Question

Four particles of equal masses of $1\text{kg}$ are connected by a light inextensible string of equal length to form a square. These are constrained to move along a frictionless ring of radius $2\text{m}$ in horizontal plane with a uniform speed speed of $2\text{m/s}$ as shown in figure. The tension in the string connecting them is

• A. $2\sqrt{2}N$
• B. $\frac{1}{\sqrt{2}}N$
• C. $\sqrt{2}N$
• D. $3\sqrt{2}N$

Answer 1

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

Free body diagram of single mass,


From FBD, a single mass will have two tensions at an angle of ${90}^{\text{o}}$ as it forms a square, so by the law of vector addition the net resultant tension$\left(T^{\prime }\right)$ will be

$T^{\prime }=\sqrt{T^2+T^2+2T^2\cos {90}^0}=T\sqrt{2}$ and the resultant tension $T^{\prime }$will provide the necessary centripetal force.
Hence $T^{\prime }=mr{\omega }^2$
$\Rightarrow T\sqrt{2}=mr{\omega }^2\Rightarrow T=\frac{mr{\omega }^2}{\sqrt{2}}$
.$T=\frac{m{v}^2}{r\sqrt{2}}=\frac{1\times (2{)}^2}{2\sqrt{2}}=\sqrt{2}\text{N}$