In the figure, the blocks are of equal mass. The pulley is fixed. In the position shown. A moves down with a speed u and `v_(B)=` the speed of B.

(i) B will never lose contact with the ground.
(ii) The downward acceleration of A is equal in magnitude to the horizontal acceleration of B
(iii) `v_(B)=ucostheta` (iv) `v_(B)=u//costheta`

Options

A. (i), (ii)
B. (ii), (iii)
C. (i),(iv)
D. (ii),(iv)

Answer

C

Explanation

`l^(2)=x^(2)+y^(2)`
`2l(dl)/(dt)+2x(dx)/(dt)=2y(dy)/(dt)`
y is cosntant
`l(dl)/(dt)=x(dx)/(dt)rArrlu=xv_(0)`
`v_(0)=(u)/(x//l)=(u)/(costheta)`

`Tsintheta+N=mg` since `Tltmg`
N is never zero.