The correct option is
A 2 tanθThe given condition is shown in the figure as
First of all we have to calculate the acceleration of the block in both the regions i.e. smooth (
aS) and rough (
aR)
Force acting down the plane in smooth region is
F=mgsinθ
∴ acceleration
aS=mgsinθm=gsinθ
Similarly, force acting down the plane in rough region is
F=mgsinθ−f=mgsinθ−μmgcosθ
∴ acceleration
aR=mg(sinθ−μcosθ)m=g(sinθ−μcosθ)
Now, applying equation of motion in both the regions:
1. For smooth plane
Let the velocity gained by block at the end of smooth surface be
v
So, from equation we have
v2−u2=2aSS
v2−(0)2=2(gsinθ)S
v2=2gSsinθ
2. For rough plane
The initial velocity of the block at the rough surface will be the velocity gained by block at the end of smooth surface.
Let the final velocity of the block at the end of the rough surface be
v1
then, from the equation we have
v21−v2=2aRS
⇒ 02−v2=2g(sinθ−μcosθ)S
⇒ −v2=2g(sinθ−μcosθ)S
⇒ −2gSsinθ=2gS(sinθ−μcosθ)
⇒ −sinθ=sinθ−μcosθ
⇒ −2sinθ=−μcosθ
⇒ μ=2sinθcosθ=2tanθ