Question

In the given figure, if ΔOAB is rotated about the x-axis, then the volume of the solid so formed is

• A. $\frac{5\pi }{3}\text{cu.units}$
  
• B. $\frac{10\pi }{3}\text{cu.units}$
  
• C. $5\pi \text{cu.units}$
  
• D. $\frac{20\pi }{3}\text{cu.units}$
  

Answer 1

The correct option is D $\frac{20\pi }{3}\text{cu.units}$


The point of intersection of lines y = 2x and x + 2y = 5 is:

2x – y = 0 …..(i)

x + 2y = 5 …..(ii)

Multiplying (ii) with 2, we get

2x + 4y = 10 …..(iii)

Subtracting (iii) from (i), we get,

(2x – y) – (2x + 4y) = 0 – 10

⇒ –5y = –10

⇒ y = 2

Substituting the value of y in (i),

2x – 2 = 0

⇒ x = 1

Thus, the coordinate of point A is (1, 2).

Point B is the intersection of the line x + 2y = 5 and x-axis.

Put y = 0 in (ii), we get

⇒ x + 2(0) = 5

⇒ x = 5

Thus, coordinate of the point B is (5, 0).

If we rotate ∆OAB about the x-axis, we get 2 cones OAAʹ and ABAʹ.

Now, radius of cone OAAʹ, $r_1$ = AC = 2 units

and height of cone OAAʹ, $h_1$ = OC = 1 unit

∴ Volume of cone OAA ' $=\frac{1}{3}\pi r_1^2{h}_1$ cubic units

$=\frac{1}{3}\pi (2{)}^2\times (1)$

$=\frac{4}{3}\pi \text{cubic units}$ ...(iv)

Also, radius of cone ABAʹ, $r_2$ = AC = 2 units

height of cone ABAʹ, $h_2$ = BC = 4 units

∴ Volume of cone ABAʹ $=\frac{1}{3}\pi r_2^2{h}_2$ cubic units

$=\frac{1}{3}\pi (2{)}^2\times 4$

$=\frac{16}{3}\pi$ cubic units ...(v)

$\therefore$ Total volume of the solid $=\frac{4}{3}\pi +\frac{16}{3}\pi$ [From (iv) and (v)]

$=\frac{20}{3}\pi$ cubic units

Hence, the correct answer is option (d).