A right triangle whose hypotenuse is √3 m long is revolved about one of its legs to generate a right circular cone. The volume of the greatest cone in cu.m. made this way is

Question

A right triangle whose hypotenuse is √3 m long is revolved about one of its legs to generate a right circular cone. The volume of the greatest cone in cu.m. made this way is (A) (2 π/3) (B) (π/3) (C) 2 π (D) none

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/1dacad7f86a9fa9dcd62949933e0c2c0-.png" /> V=(1/3) π x2 y and x2+y2=3 V =(1/3) π y (3- y 2) V prime( y )=(π/3)(3-3 y 2)=0 ⇒ y =1

Q. A right triangle whose hypotenuse is $3 m$ long is revolved about one of its legs to generate a right circular cone. The volume of the greatest cone in cu.m. made this way is

$\frac{2 π}{3}$

$\frac{π}{3}$

$2 π$

none

$V = \frac{1}{3} π x^{2} y$ and $x^{2} + y^{2} = 3$
$V = \frac{1}{3} π y (3 - y^{2})$
$V^{'} (y) = \frac{π}{3} (3 - 3 y^{2}) = 0 \Rightarrow y = 1$

Q. A right triangle whose hypotenuse is 3​m long is revolved about one of its legs to generate a right circular cone. The volume of the greatest cone in cu.m. made this way is

32π​

3π​

2π