hand side, we get
\[ \Rightarrow \pi \times {h^2} \times \dfrac{4}{4} + \pi \left( {\dfrac{{{h^2}}}{4}} \right) = 616\]
\[ \Rightarrow \dfrac{{4\pi {h^2}}}{4} + \dfrac{{\pi {h^2}}}{4} = 616\]
Adding the terms, we get
\[ \Rightarrow \dfrac{{5\pi {h^2}}}{4} = 616\]
Substituting the value of \[\pi = \dfrac{{22}}{7}\], we get
\[ \Rightarrow \dfrac{5}{4} \times \dfrac{{22}}{7} \times {h^2} = 616\]
Rewriting the equation, we get
\[ \Rightarrow {h^2} = 616 \times \dfrac{7}{{22}} \times \dfrac{4}{5}\]
Multiplying the terms, we get
\[ \Rightarrow {h^2} = \dfrac{{28 \times 28}}{5}\]
Taking square root on both the sides, we get
\[ \Rightarrow h = \dfrac{{28}}{{\sqrt 5 }}{\rm{cm}}\]
Now, we have to find the volume of a cylinder or the capacity of the container.
Substituting \[r = \dfrac{h}{2}\] in the formula \[V = \pi {r^2}h\], we get
\[ \Rightarrow V = \pi {\left( {\dfrac{h}{2}} \right)^2}h\]
Computing the square, we get
 \[ \Rightarrow V = \pi \times \dfrac{1}{4}{h^2} \times h\]
Substituting \[h = \dfrac{{28}}{{\sqrt 5 }}\] and \[\pi = \dfrac{{22}}{7}\] in the above equation, we get
\[ \Rightarrow V = \dfrac{{22}}{7} \times \dfrac{1}{4} \times \dfrac{{28 \times 28}}{5} \times \dfrac{{28}}{{\sqrt 5 }}\]
\[ \Rightarrow V = 22 \times \dfrac{{28}}{5} \times \dfrac{{28}}{{\sqrt 5 }}\]
Simplifying the expression, we get
\[ \Rightarrow V = \dfrac{{17248}}{{5 \times \sqrt 5 }}\]
\[ \Rightarrow V = 1542.7\]
We get the volume as \[1542.7{\rm{c}}{{\rm{m}}^3}\].
We will convert the cubic units to liters, for that we have to divide the volume by 1000 because \[1{\rm{L}} = 1000{\rm{c}}{{\rm{m}}^3}\] . Therefore, we get
\[ \Rightarrow V = \dfrac{{1542.7}}{{1000}}{\rm{L}}\]
\[ \Rightarrow V = 1.54{\rm{L}}\]
Therefore, 1.5 Liters of milk can be approximately contained in the container.

Thus Option (B) is the correct answer.

Note:
We should always keep in mind that the amount of a substance in the container or the capacity of the container is nothing but the volume of the container. We also know that volume will always be represented in cubic units. But the given substance milk is in the container in the liquid state, so the volume should be represented in liters. It is essential for us to convert the units into the SI units.