Question

Three beads A, B and C each of mass $m=2kg$ are connected by means of strings of equal lengths. They are constrained to move on a frictionless ring with an angular speed of $\omega =10rad/s$ on a horizontal circular path of radius $r=10cm$ as shown in the figure. The tension in the strings connecting them is

• A. $\frac{20}{\sqrt{3}}N$
• B. $\frac{40}{\sqrt{3}}N$
• C. $10\sqrt{3}N$
• D. $5\sqrt{3}N$

Answer 1

The correct option is A $\frac{20}{\sqrt{3}}N$

FBD of a single mass is given below


All the strings will have same tension as all the strings are identical and connected to the same mass at same distance.

From the FBD, by the law of vector addition, the net resultant tension$\left(T^{\prime }\right)$ will be

$T^{\prime }=\sqrt{T^2+T^2+2T^2\cos {60}^{\circ }}$
$=\sqrt{3}T$

The resultant tension will provide the necessary centripetal force $\left(m{\omega }^{2}r\right)$.
$T^{\prime }=mr{\omega }^2$
$\Rightarrow \sqrt{3}T=mr{\omega }^2$
$\therefore T=\frac{mr{\omega }^2}{\sqrt{3}}=\frac{2\times 10\times {10}^{-2}\times {10}^2}{\sqrt{3}}=\frac{20}{\sqrt{3}}N$