Ashish Kumar
Last Activity: 7 Years ago
The answer is 4) 3:5:6
Assume mass of each ball is m.
For the ball C, Vo =ω3L, from here ω = Vo / 3L.
For ball A force due to rotation on it = mω2L=mVo / 9L
For ball B force due to rotation on it = mω22L= 2mVo / 9L
For ball C force due to rotation on it = mω23L= mVo / 3L
By drawing free body diagrams of each ball you will get-
Tension in the string joining the ball B and C = mVo / 3L =T1
Tension in the string joining the ball A and B =2mVo / 9L + mVo / 3L= 5mVo/9L = T2
Tension in the string joining the ball O and A = mVo / 9L + 2mVo / 9L + mVo / 3L= 2mVo/3L = T3
Now T1 : T2 : T3 = mVo / 3L : 5mVo/9L : 2mVo/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 Answer
