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) =`
`:. 2 x x, + 2 yy' = 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) = 4 m//s]`
from `v_(B) = v_(A) tan theta`
We can see that `v_(B) lt v_(A)`
for `0 le theta le (pi)/(4)`
`:.` from `theta = 0` to `theta = (pi)/(4)`
distance moved by 'B' is
`d=1-x=1 -(1)/(sqrt(2))=((sqrt(2)-1)/(sqrt(2)))`
`["as" x = (1)/(sqrt(2)) "at" theta = (pi)/(4)]`.