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Question

A particle suspended by a thread of length l is projected horizontally with a velocity 3gl at the lowest point. The height from the bottom at which the tension in the string becomes zero is :

A
4l3
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B
2l3
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C
5l3
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D
l3
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Solution

The correct option is B 2l3

When no external force is acting on the particle then,

ΔK.E.+ΔP.E.=0

Let at heighth the tension in thread becomes zero and the radial force acting at this height is given as and the height at which particle makes angleθis given as

h=l+lcosθ

Fnet=T+mgcosθ

LetvBbe speed of the particle whenT=0,

Fr=mv2Bl

Now on equating above two equations we get

v2B=glcosθ

By using law of conservation we get

12mglcosθ+32mglmgl(1+cosθ)=0

cosθ=13

By substitutingcosθ=1inh=l+lcosθwe get

h=2l3


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