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 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 $3:5:6$

Assume mass of each ball is m.

For the ball C, V_o =ω3L, from here ω = V_o / 3L.

For ball A force due to rotation on it = mω²L=mV_o / 9L

For ball B force due to rotation on it = mω²2L= 2mV_o / 9L

For ball C force due to rotation on it = mω²3L= mV_o / 3L

By drawing free body diagrams of each ball you will get-

Tension in the string joining the ball B and C = mV_o / 3L =T₁

Tension in the string joining the ball A and B =2mV_o / 9L + mV_o / 3L= 5mV_o/9L = T₂

Tension in the string joining the ball O and A = mV_o / 9L + 2mV_o / 9L + mV_o / 3L= 2mV_o/3L = T₃

Now T₁ : T₂ : T₃ = mV_o / 3L : 5mV_o/9L : 2mV_o/3L = 1/3 : 5/9 : 2/3

Now in the first and third fraction multiply 3 to the numerator and denominator to make the denominator of each fraction = 9

=>T₁ : T₂ : T₃ =1x3/3x3 : 5/9 : 2x3/3x3 = 3/9 : 5/9 : 6/9 , Now since the denominator of each fraction is same so it will be cancelled from all the three fractions so you will get T₁ : T₂ : T₃ = 3 : 5 : 6

Hence,

option $\left(D\right)$ is correct answer.