Question

2. An open cylindrical tank has a diameter of 3.5m,
and height of 7 meters. How much will it cost to
paint the curved surface and the base of the tank,
if the rate of painting is 5.00 per m?​

Answer 1

Step-by-step explanation:

cost of painting the tank is 577.5

Answer 2

$\huge\mathbb{\underline{SOLUTION:-}}$

$\bold{Given}\begin{cases}\sf{Diameter=3.5m}\\ \sf{radius=\frac{3.5}{2}}\\ \sf{Height=7m}\end{cases}$

• We have to find its curve surface area and area of its base

$\boxed{\sf{Curve\: surface\:area=2\pi r h}}$

$\boxed{\sf{Area\:of\:base=\pi r{}^{2}}}$

$\boxed{\sf{\red{Total\:area=2\pi r h+\pi r{}^{2}}}}$

• Now area :-

$\sf\implies C.S.A=2\times \frac{\cancel{22}}{\cancel7}\times \frac{3.5}{\cancel2}\times \cancel{7}\\ \\ \sf\implies C.S.A=2\times 11\times 3.5\\ \\ \sf\implies C.S.A=22\times 3.5\\ \\ \sf\implies C.S.A=77m{}^{2}$

• $\boxed{\sf{Curve\: surface\:area=77m{}^{2}}}$

• Now area of Base

$\sf\implies Area\:of\:base=\pi r{}^{2}\\ \\ \sf\implies Area\:of\:base=\frac{\cancel{22}}{\cancel7}\times \frac{\cancel{3.5}}{2}\times \frac{3.5}{\cancel2}\\ \\ \sf\implies Area\:of\:base=\frac{11\times 0.5\times 3.5}{2}\\ \\ \sf\implies Area\:of\:base=\cancel{\frac{19.25}{2}}\\ \\ \sf\implies Area\:of\:base=9.625m{}^{2}$

• $\boxed{\sf{Area\:of\:base=9.625m{}^{2}}}$

• Now total area

$\sf\implies Total\:area=77+9.625\\ \\ \sf\implies 86.625$

• Now the rate of painting is Rs 5 per metre

$\sf\implies Rate\:of\: painting = 86.625m{}^{2}\times 5\\ \\ \sf\implies Rs.433.125$

$\boxed{\sf{\purple{Cost\:of\: painting=Rs.433.125m{}^{2}}}}$

#BAL

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