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Question: A vessel is in a conical shape. If its volume is 33.264 liters and height is 72 cm, the cost of repa...

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

Explanation

Solution

Use the relation, 1 liter = 1000 cm3c{{m}^{3}} and convert the volume of the conical vessel into cm3c{{m}^{3}} . Assume that the radius of the vessel is rr cm. Now, use the height given in the question and calculate the volume of the conical vessel by using the formula, 13πr2h\dfrac{1}{3}\pi {{r}^{2}}h . Compare it with the given volume and get the value of rr . Now, use the formula l=r2+h2l=\sqrt{{{r}^{2}}+{{h}^{2}}} and curved surface area of a cone = πrl\pi rl . Convert the curved surface area into m2{{m}^{2}} using 10000cm2=1m21cm2=110000m210000\,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 cm3c{{m}^{3}} , 1 liter = 1000 cm3c{{m}^{3}} …………………………………(3)
Now, from equation (1) and equation (3), we get
The volume of the conical vessel = 33.264×1000cm3=33264cm333.264\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 rr cm ………………………………………(5)

We also know the formula for the volume of the conical vessel = 13πr2h\dfrac{1}{3}\pi {{r}^{2}}h , where rr and hh 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 = 13×227×r2×72=22×247r2\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

& \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{aligned}$$ $$\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.