$A$ solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is:

An inclined plane makes an angle $30^{\circ}$ with the horizontal. $A$ solid sphere rolling down an inclined plane from rest without slipping has a linear acceleration (where $g$ is the acceleration due to gravity and $\sin 30^{\circ} = 0.5$).

$A$ solid sphere rolls down from the top of an inclined plane,$7 \ m$ high,without slipping. Its linear speed at the foot of the plane is $\left(g = 10 \ m/s^2\right)$

$A$ solid sphere of mass $2 \,kg$ rolls on a smooth horizontal surface at $10 \,m/s$. It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g=10 \,m/s^2$].

$A$ ring,sphere,and disc are rolling down an inclined plane from the same height. Find the wrong statement: (where $t$ is the time of descent,$a$ is the acceleration,and $v$ is the speed at the bottom).

$A$ solid sphere and a solid cylinder of identical radii approach an incline with the same linear velocity (see figure). Both roll without slipping throughout the motion. The two climb maximum heights $h_{sph}$ and $h_{cyl}$ on the incline. The ratio $\frac{h_{sph}}{h_{cyl}}$ is given by