A body is sliding down a rough inclined plane. The coefficient of friction between the body and the plane is 0.5. The ratio of the net force required for the body to slide down and the normal reaction on the body is 1:2. Then, the angle of the inclined plane is

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

(c)

Explanation

`mu=0.5`
From the figure,
`F=mgsintheta-f`
`F=mgsintheta-muN`
`F=mgsintheta-mumgcostheta (becauseN=mgcostheta)`
`F:R=1:2 (given)`
`(F)/(R)=(1)/(2)`
rArr `(mgsintheta-mumgcostheta)/(mgcostheta)=(1)/(2)`
`rArr 2mgsintheta-2mumgcostheta=mgcostheta`
` rArr 2sintheta -2mucostheta =costheta`
`rArr 2tantheta-2mu=1`
` rArr tantheta -mu=(1)/(2)`
`tantheta=mu+(1)/(2)=0.5+(1)/(2)=1`
`tantheta=tan45^(@)`
`because theta = 45^(@)`