Question

Three identical particles are joined by a thread as shown in the 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 the tension 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, Mass=$m$

For ball C, $V_o=\omega \times 3l\Rightarrow \omega =V_o/3L$

For ball A, due to rotation= $m{\omega }^{2}L=m{V}_o/9L$

For ball B=$m{\omega }^{2}2L=2m{V}_o/9L$

For ball C=$m{\omega }^{2}3L=m{V}_o/3L$

For tension, ball B & C=$m{V}_o/3L=T_1$

ball A & B= $2m{V}_o/9L+m{V}_o/3L=5m{V}_o/9L=T_2$

ball C & A=$m{V}_o/9L+2m{V}_o/9L+m{V}_o/3L=2m{V}_o/3L=T_3$

Now, $T_1:T_2:T_3=1/3:5/9:2/9$

$=3:5:6$