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The ring shown in fig. is given a consta...

The ring shown in fig. is given a constant horizontal acceleration `(a_(0)=gsqrt(3))`. The maximum deflection of the string from the vertical is `theta_(0)`. Then

A

`theta_(0) = 30^(@)`

B

`theta_(0) = 60^(@)`

C

At maximum deflection, tension in string is equal to mg.

D

At maximum deflection, tension in string is equal to `(2mg)/(sqrt(3))`

Text Solution

Verified by Experts

The correct Answer is:
A, D
`T cos theta_(0)=mg` ..(i)
`T sin theta_(0)=ma_(0)` …(ii)
Dividing Eq, (ii) by (i) we get
`tan theta_(0) =a/g implies theta_(0)=30^(@)` ltbtgt `T=(mg)/(cos 30^(@))=(2mg)/(sqrt(3))`
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