Question

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
(a) 2u cos θ
(b) u/cos θ
(c) 2u/cos θ
(d) u cos θ
Figure

Answer 1

Answer:

B

Solution :

In the right angle `Lambda PQR`

`l^(2) =c^(2)+y^(2)`

Differentiating this equation with respect to time, we get

`2l(dl)/(dt) =0+2y(dy)/(dt)` or `((-dy)/(dt))=(l)/(y) ((-dl)/(dt))`

Here ` -(dy)/(dt)=u_(M),(l)/(y)=(1)/(cos theta)` and `-dl//dt =U`

Hence `u_(M) =(U)/(cos theta)`

`:.` the corrent option is (b)

Answer 2

The mass M moves upwards with a speed of $\frac{u}{\cos \theta}$

Option (b) $\frac{u}{\cos \theta}$

Explanation:

Speed of Q is "u" descending so in unit time (t)  it goes descending. Since the strings are unadaptable the other end of the string near mass M will shorten by a length (L)  of u.

                    BY = BX’  

             ∠ XX’Y = 90

Here,   ( u = XX’ = XY Cosθ )    

                   $X Y=\frac{u}{\cos \theta}$

When X should move to X' , Q moves to Q' so that QQ'=XX'=u. But the mass moves higher, so X will actually move higher  to Y so as BY = BX'. For the small angle X' by  the line X'Y perpendicular XX'. So ΔXX'Y is a right angled triangle.

                    $\frac{X X^{\prime}}{X Y}=\cos \theta$

                    $\frac{X X^{\prime}}{\cos \theta}=X Y=\frac{u}{\cos \theta}$

It is the distance traveled by mass M in unit time ie, speed of M.  

Therefore, the speed is $\frac{u}{\cos \theta}$ .