A weightless thread can bear tension up to 3.7 kg wt. A stone of mass 500 gram is tied to it and revolved in circular path of radius 4 m in vertical plane. Maximum angular velocity of the stone will be (acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

Maximum angular velocity occurs at the lowest point where tension is greatest due to the combined effect of weight and centripetal force.

The tension formula at the lowest point is $T = mg + m\omega^2 r$.

Given: $T_{\text{max}} = 3.7 \, \text{kg wt} = 37 \, \text{N}$, $m = 0.5 \, \text{kg}$, $r = 4 \, \text{m}$, $g = 10 \, \text{m/s}^2$.

Substituting known values:
$37 = 0.5 \times 10 + 0.5 \times \omega^2 \times 4$
$37 = 5 + 2\omega^2$
$32 = 2\omega^2$
$\omega^2 = 16$

The angular velocity is $\omega = 4 \, \text{rad/s}$.

Answer: B