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A mass $m$ is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:
Options:
Solution:
2625 Upvotes
Verified Answer
The correct answer is:
the mass is at the lowest point.
We know that,
$T-m g=\frac{m u^2}{l}$
$\Longrightarrow T=m g+\frac{m u^2}{l}$
The tension is maximum at the lowest position of mass, so the chance of breaking is maximum.
Hence option 4 is the correct answer
$T-m g=\frac{m u^2}{l}$
$\Longrightarrow T=m g+\frac{m u^2}{l}$
The tension is maximum at the lowest position of mass, so the chance of breaking is maximum.
Hence option 4 is the correct answer
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