Answer
Verified
422.7k+ views
Hint: First of all find the outer surface by dividing the total cost of the painting by cost per sq. dm. Then use the formula of the total surface area of the cylinder that is \[2\pi r\left( r+h \right)\] to find the radius and height of the tank. Use this radius and height to find the volume that is \[\pi {{r}^{2}}h\].
Complete step-by-step answer:
We are given that the cost of painting the total outside surface of the closed cylindrical oil tank at 60 paise per sq. dm is Rs. 237.60. Also, the height of the tank is 6 times the radius of the base of the tank. We have to find the volume of the cylinder.
Let us consider the cylindrical tank diagrammatically.
Let us consider the total outside surface area of the cylindrical tank as S \[\text{d}{{\text{m}}^{2}}\].
We are given that the cost of painting per sq dm = 60 paise
So, we get the cost of painting the total outside surface area of the cylinder (S) = 60 S paise
We know that 1 paise \[=\text{Rs}\text{.}\dfrac{1}{100}\]
So, we get, 60 S paise \[=\text{Rs}\text{.}\dfrac{60S}{100}\]
Also, we are given that the total cost of the painting = Rs. 237.60
So, we get,
\[\text{Rs}\text{.}\dfrac{60S}{100}=\text{Rs}.237.6\]
By multiplying \[\dfrac{100}{60}\] on both the sides, we get,
\[S=\left( \dfrac{100}{60} \right)\times \left( 237.6 \right)\]
\[S=396\text{ }d{{m}^{2}}\]
Now, we know that the surface area of the cylinder \[=2\pi r\left( r+h \right)\] where r and h are the radius and height of the cylinder respectively.
We are given that the height is 6 times the radius of the base of the tank. So, we get,
h = 6r
So, we get the total outer surface area of the tank \[=2\pi r\left( r+6r \right)\]
\[\begin{align}
& =2\pi r.7r \\
& =14\pi {{r}^{2}} \\
\end{align}\]
We have already found that the total outer surface area of the tank \[=396\text{ d}{{\text{m}}^{2}}\]. So, we get,
\[14\pi {{r}^{2}}=396\text{ d}{{\text{m}}^{2}}\]
Or, \[{{r}^{2}}=\dfrac{396}{14\pi }\]
By substituting \[\pi =\dfrac{22}{7}\], we get,
\[{{r}^{2}}=\dfrac{396\times 7}{14\times 22}\]
By simplifying the above equation, we get,
\[{{r}^{2}}=9\]
By taking square root on both the sides, we get,
\[r=\sqrt{9}=3\text{ dm}\]
So, we get, h = 6r = 6 x 3 = 18 dm.
Now, we know that the volume of the cylinder \[=\pi {{r}^{2}}h\]
So, we also get the volume of the cylindrical tank \[=\pi {{r}^{2}}h\]
By substituting r = 3 dm and h = 18 dm, we get,
The volume of the cylindrical tank \[=\left( \dfrac{22}{7} \right)\left( {{3}^{2}} \right)\left( 18 \right)\]
\[=\dfrac{22}{7}\times 9\times 18\]
\[=509.14\text{ d}{{\text{m}}^{3}}\]
Hence, we get the volume of the cylindrical tank as \[509.14\text{ d}{{\text{m}}^{3}}\].
Note: In this question, students often make mistakes in the conversion of units and sometimes even forget to convert one unit into another. So, these mistakes must be avoided and students should properly consider the units as well with magnitudes of each quantity. Also, students must remember the basic conversions like Re. 1 = 100 paise.
Complete step-by-step answer:
We are given that the cost of painting the total outside surface of the closed cylindrical oil tank at 60 paise per sq. dm is Rs. 237.60. Also, the height of the tank is 6 times the radius of the base of the tank. We have to find the volume of the cylinder.
Let us consider the cylindrical tank diagrammatically.
Let us consider the total outside surface area of the cylindrical tank as S \[\text{d}{{\text{m}}^{2}}\].
We are given that the cost of painting per sq dm = 60 paise
So, we get the cost of painting the total outside surface area of the cylinder (S) = 60 S paise
We know that 1 paise \[=\text{Rs}\text{.}\dfrac{1}{100}\]
So, we get, 60 S paise \[=\text{Rs}\text{.}\dfrac{60S}{100}\]
Also, we are given that the total cost of the painting = Rs. 237.60
So, we get,
\[\text{Rs}\text{.}\dfrac{60S}{100}=\text{Rs}.237.6\]
By multiplying \[\dfrac{100}{60}\] on both the sides, we get,
\[S=\left( \dfrac{100}{60} \right)\times \left( 237.6 \right)\]
\[S=396\text{ }d{{m}^{2}}\]
Now, we know that the surface area of the cylinder \[=2\pi r\left( r+h \right)\] where r and h are the radius and height of the cylinder respectively.
We are given that the height is 6 times the radius of the base of the tank. So, we get,
h = 6r
So, we get the total outer surface area of the tank \[=2\pi r\left( r+6r \right)\]
\[\begin{align}
& =2\pi r.7r \\
& =14\pi {{r}^{2}} \\
\end{align}\]
We have already found that the total outer surface area of the tank \[=396\text{ d}{{\text{m}}^{2}}\]. So, we get,
\[14\pi {{r}^{2}}=396\text{ d}{{\text{m}}^{2}}\]
Or, \[{{r}^{2}}=\dfrac{396}{14\pi }\]
By substituting \[\pi =\dfrac{22}{7}\], we get,
\[{{r}^{2}}=\dfrac{396\times 7}{14\times 22}\]
By simplifying the above equation, we get,
\[{{r}^{2}}=9\]
By taking square root on both the sides, we get,
\[r=\sqrt{9}=3\text{ dm}\]
So, we get, h = 6r = 6 x 3 = 18 dm.
Now, we know that the volume of the cylinder \[=\pi {{r}^{2}}h\]
So, we also get the volume of the cylindrical tank \[=\pi {{r}^{2}}h\]
By substituting r = 3 dm and h = 18 dm, we get,
The volume of the cylindrical tank \[=\left( \dfrac{22}{7} \right)\left( {{3}^{2}} \right)\left( 18 \right)\]
\[=\dfrac{22}{7}\times 9\times 18\]
\[=509.14\text{ d}{{\text{m}}^{3}}\]
Hence, we get the volume of the cylindrical tank as \[509.14\text{ d}{{\text{m}}^{3}}\].
Note: In this question, students often make mistakes in the conversion of units and sometimes even forget to convert one unit into another. So, these mistakes must be avoided and students should properly consider the units as well with magnitudes of each quantity. Also, students must remember the basic conversions like Re. 1 = 100 paise.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The cell wall of prokaryotes are made up of a Cellulose class 9 biology CBSE
What organs are located on the left side of your body class 11 biology CBSE
Select the word that is correctly spelled a Twelveth class 10 english CBSE
a Tabulate the differences in the characteristics of class 12 chemistry CBSE