Question

Three identical particles are joined together by a thread as shown in figure.
All the three particles are moving in circular path 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 is

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

Answer 1

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

$\text{Step 1: Drawing free body diagrams}$ $\text{[Ref. Fig 1 \& 2]}$

Angular velocity $\left(\omega \right)$ will be same for all

$\text{Step 2: Writing force equation for A, B and C}$

Solving in ground frame:

Applying Newton's Second Law on A, B, and C along the centripetal direction

(Considering direction towards center as positive)

$\sum F=m{a}_c$

Where, $a_c=$ Centripetal acceleration $={\omega }^{2}r$, $r=$ radius

$\underset{–}{ForC}$

$T_1=m{a}_C=m{\omega }^{2}3l$ $....\left(1\right)$

$\underset{–}{ForB}$

$T_2-T_1=m{a}_B$

$\Rightarrow$ $T_2=m{\omega }^{2}3l+m{\omega }^{2}2l=m{\omega }^{2}5l$ $....\left(2\right)$

$\underset{–}{ForA}$

$T_3-T_2=m{a}_A$

$\Rightarrow T_3=m{\omega }^{2}5l+m{\omega }^{2}l=m{\omega }^{2}6l$ $....\left(3\right)$

$\text{Step 3: Ratio calculation}$

From eqn. (1), (2) and (3)

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

Hence option D is correct