Question

Three identical particles are joined together by a thread as shown in Fig. All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is $v_0$, then the ratio of tensions in the three sections of the string ($T_1:T_2:T_3=?$) is

• A. 6: 5: 3
• B. 3: 5: 6
• C. 3: 4: 5
• D. None of these

Answer 1

The correct option is A 6: 5: 3

Angular velocity will be same for all three particles -

$T_1-T_2=m{w}^2\left(l\right)\to \left(1\right)$ [Ref. image 1]

$T_2-T_3=m{w}^2\left(2l\right)\to \left(2\right)$ [Ref. image 2]

$T_3=m{w}^2\left(3l\right)\to \left(3\right)$ [Ref. image 3]

eq $\left(2\right)+\left(3\right):-T_2=m{w}^2\left(5l\right)$

eq $\left(1\right)+\left(2\right)+\left(3\right):T_1=m{w}^2\left(6l\right)$

$\Rightarrow T_1:T_2:T_3=6:5:3.$

Hence, the answer is $6:5:3.$