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4-1.Newton's Laws of Motion
normal
Three identical particles are joined together by a thread as shown in the figure. All the three particles are moving on a smooth horizontal plane about point $O$. If the speed of the outer most particle is $v_0$ then the ratio of tension in the three sections of the thread is
A
$3 : 5 : 7$
B
$3 : 4 : 5$
C
$7 : 11 : 6$
D
$3 : 5 : 6$
Solution
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 $"W"$ by each strings length.
in case of particle $C=T_{3}=m W^{2} 3 l$
in case of particle $B=T_{2}-T_{3}=m W^{2} 3 l .$ so $T_{2}=m W^{2} 5 l$
in case of particle $A=T_{1}-T_{2}=m W^{2} 2 l \quad$ so $T_{1}=m W^{2} 6 l$
Hence the ratio will $3: 5: 6$
Standard 11
Physics
