Question

A vessel is in a conical shape. If its volume is 33.264 liters and height is 72 cm, the cost of repairing its CSA at Rs. 12 sq.m is
(A) 5.94
(B) 6.94
(C) 7.95
(D) None of the above

Answer 1

Hint: Use the relation, 1 liter = 1000 \[c{{m}^{3}}\] and convert the volume of the conical vessel into \[c{{m}^{3}}\] . Assume that the radius of the vessel is \[r\] cm. Now, use the height given in the question and calculate the volume of the conical vessel by using the formula, \[\dfrac{1}{3}\pi {{r}^{2}}h\] . Compare it with the given volume and get the value of \[r\] . Now, use the formula \[l=\sqrt{{{r}^{2}}+{{h}^{2}}}\] and curved surface area of a cone = \[\pi rl\] . Convert the curved surface area into \[{{m}^{2}}\] using \[10000\,c{{m}^{2}}=1\,{{m}^{2}}\Rightarrow 1\,c{{m}^{2}}=\dfrac{1}{10000}{{m}^{2}}\] . We are given that the cost of repairing the curved surface area is Rs. 12 per meter square and calculate the cost of repairing the curved surface area of the vessel.

Complete answer:
According to the question, we are given that a vessel is in the conical shape.
The volume of the conical vessel = 33.264 liters………………………………………(1)
The height of the conical vessel = 72 cm ……………………………………(2)
We know the relation between liter and \[c{{m}^{3}}\] , 1 liter = 1000 \[c{{m}^{3}}\] …………………………………(3)
Now, from equation (1) and equation (3), we get
The volume of the conical vessel = \[33.264\times 1000\,c{{m}^{3}}=33264\,c{{m}^{3}}\] …………………………………(4)
Here, let us assume that the radius of the circular base of the conical base is \[r\] cm ………………………………………(5)

We also know the formula for the volume of the conical vessel = \[\dfrac{1}{3}\pi {{r}^{2}}h\] , where \[r\] and \[h\] is the radius and height of the vessel respectively ………………………………(6)
Now, from equation (2), equation (5), and equation (6), we get
The volume of the conical vessel = \[\dfrac{1}{3}\times \dfrac{22}{7}\times {{r}^{2}}\times 72=\dfrac{22\times 24}{7}{{r}^{2}}\] …………………………………..(7)
On comparing equation (4) and equation (7), we get
\[\begin{align}
  & \Rightarrow \dfrac{22\times 24}{7}{{r}^{2}}=33264 \\
 & \Rightarrow {{r}^{2}}=\dfrac{33264\times 7}{22\times 24} \\
 & \Rightarrow {{r}^{2}}=63\times 7 \\
 & \Rightarrow r=\sqrt{441} \\
\end{align}\]
\[\Rightarrow r=21\] ……………………………..(8)
We know the formula, \[l=\sqrt{{{r}^{2}}+{{h}^{2}}}\] where l is the slant height, r is the radius and h is the height ……………………………………………(9)
Now, from equation (2), equation (8), and equation (9), we get
The slant height of the conical vessel = \[\sqrt{{{\left( 21 \right)}^{2}}+{{\left( 72 \right)}^{2}}}\] cm = 75 cm ……………………………(10)
We also know the formula for the curved surface area of a cone = \[\pi rl\] , where r is the radius and l is the slant height ………………………………….(11)
Now, from equation (8), equation (10), and equation (11), we get
The curved surface area of the conical vessel = \[\dfrac{22}{7}\times 21\times 75\,c{{m}^{2}}=22\times 3\times 75\,c{{m}^{2}}=4950\,c{{m}^{2}}\] …………………………………………………(12)
We also know the relation, \[10000\,c{{m}^{2}}=1\,{{m}^{2}}\Rightarrow 1\,c{{m}^{2}}=\dfrac{1}{10000}{{m}^{2}}\] ……………………………….(13)
Now, from equation (12) and equation (13), we get
The curved surface area of the conical vessel = \[\dfrac{4950}{10000}\,{{m}^{2}}=0.495\,{{m}^{2}}\] ……………………………….(14)
It is also given in the question that the cost of repairing the curved surface area of the vessel is Rs 12 per meter square ……………………………………..(15)
Now, from equation (14) and equation (15), we get
The cost of repairing the curved surface area of the vessel = Rs. \[12\times 0.495\] = Rs. 5.94.

Hence, the correct option is (A).

Note:
For this type of question, the conversion is base. A silly mistake in conversion can make the whole solution incorrect. For instance, the conversion that we used here is 1 liter = 1000 \[c{{m}^{3}}\] and \[10000\,c{{m}^{2}}=1\,{{m}^{2}}\Rightarrow 1\,c{{m}^{2}}=\dfrac{1}{10000}{{m}^{2}}\] . Therefore, be careful while applying the conversion.