A truck is travelling in a straight line on level ground, and is accelerating uniformly with an acceleration of magnitude a. A rope (which of course is massless and inextensible) is tied to the back of the truck. The other end of the rope is tied to a bucket of mass M. The bucket tosses widly when the truck starts to accelerate, but due to friction it soon settles into a position at a fixed distance behind the truck, with the rope hanging straight at a fixed angle, as shown in the diagram. Although friction is needed to cause the rope to settle to an equillibrium position, we will assume that is can otherwise be neglected.

Now suppose that the truck comes to a downhill section of road, at an angle `alpha` relative to the horizontal, as shown in the diagram. Suppose that the truck continues to accelerate with an acceleration of magnitude a. Once again the rope settles to a fixed angle `theta` relative to the truck and the new tension in the rope is T'

Options

A. The value of `theta` is `tan^(-1).(g)/(a)`
B. The value of T is `msqrt(g^(2)-a^(2))`
C. The value of `theta` is `tan^(-1){(a-gsinalpha)/(gcosalpha)}`
D. The value of T' is `Msqrt(g^(2)+a^(2)-2agsinalpha)`

Answer

A

Explanation

a of block = `(2F)/(m)`, acceleration of pulley at extreme right (pulley of point P) is = `(2F)/(m),a_(P)=2a=(4F)/(m)`