A flexible chain of weight W hangs between two fixed points A and B at the same level. The inclination of the chain with the horizontal at the two points of support is `theta = 60^(@)`. What is the tension of the chain at the point P where inclination of chain with horizonrtal is `phi = 30^(@)`

Options

A. `(2W)/(3)`
B. `(W)/(2)`
C. `W`
D. `(W)/(3)`

Answer

D

Explanation

`2T sin theta = W`
`T = (W)/(2) cosec theta`

`T cos 60^(@)=T^(1) cos 30^(@)`
`T^(1) = (W cosec 60^(@))/(2 cos 30^(@)) cos 60^(@) = (W)/(3)`.