Two stones are projected with the same magnitude of velocity, but making different angles with horizontal. The angle of projection of one is π/3 and its maximum height is Y, the maximum height attained by the other stone with as π/6 angle of projection is

The correct option is Y/3.

To find the maximum height attained by the second stone, let us recall the formula for the maximum height ($H$) reached by a projectile projected with an initial velocity $u$ at an angle $\theta$ with the horizontal:
$H=\frac{u^2{\sin }^2\theta }{2g}$
where $g$ is the acceleration due to gravity.

For the first stone, the angle of projection is ${\theta }_1=\frac{\pi }{3}$ and its maximum height is $H_1=Y$. Since the stones are projected with the same magnitude of velocity, let the initial velocity of both stones be $u$. Therefore, we can write:
$Y=\frac{u^2{\sin }^2\left(\frac{\pi }{3}\right)}{2g}$
Since $\sin \left(\frac{\pi }{3}\right)=\frac{\sqrt{3}}{2}$, we have:
${\sin }^2\left(\frac{\pi }{3}\right)={\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{3}{4}$
Substituting this value back into the equation for $Y$:
$Y=\frac{u^2}{2g}·\frac{3}{4}$ --- (Equation 1)

For the second stone, the angle of projection is ${\theta }_2=\frac{\pi }{6}$. Let its maximum height be $H_2$:
$H_2=\frac{u^2{\sin }^2\left(\frac{\pi }{6}\right)}{2g}$
Since $\sin \left(\frac{\pi }{6}\right)=\frac{1}{2}$, we have:
${\sin }^2\left(\frac{\pi }{6}\right)={\left(\frac{1}{2}\right)}^2=\frac{1}{4}$
Substituting this value back into the equation for $H_2$:
$H_2=\frac{u^2}{2g}·\frac{1}{4}$ --- (Equation 2)

Now, we can find the ratio of $H_2$ to $Y$ by dividing Equation 2 by Equation 1:
$\frac{H_2}{Y}=\frac{\frac{u^2}{2g}·\frac{1}{4}}{\frac{u^2}{2g}·\frac{3}{4}}$
The common terms $\frac{u^2}{2g}$ and the denominators $4$ cancel out, leaving:
$\frac{H_2}{Y}=\frac{1}{3}$
Solving for $H_2$:
$H_2=\frac{Y}{3}$

Thus, the maximum height attained by the other stone is Y/3.