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 v₀, 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

Answer: (D)

3 : 5 : 6

Answer 2

Let the angular speed of the thread is ω

For particle ‘C’$\Rightarrow$ $T_{3} = m\omega^{2}3l$

For particle ‘B’ $T_{2} - T_{3} = m\omega^{2}2l$ $\Rightarrow$ $T_{2} = m\omega^{2}5l$

For particle ‘C’ $T_{1} - T_{2} = m\omega^{2}l$ $\Rightarrow$ $T_{1} = m\omega^{2}6l$

∴ $T_{3}:T_{2}:T_{1} = 3:5:6$