A meter stick `AB` of length 1 meter rests on a frictionless floor in horizontal position with end `A` attached to the string as shown. Assume that string connecting meter stick with pulley always remains vertical.

If block 1 and 2 are given constant speeds as shown then the distance moved by end `B` over the floor in the period for which speed of `B` is less than `A`.

Options

A. `((sqrt(2)+1)/(sqrt(2)))m`
B. `((sqrt(2)-1)/(sqrt(2)))m`
C. `(1)/(sqrt(2))m`
D. `(1)/(2)m`

Answer

B

Explanation

for any angle `'theta'`
`x^(2)+y^(2)=l^(2)`
`:. 2x x'+2y y'=0`
`:. X(-v_(B))+y(v_(A))=0 i.e. v_(B)=v_(A) tan theta`
or `v_(B)=4 tan theta ….(i)`
`[as v_(A)=(3+5)/(2)=4m//s]`
from `v_(B)=v_(A) tan theta`
we can see that `v_(B)ltv_(A)` for `0lethetale(pi)/(4)`
distance moves by `'B'` is
`d=1-x=1-(1)/(sqrt(2))=((sqrt(2)-1))/(sqrt(2))`
`[as x=(1)/(sqrt(2))` at `theta=(pi)/(4)]`