Question

Three identical particles $\text{A},\text{B}$ and $\text{C}$ are joined together by a thread as shown in figure. All the particles are moving in a horizontal plane around point $O$. If the velocity of the outermost particle is $v_0$ then the ratio of tensions in the three sections of the string is

• A. $3:5:7$
• B. $3:4:5$
• C. $7:11:6$
• D. $3:5:6$

Answer 1

The correct option is D $3:5:6$

Let $\omega$ be the angular speed during revolution.


Applying equation of circular dynamics in horizontal plane for different particle,

For particle $\text{C}$:

$T_3=m{\omega }^2\left(3l\right)=3m{\omega }^{2}l$

For particle $\text{B}$:

$T_2-T_3=m{\omega }^2\left(2l\right)$

$\Rightarrow T_2=2m{\omega }^{2}l+3m{\omega }^{2}l=5m{\omega }^{2}l$

For particle $\text{A}$:

$T_1-T_2=m{\omega }^{2}l$

$\Rightarrow T_1=5m{\omega }^{2}l+m{\omega }^{2}l=6m{\omega }^{2}l$

Thus taking the ratio of tensions,

$T_3:T_2:T_1=3:5:6$

Therefore, option $\left(d\right)$ is right choice.