In the arrangement shown in figure the ends P and Q of an inextensible string move downwards with uniform speed u. Pulleys A and B are fixed. The mass M moves upwards with a speed

Explanation

`CE=l,DE=x,CD=y`
`x^(2)+y^(2)=l^(2)`
Differentiating w.r.t. time t, we get
`2x(dx)/(dt)+2y(dy)/(dt)=2l(dl)/(dt)`
`(dx)/(dy)=0`, since x is constant
`y(dy)/(dt)=2l(dl)/(dt)`
`(dy)/(dt)=v',(dl)/(dt)=-v`
[As t increases, both y and l decreases]
`yv'=lv`
`v'=(lv)/(y)=(v)/(y//l)=(v)/(costheta)`
OR
Since a string is inextensible, the velocity of any two points along the length of the string is same.

`v'costheta=v`
`v'=(v)/(costheta)`