A bar of mass m resting on a smooth horizontal plane starts moving due to force
`|F| =(mg)/9`. The magnitude of force remains
constant with time. The force vector makes an angle `theta` with the horizontal which varies with the distance covered as `theta =Cx`. if the
constant , `C=10 (("degree")/("meter"))`, then the speed of
the bar , when `theta` becomes equal to `30^@` for the first time is, (Take , `g =10 m//s^2`)

Options

A. 0.33 m/s
B. 0.50 m/s
C. 1.0 m/s
D. 0.8 m/s

Answer

A

Explanation

Figure shows a force acting on a bar,

Since it is given that the minimum force to move the block ,
`F_(min) =(mg)/9` ………(i)
So, the force acting on the bar
`F_H =F cos theta =F cos 10 x`
Here, `theta =cx and c =10^@ m^(-1) , theta = 30^@`
Hence at an angle `30^@` ,
`theta = 10x rArr 30^@ = 10x rArr x = 3m` ........(ii)
Given that at `30^@` block start moving .
Hence , from Eq. (i) `F cos 30^@ = (mg)/9`
`F = (2mg)/(9sqrt3)` ..........(iii)
From energy conservation law,
`1/2 mv^2|int . cos theta ""dx|`
Substituting the value F from Eq. (iii) we get
`rArr 1/2 mv^2 =(2mg)/(9sqrt3) |overset(3)underset(0)int cos 10x dx |`
`1/2 mv^2 =(20)/(9sqrt3) mxx (sin 30^@)/(10) |`
`rArr v^2 =2/(9sqrt3) rArr v = 0.35 ~~ 0.33 m//s`