Question

Two beads connected by a light inextensible string are placed over fixed vertical ring as shown in figure. Mass of each bead is $100\text{gm}$ and all surfaces are frictionless. If the tension is $T$ in the string just after the beads are released from the shown position, then find $100T^2$ (in ${\text{N}}^2)$ and take $g=10m{s}^{-2}$

Answer 1

From FBD,
$N_1=\frac{T}{\sqrt{2}}+mg....\left(1\right)$ and
$\frac{T}{\sqrt{2}}=ma....\left(2\right)$
For $2$nd bead,
$\frac{T}{\sqrt{2}}=N_2$
$mg-\frac{T}{\sqrt{2}}=ma.....\left(3\right)$
By using eq. $\left(2\right)$,
$mg-\frac{T}{\sqrt{2}}=\frac{T}{\sqrt{2}}$
$\Rightarrow mg-\frac{T}{\sqrt{2}}=\frac{T}{\sqrt{2}}\Rightarrow mg=T\sqrt{2}$
$\Rightarrow T=\frac{mg}{\sqrt{2}}=\frac{\left(0.1\right)\left(10\right)}{\sqrt{2}}=\frac{1}{\sqrt{2}}\text{N}.$
$\therefore 100T^2=100\left(\frac{1}{2}\right)=50$
$\Rightarrow 100T^2=50$