Question

Three identical particles are joined together by a thread as shown in figure. 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 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

$T_3=m{\omega }^{2}3l$
$T_2-T_3=m{\omega }^{2}2l\Rightarrow T_2=m{\omega }^{2}5l$
$T_1-T_2=m{\omega }^{2}l\Rightarrow T_1=m{\omega }^{2}6l$
$T_3:T_2:T_1=3:5:6$