As shown in the figure,two strings of length | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let $	heta$ be the angle that each string makes with the vertical axis. Since the triangle $ABC$ is equilateral (all sides are $l$),$	heta = 60^

Vedclass · Accepted Answer

$T_1 - T_2 = 2mg$

Vedclass · Answer

(D) Let $	heta$ be the angle that each string makes with the vertical axis. Since the triangle $ABC$ is equilateral (all sides are $l$),$	heta = 60^\circ$.Resolving forces on mass $m$:Vertical direction: $(T_1 - T_2) \cos 	heta = mg \implies T_1 - T_2 = \frac{mg}{\cos 60^\circ} = 2mg$.Horizontal direction: $(T_1 + T_2) \sin 	heta = m \omega^2 r$,where $r = l \sin 60^\circ = l \frac{\sqrt{3}}{2}$.$(T_1 + T_2) \frac{\sqrt{3}}{2} = m \omega^2 l \frac{\sqrt{3}}{2} \implies T_1 + T_2 = m \omega^2 l$.Solving these equations: $T_1 = \frac{m \omega^2 l}{2} + mg$ and $T_2 = \frac{m \omega^2 l}{2} - mg$.For string $BC$ to remain taut,$T_2 \geq 0 \implies \frac{m \omega^2 l}{2} \geq mg \implies \omega \geq \sqrt{2g/l}$.Thus,option $D$ is correct.

Similar Questions

$A$ body of mass $M = 100 \; g$ is rotating with an angular frequency of $2/\pi \; \text{rev/sec}$ as shown in the figure. What is the angle made by the string with the vertical? $(g = 10 \; m/s^2)$

When a vehicle of mass $m$ is moving with a velocity $v$ over a concave over-bridge of radius of curvature $r$,the thrust on the road at the lowest point on the bridge will be

If the radius of curvature of the path of two particles of same masses are in the ratio $1 : 2$,then in order to have constant centripetal force,their velocity,should be in the ratio of

Vedclass Test Series

Exam Paper Generator

Online Exam Module