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The bob of a 0.2 m pendulum describes an...

The bob of a `0.2` m pendulum describes an arc of circle in a vertical plane. If the tension in the cord is `sqrt3` times the weight of the bob when the cord makes an angle `30^(@)` with the vertical, the acceleration of the bob in that position is

A

g

B

`(g)/(2)`

C

`(sqrt3g)/(2)`

D

`(g)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
A
`a=sqrt(a_(n)^(2)+a_(t)^(2))=sqrt(((T-"mg cos 30"^(@))/(m))^(2)+("g sin 30")^(2))`
`=gsqrt((sqrt3-(sqrt3)/(2))^(2)+((1)/(4)))=g`
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Knowledge Check

  • A bob of mass 1 kg is suspended from an inextensible string of length 1 m. When the string makes an angle 60^(@) with vertical, speed of the bob is 4 m/s Net acceleration of the bob at this instant is

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