Question

Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane in circular path with $O$ as centre. If the velocity of the outermost particle $v_0$, 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 $\omega$ is the angular speed of revolution

From FBD of $C$
$T_3=m{\omega }^{2}3l$
From FBD of $B$
$T_2-T_3=m{\omega }^{2}2l\Rightarrow T_2=m{\omega }^{2}5l$
From FBD of $A$
$T_1-T_2=m{\omega }^{2}l\Rightarrow T_1=m{\omega }^{2}6l$

Ratio of tension in the strings is
$T_3:T_2:T_1=m{\omega }^{2}3l:m{\omega }^{2}5l:m{\omega }^{2}6l=3:5:6$