`cos 405^@` equal to ______

Answer

A, B, C

Explanation

The block m will slide down the inclined pklane of mass M with aceleration `a_1=g sin alpha` (relative) to the inclined plane.
The horizontal component of `a_1` will be `a_x=g sin alpha cos alpha, for which the blcok M will accelerate towards left, let teh acceleration be `a_2.`
According to the concept of centre of mass, (in the horizontl direcrtion exterN/Al force is zero)
`ma_x=(M+m)a_2`
`rarr a_2=(ma_x)/(M+m)`
`=(mgsiN/Alpha cosalpha)/(M+m)`......i ltbr.gt so the absolute (resulstant) acceleration of m on the block M along the direction of the incline will be
`a=gsiN/Alpha-a_2cosalpha`
`=gsiN/Alpha- (mgsiN/Alphacos^2alpha)/(M+m)`
`=gsiN/Alpha[1-(mcos^2alpha)/(M+m)]`
`=g sin alpha[(M+m-mcos^2alpha)/(M+m)]`
So, `a=g siN/Alpha [(M+msin^2alpha)/(M+m)]`...ii
Let the time taken by the block m to reach the bottom end be t
`Now, S=ut+(1/2)at^2`
`rarrh/(siN/Alpha)=(1/2)at^2`
`rarr t=(sqrt(2h)/(a siN/Alpha))`
So, the velocity of the bigger block after time t will be
`v_m=u=a_2t`
`=(mgsiN/Alphacosalpha)/(M+m)`
(sqrt(2h)/(alphasiN/Alpha))`
`=[(2m^2g^2hsin^2alphacos^2alpha)/((M+m)^2a siN/Alpha)]^(1/2)`
now substituting the value of a from equtioin ii we get
` V_m=[(2m^2g^2hsin^2alpha)/((M+m)^2siN/Alpha)xx(cos^2alpha)/(gsiN/Alpha)((M+m))/((M+msin^2alpha))]`
or ` V_m=[(2m^2ghcos^2alpha)/((M+m)(M+msin^2alpha)]^(1/2)`