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`        A monkey pulls the midpoint of a 10cm long light inextnsible string connecting two identical objects A and B lying on the smooth table of masses 0.3kg continously along the perpendicular bisector of line joining the masses. the masses are found to approach each other at a relative acceleratioin of 5m/s2 when they are 6cm apart. the constsnt force applied by monkey is............................................................`
7 years ago

AYUSH SHARMA
33 Points
```										hiiiiiiiiiiiiiiiiiiiiii dude please approve this
```
7 years ago
510 Points
```										let length of string is 2l
monkey is pulling from mid point , let after some time its particles are at 2x distance from each other then
there will be similar  triangles whose hypotenuse is l & one of its side is x....
other side will be (l2-x2)1/2
2Tcos@ = F      (force applied by monkey)
T = F/2cos@
ma = F/2cos@
a = F/2mcos@............1
from figure cos@ = (l2-x2)/l               (@ is angle bw perpendicular bisector & tension acting on any of particle)
a = Fl/2m(l2-x2)
2l = 10 , l =5cm
2x = 6 ,x =3
putting these values
a = 5F/8m
5 = 5F/8m                      (a = 5m/s2 (given))
F = 8m = 2.4N
```
7 years ago
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