Part B: Two Ends of an Inextensible String ......

Two ends of an inextensible string passing over two fixed pulleys with midpoint connected to a block are pulled down with constant speed ve making the block to move up. Then which of the following options is correct?

 
A) $v_m = v_e $
B) $v_m > v_e $
C) $v_m < v_e $
D) Data is insufficient to conclude

Solution

Let's say in certain small time (say, $\Delta t$) the block moves up by $y_2 - y_1 = \Delta y $ distance.

Since the length of the string is constant (say, l), the end point goes down by distance $\left( {l - \sqrt {{a^2} + y_1^2} } \right) - \left( {l - \sqrt {{a^2} + y_2^2} } \right)$ in the same time. 

${v_m} = \frac{{\Delta y}}{{\Delta t}}$ and ${v_e} = \frac{{\left( {l - \sqrt {{a^2} + y_1^2} } \right) - \left( {l - \sqrt {{a^2} + y_2^2} } \right)}}{{\Delta t}}$

${v_e} = \frac{{\sqrt {{a^2} + y_2^2}  - \sqrt {{a^2} + y_1^2} }}{{\Delta t}}$

$ \Rightarrow {v_e} = \frac{{\sqrt {{a^2} + {{({y_1} + \Delta y)}^2}}  - \sqrt {{a^2} + y_1^2} }}{{\Delta t}}$

$ \Rightarrow {v_e} \approx \frac{{\sqrt {{a^2} + y_1^2 + 2{y_1}\Delta y}  - \sqrt {{a^2} + y_1^2} }}{{\Delta t}}$ (assuming $\Delta t$ to be very small which means $\Delta y$ is also very small)

$ \Rightarrow {v_e} \approx \frac{{\sqrt {{a^2} + y_1^2} {{\left( {1 + \frac{{2{y_1}\Delta y}}{{{a^2} + y_1^2}}} \right)}^{1/2}} - \sqrt {{a^2} + y_1^2} }}{{\Delta t}}$

$ \Rightarrow {v_e} \approx \frac{{\sqrt {{a^2} + y_1^2} \left( {1 + \frac{1}{2}.\frac{{2{y_1}\Delta y}}{{{a^2} + y_1^2}}} \right) - \sqrt {{a^2} + y_1^2} }}{{\Delta t}}$

$ \Rightarrow {v_e} \approx \frac{{\frac{{{y_1}\Delta y}}{{\sqrt {{a^2} + y_1^2} }}}}{{\Delta t}}$

$ \Rightarrow {v_e} \approx \frac{{{y_1}}}{{\sqrt {{a^2} + y_1^2} }}\frac{{\Delta y}}{{\Delta t}}$

$ \Rightarrow {v_e} \approx \frac{{{y_1}}}{{\sqrt {{a^2} + y_1^2} }}{v_m}$

$ \Rightarrow {v_e} < {v_m}$

Hence, option (B).