MCQ
A mass is performing vertical circular motion $($see figure$)$. If the average velocity of the particle is increased, then at which point is the maximum breaking possibility of the string:
- A$A$
- ✓$B$
- C$C$
- D$D$
✓
Answer
Correct option: B.
$B$
Tension at any point in vertical motion is given by:
$\text{T}=\frac{\text{min}^2}{1}+\text{mgcos}\theta$
where $I =$ angular displacement from lowest point,
$l =$ length of string
$m =$ mass of string
It is clear that tension at the lowest point $(B)$ is greatest than at other points $(A, C, D).$ If we increase average velocity, tension will increase at lowest point, therefore at point $B,$ string has maximum possibility of break.
$\text{T}=\frac{\text{min}^2}{1}+\text{mgcos}\theta$
where $I =$ angular displacement from lowest point,
$l =$ length of string
$m =$ mass of string
It is clear that tension at the lowest point $(B)$ is greatest than at other points $(A, C, D).$ If we increase average velocity, tension will increase at lowest point, therefore at point $B,$ string has maximum possibility of break.
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