
A military tent is in the form of a right circular cone 21dm in height the diameter of the base being 4m. If 16 men sleep in it find the average volume in cu.dm of air per man:
A. $230$
B. $550$
C. $350$
D. $250$
Answer
581.7k+ views
Hint: Volume of any object is the amount of space occupied by it. The formula to calculate volume of a right circular cone is
$V = \dfrac{1}{3}\pi {r^2}h$
Where, $r$ is the radius of base of the cone
$h$ is the height of right circular cone
$V$ is the volume of cone
The diameter $\left( d \right)$ of any circle is twice the radius of that circle i.e $d = 2r$.
Now total volume of air inside the right circular cone shaped tent is the volume of tent itself and if some men are inside a right circular cone shaped tent and the amount of air present inside tent is uniformly distributed for each man then
$Volume{\text{ }}of{\text{ }}air{\text{ }}per{\text{ }}man{\text{ }} = {\text{ }}\dfrac{{Volume{\text{ }}of{\text{ }}Cone}}{{Number{\text{ }}of{\text{ }}men}}$
Complete Answer:
We have diameter of base of cone, $d = 4m$
$
\therefore Radius,r = \dfrac{d}{2} \\
{\text{ }} = \dfrac{{4m}}{2} \\
{\text{ }} = 2m \\
{\text{ }} = 2 \times 10dm \\
{\text{ }} = 20dm \\
$
Now we height of cone is, $h = 21dm$
The Right Circular Cone shaped tent is
Now we find the volume of air inside cone which is equal to the volume of cone
$
V = \dfrac{1}{3}\pi {r^2}h \\
= \dfrac{1}{3} \times \dfrac{{22}}{7} \times {\left( {20dm} \right)^2} \times 21dm \\
= \dfrac{{22 \times 400 \times 21}}{{3 \times 7}}d{m^3} \\
= 8800d{m^3} \\
$
Now we have 16 men sleeping in a tent. So, we have to divide Volume of air inside the tent by the number of men present in the tent.
$
Volume{\text{ }}of{\text{ }}air{\text{ }}per{\text{ }}man{\text{ }} = {\text{ }}\dfrac{{Volume{\text{ }}of{\text{ }}Cone}}{{Number{\text{ }}of{\text{ }}men}} \\
= \dfrac{{8800d{m^3}}}{{16}} \\
= 550d{m^3} \\
$
Hence the average volume in cu.dm of air per man is $550\,d{m^3}$.
Hence Option B is correct.
Note: In these kinds of questions, check the units of all given parameters, if the units are different convert them into one form. Meter and Decimeter are the two units of distances and
$1\,m = 10\,\,dm$.
$V = \dfrac{1}{3}\pi {r^2}h$
Where, $r$ is the radius of base of the cone
$h$ is the height of right circular cone
$V$ is the volume of cone
The diameter $\left( d \right)$ of any circle is twice the radius of that circle i.e $d = 2r$.
Now total volume of air inside the right circular cone shaped tent is the volume of tent itself and if some men are inside a right circular cone shaped tent and the amount of air present inside tent is uniformly distributed for each man then
$Volume{\text{ }}of{\text{ }}air{\text{ }}per{\text{ }}man{\text{ }} = {\text{ }}\dfrac{{Volume{\text{ }}of{\text{ }}Cone}}{{Number{\text{ }}of{\text{ }}men}}$
Complete Answer:
We have diameter of base of cone, $d = 4m$
$
\therefore Radius,r = \dfrac{d}{2} \\
{\text{ }} = \dfrac{{4m}}{2} \\
{\text{ }} = 2m \\
{\text{ }} = 2 \times 10dm \\
{\text{ }} = 20dm \\
$
Now we height of cone is, $h = 21dm$
The Right Circular Cone shaped tent is
Now we find the volume of air inside cone which is equal to the volume of cone
$
V = \dfrac{1}{3}\pi {r^2}h \\
= \dfrac{1}{3} \times \dfrac{{22}}{7} \times {\left( {20dm} \right)^2} \times 21dm \\
= \dfrac{{22 \times 400 \times 21}}{{3 \times 7}}d{m^3} \\
= 8800d{m^3} \\
$
Now we have 16 men sleeping in a tent. So, we have to divide Volume of air inside the tent by the number of men present in the tent.
$
Volume{\text{ }}of{\text{ }}air{\text{ }}per{\text{ }}man{\text{ }} = {\text{ }}\dfrac{{Volume{\text{ }}of{\text{ }}Cone}}{{Number{\text{ }}of{\text{ }}men}} \\
= \dfrac{{8800d{m^3}}}{{16}} \\
= 550d{m^3} \\
$
Hence the average volume in cu.dm of air per man is $550\,d{m^3}$.
Hence Option B is correct.
Note: In these kinds of questions, check the units of all given parameters, if the units are different convert them into one form. Meter and Decimeter are the two units of distances and
$1\,m = 10\,\,dm$.
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