Two objects A and B, each of mass m, are connected by a light inextensible string. They are restricted to move on a frictionless ring of radius R in a vertical plane (as shown in the figure). The objects are released from rest at the position shown. Then the tension in the cord just after release is

Question

Two objects A and B, each of mass m, are connected by a light inextensible string. They are restricted to move on a frictionless ring of radius R in a vertical plane (as shown in the figure). The objects are released from rest at the position shown. Then the tension in the cord just after release is (A) Zero (B) m g (C) √2 m g (D) m g / √2

Tardigrade Admin · Accepted Answer

As shown in Figs. (a) and (b) from FBD of A, T cos 45°=m a ...(1) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9e709fe8bd9c737d71021e25a0491d3a-.png" /> From FBD of B: M g-T cos 45°=m a ...(2) From (i) and (ii), we get T=m g / √2

Zero

$m g$

$2 m g$

$m g / 2$

As shown in Figs. $(a)$ and $(b)$ from $FB D$ of $A$ ,
$T cos 4 5^{\circ} = ma$ ...(1)

From $FB D$ of $B$ :
$M g - T cos 4 5^{\circ} = ma$ ...(2)
From (i) and (ii),
we get $T = m g / 2$

Zero

mg

2​mg

mg/2​

As shown in Figs. (a) and (b) from FBD of A, Tcos45∘=ma ...(1) From FBD of B : Mg−Tcos45∘=ma ...(2) From (i) and (ii), we get T=mg/2​

$m g$

$2 m g$

$m g / 2$

As shown in Figs. $(a)$ and $(b)$ from $FB D$ of $A$ ,
$T cos 4 5^{\circ} = ma$ ...(1)

From $FB D$ of $B$ :
$M g - T cos 4 5^{\circ} = ma$ ...(2)
From (i) and (ii),
we get $T = m g / 2$