Question

Three identical particles are joined together by a thread as shown in figure. All the three particles are moving on a smooth horizontal plane about point $O$. If the speed of the outermost particle is $v_0$, then the ratio of tensions in the three section of the string is: (Assume that the string remains straight)

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

Answer 1

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

According to the question we know that we have to find the ratio of tension in the three section of string.
So let's suppose angular speed of thread is equal to $\omega$ by each strings length.
In case of particle$C=T_3=m{\omega }^{2}3l.$
In case of particle $B=T_2-T_3=m{\omega }^{2}3l.so{T}_2=m{\omega }^{2}5l$
In case of particle $A=T_1-T_2=m{\omega }^{2}2lso{T}_1=m{\omega }^{2}6l$.
Hence the ratio will 3:5:6.