A small block of mass m is placed at the highest point of a smooth...

At the point where the block loses contact with the sphere, the normal reaction $N$ becomes zero.
The equation of motion along the radial direction is given by:
$mg \cos \alpha - N = \dfrac{m v_2^2}{R}$

Setting $N = 0$, we get:
$mg \cos \alpha = \dfrac{m v_2^2}{R} \Rightarrow v_2^2 = gR \cos \alpha$

Applying the principle of conservation of mechanical energy between the highest point and the point of losing contact:
$\dfrac{1}{2} m v_1^2 + mgR = \dfrac{1}{2} m v_2^2 + mgR \cos \alpha$

Multiplying by $\dfrac{2}{m}$ throughout:
$v_1^2 + 2gR = v_2^2 + 2gR \cos \alpha$
$v_1^2 = v_2^2 - 2gR(1 - \cos \alpha)$

From the radial equation, we have $gR = \dfrac{v_2^2}{\cos \alpha}$. Substituting this into the energy equation:
$v_1^2 = v_2^2 - 2 \left( \dfrac{v_2^2}{\cos \alpha} \right) (1 - \cos \alpha)$

$v_1^2 = v_2^2 \left[ 1 - \dfrac{2(1 - \cos \alpha)}{\cos \alpha} \right]$

$v_1^2 = v_2^2 \left[ \dfrac{\cos \alpha - 2 + 2 \cos \alpha}{\cos \alpha} \right]$

$v_1^2 = v_2^2 \left( \dfrac{3 \cos \alpha - 2}{\cos \alpha} \right)$

Rearranging to find the ratio $\dfrac{v_2}{v_1}$:
$\dfrac{v_2^2}{v_1^2} = \dfrac{\cos \alpha}{3 \cos \alpha - 2}$

$\dfrac{v_2}{v_1} = \left( \dfrac{\cos \alpha}{3 \cos \alpha - 2} \right)^{\dfrac{1}{2}}$

Answer: \left( \dfrac{\cos \alpha}{3 \cos \alpha - 2} \right)^{\frac{1}{2}}