Correct Answer - A
The initial velocity is `15m//s` acceleration of practicle is `(g sin theta + mu g cos theta )`
downwards along the tube.
`:. A=10 sin 37 +(3)/(8) xx10 xx cos 37=10xx(3)/(5)+(3)/(8)xx 10 xx(4)/(5)=9m//s^(2)`
` :. V ^(2)=u^(2)+2as`
`V^(2)=15^(2)-2xx9xx12.5=0`
`:. V^(2)=0`
Given that the tube is slightly less than `12.5m`. It means the particle will just drop from tube. Hence `K.E.` at the tube end `=0`