Question

Two beads $\text{A}$ and $\text{B}$ of equal masses $2\text{kg}$ each are connected by a light rod, which acts as a chord for the ring. They are constrained to move on the frictionless ring in a vertical plane. The beads are released from rest as shown in figure. The tension in the chord just after release will be (Use $g=10{\text{m/s}}^2)$

• A. $12.5\text{N}$
  
• B. $25\text{N}$
• C. $25\sqrt{2}\text{N}$
  
• D. $\frac{25}{\sqrt{2}}\text{N}$

Answer 1

The correct option is C $25\sqrt{2}\text{N}$


Just after the release, bead $\text{B}$ moves vertically downwards and bead $\text{A}$ moves leftwards horizontally with the same acceleration, i.e $a$.


From $\text{FBD}$ of $\text{A}$, applying equation of circular motion along tangential direction (i.e, direction of acceleration)

$T\cos {45}^{\circ }=ma$

$\therefore \frac{T}{\sqrt{2}}=ma.........\left(1\right)$


Similarly, for bead $\text{B}:$


$mg-T\cos {45}^{\circ }=ma.........\left(2\right)$


Substituting $T$ from equation (1) in equation (2),

$\Rightarrow mg-ma=ma$

$\therefore a=\frac{g}{2}.........\left(3\right)$

Now, from equations $\left(3\right)$ and $\left(1\right)$, we get

$T=ma\sqrt{2}=\frac{mg}{\sqrt{2}}=\frac{5\times 10}{\sqrt{2}}=25\sqrt{2}\text{N}$

Hence, option (c) is the correct answer.