
A friction clutch is in the form of a frustum of a cone, the diameter of the ends being 32 cm and 20 cm and length 8 cm. Find its bearing surface and volume.
Answer
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Hint:In the above given question, use the given data for the pictorial representation, it will simplify the calculations. Then substitute the values in the formula for the bearing surface and the volume of the frustum and further simplify them to obtain the solution.
Complete step-by-step answer:
Let ABB'A be the friction clutch of slant height $ l $ cm.
We have the given values as
$ {r_1} = 16cm, {r_2} = 10cm $ and $ h = 8cm $
Now, we know the formula
$ {l^2} = {h^2} + {r^2} $
Here, in case of a frustum, $ r = {r_1} - {r_2} $
Therefore, we get,
$ {l^2} = {h^2} + {\left( {{r_1} - {r_2}} \right)^2} $
$ \Rightarrow {l^2} = {8^2} + {\left( {16 - 10} \right)^2} $
\[ \Rightarrow {l^2} = 64 + {\left( 6 \right)^2}\]
$ \Rightarrow {l^2} = 64 + 36 $
$ \therefore l = 10cm $
Now, the bearing surface of the clutch = lateral surface of the frustum
And we know that the lateral surface of the frustum is given as
$ \begin{gathered}
= \pi \left( {{r_1} + {r_2}} \right)l \\
\\
\end{gathered} $
$ = \left[ {\dfrac{{22}}{7} \times \left( {16 + 10} \right) \times 10} \right]c{m^2} $
$ = 817.14c{m^2} $
The volume of a frustum cone $ = \dfrac{1}{3}\pi \left( {r_1^2 + r_2^2 + {r_1}{r_2}} \right)h $
Therefore, the volume of the friction clutch which is of the form of a frustum cone
\[ = \left[ {\dfrac{1}{3} \times \dfrac{{22}}{7} \times 8 \times \left( {{{16}^2} + 16 \times 10 + {{10}^2}} \right)} \right]c{m^3}\]
\[ = \left[ {\dfrac{{176}}{{21}}\left( {256 + 160 + 100} \right)} \right]c{m^3}\]
$ = \left[ {\dfrac{{176}}{{21}} \times 516} \right]c{m^3} $
$ = 4324.57c{m^3} $
Hence, the bearing surface of the friction clutch is $ 817.14c{m^2} $ and the volume is $ 4324.57c{m^3} $ respectively.
Note: To solve these kinds of problems, you must have an efficient knowledge of the solid shapes. Make sure you calculate the radius of the frustum before calculating the slant height. Similarly, while calculating the lateral surface area, make sure you have used the correct radius values.
Complete step-by-step answer:
Let ABB'A be the friction clutch of slant height $ l $ cm.
We have the given values as
$ {r_1} = 16cm, {r_2} = 10cm $ and $ h = 8cm $
Now, we know the formula
$ {l^2} = {h^2} + {r^2} $
Here, in case of a frustum, $ r = {r_1} - {r_2} $
Therefore, we get,
$ {l^2} = {h^2} + {\left( {{r_1} - {r_2}} \right)^2} $
$ \Rightarrow {l^2} = {8^2} + {\left( {16 - 10} \right)^2} $
\[ \Rightarrow {l^2} = 64 + {\left( 6 \right)^2}\]
$ \Rightarrow {l^2} = 64 + 36 $
$ \therefore l = 10cm $
Now, the bearing surface of the clutch = lateral surface of the frustum
And we know that the lateral surface of the frustum is given as
$ \begin{gathered}
= \pi \left( {{r_1} + {r_2}} \right)l \\
\\
\end{gathered} $
$ = \left[ {\dfrac{{22}}{7} \times \left( {16 + 10} \right) \times 10} \right]c{m^2} $
$ = 817.14c{m^2} $
The volume of a frustum cone $ = \dfrac{1}{3}\pi \left( {r_1^2 + r_2^2 + {r_1}{r_2}} \right)h $
Therefore, the volume of the friction clutch which is of the form of a frustum cone
\[ = \left[ {\dfrac{1}{3} \times \dfrac{{22}}{7} \times 8 \times \left( {{{16}^2} + 16 \times 10 + {{10}^2}} \right)} \right]c{m^3}\]
\[ = \left[ {\dfrac{{176}}{{21}}\left( {256 + 160 + 100} \right)} \right]c{m^3}\]
$ = \left[ {\dfrac{{176}}{{21}} \times 516} \right]c{m^3} $
$ = 4324.57c{m^3} $
Hence, the bearing surface of the friction clutch is $ 817.14c{m^2} $ and the volume is $ 4324.57c{m^3} $ respectively.
Note: To solve these kinds of problems, you must have an efficient knowledge of the solid shapes. Make sure you calculate the radius of the frustum before calculating the slant height. Similarly, while calculating the lateral surface area, make sure you have used the correct radius values.
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