A uniform cylinder of mass `M` and radius `R` is released from rest on a rough inclined surface of inclination `theta` with the horizontal as shown in figure. As the cylinder rolls down the inclined surface, the maximum elongation it the spring stiffness `k` is

Options

A. `(3)/(4)(Mgsintheta)/(k)`
B. `(2Mgsintheta)/(k)`
C. `(Mgsintheta)/(k)`
D. none of these

Answer

B

Explanation

From energy conservation principle, maximum spring potential energy.
`=` maximum decrease in gravitational potential energy
`therefore(1)/(2)kx_(m)^(2)=Mgh=Mg(x_(m)sintheta)`
`thereforex_(m)=(2Mgsintheta)/(k)`