the corect relation b/w V1 & V2 a)V2=V1 b)V2=V1sin? c)2V2sin?= V1 d)V2(1+sin?)=V1

							
hi,
at any time ,
let the length of string between the wedge & pully(which is stationary ) = x2
 length of string between the block  & pully (to which it is hanged ) = x1
 length of string between the both pullys =  x2/ cos(90-θ)  = x2 / sinθ
since the length of strig would be const. so the sum of upper mentioned lenght would be const.
x1 + x2 + x2/ sinθ  = const.
d/dt x1  + d/dt x2  + d/dt ( x2/ sinθ  ) = 0
v1  - v2  - v2 /sin θ   -  x2 cosθ /sin^θ  (d/dt θ) =0 .............(1)
let the height of the wedge be l,then
  tan (90 -θ) =  l / x2
 
x2 cot θ = l
 
 cotθ d/dt x2 + x2 d/dt cot θ = 0
 
-cotθ v2   - x2 cosec^2 θ (d/dt θ) =0
 
cosθ sinθ v2 + x2 (d/dt θ) = 0
 
(d/dt θ)  =  - cosθ sinθ v2 /x2 ..............put in eq . (1)
 
 
 
 v1  - v2  - v2 /sin θ   -  x2 cosθ /sin^θ  (d/dt θ) =0
 v1  - v2  - v2 /sin θ   +  x2 cosθ /sin^θ  cosθ sinθ v2 /x2  =0

 v1  - v2  - v2 /sin θ   + v2 cos^θ / sinθ   =0 
v1- v2 - v2 /sinθ( 1 - cos^2θ ) = 0 
v1 -v2  - v2 sinθ =0
v1 = v2 (1 + sinθ )     (d )