A) 3 : 5 : 7
B) 3 : 4 : 5
C) 7 : 11 : 6
D) 3 : 5 : 6
Correct Answer: D
Solution :
Let \[\omega \] is the angular speed of revolution \[{{T}_{3}}=m{{\omega }^{2}}3l\] \[{{T}_{2}}-{{T}_{3}}=m{{\omega }^{2}}2l\] Þ \[{{T}_{2}}=m{{\omega }^{2}}5l\] \[{{T}_{1}}-{{T}_{2}}=m{{\omega }^{2}}l\Rightarrow {{T}_{1}}=m{{\omega }^{2}}6l\] \[{{T}_{3}}:{{T}_{2}}:{{T}_{1}}=3:5:6\]You need to login to perform this action.
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