$A$ solid disc of radius $a$ and mass $m$ rolls down without slipping on an inclined plane making an angle $\theta$ with the horizontal. The acceleration of the disc will be $\frac{2}{b} g \sin \theta$ where $b$ is $........$. (Round off to the Nearest Integer)
($g =$ acceleration due to gravity)
($\theta =$ angle as shown in figure)

$A$ uniform solid spherical ball is rolling down a smooth inclined plane from a height $h$. The velocity attained by the ball when it reaches the bottom of the inclined plane is $v$. If the ball is now thrown vertically upwards with the same velocity $v$, the maximum height to which the ball will rise is

$A$ ring takes time $t_1$ in slipping down an inclined plane of length $L$ and takes time $t_2$ in rolling down the same plane. The ratio $\frac{t_1}{t_2}$ is

$A$ solid sphere is rolling on a frictionless surface,as shown in the figure,with a translational velocity $v \, m/s$. If it is to climb the inclined surface to a height $h$,then $v$ should be:

$A$ rolling wheel of $12 \, kg$ is on an inclined plane at position $P$ and connected to a mass of $3 \, kg$ through a string of fixed length and a pulley as shown in the figure. Consider $PR$ as a friction-free surface. The velocity of the centre of mass of the wheel when it reaches the bottom $Q$ of the inclined plane $PQ$ will be $\frac{1}{2} \sqrt{xgh} \, m/s$. The value of $x$ is.............

$A$ solid sphere rolls down without slipping from the top of an inclined plane of height $28 \text{ m}$ and angle of inclination $30^{\circ}$. The velocity of the sphere, when it reaches the bottom of the plane is (Acceleration due to gravity $= 10 \text{ ms}^{-2}$)