Question

The diagram shows a cylindrical water tank of diameter 4.2 m and height 2.5 m

The curved surface and the top of the tank need to be painted.

If one liter of paint covers 8 m2 and paint is sold in 2 liter cans, how many cans of paint are needed to paint the tank?
(5 Points)​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

• Diameter of cylindrical water tank, d = 4.2 m

• Radius of cylindrical water tank, r = 2. 1 m

• Height of cylindrical water tank, h = 2.5 m

We know,

☆ Curved Surface Area of cylinder is given by

$\green{ \boxed{ \bf \: CSA_{(Cylindrical \: tank)} \: = \: 2 \: \pi \: rh }}$

and

☆ Top area of cylindrical tank,

$\green{ \boxed{ \bf \: Area_{(top \: of \: Cylindrical \: tank)} \: = \: \pi \: {r}^{2} }}$

Now,

$\red{\bf :\longmapsto\:Area_{(to \: be \: painted)}}$

$\rm \:= \: \:CSA_{(Cylindrical \: tank)} + Area_{(top \: of \: Cylindrical \: tank)}$

$\rm \:= \: \:2\pi \: rh + \pi \: {r}^{2}$

$\rm \:= \: \:\pi \: r(2h + r)$

$\rm \:= \: \:\dfrac{22}{7} \times 2.1 \times (5 + 2.1)$

$\rm \:= \: \:46.86 \: {m}^{2}$

Now,

It is given that,

• I liter of paint can covered an area of 8 sq. m.

So,

• 2 liter of paint can covered an area of 16 sq. m

It means,

• 1 can is required to covered an area of 16 sq. m.

So,

$\rm :\longmapsto\:\sf \: To \: paint \: an \: area \: of \: 46.86 \: {m}^{2}$

$\rm :\longmapsto\:Number \: of \: cans \: required = \dfrac{46.86}{16} = 2.93$

$\bf\implies \:\:Number \: of \: cans \: required = 3$

Additional Information :-

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²