If $M_1 = M_2$ and $\theta = 30^o$,then the acceleration is ........... $ms^{-2}$.

$A$ block of mass $m_1 = 2\,kg$ on a smooth inclined plane at an angle of $30^\circ$ is connected to a second block of mass $m_2 = 3\,kg$ by a cord passing over a frictionless pulley as shown in the figure. The acceleration of each block is ....... $m/s^2$ (assume $g = 10\,m/s^2$).

Two masses $M_1$ and $M_2$ are attached to the ends of a string which passes over a pulley attached to the top of a double inclined plane of angles of inclination $\alpha$ and $\beta$. If the system is released,the acceleration $a$ of the system is given by:

$A$ body is slipping from an inclined plane of height $h$ and length $l$. If the angle of inclination is $\theta$,the time taken by the body to come from the top to the bottom of this inclined plane is

$A$ block is kept on a fixed smooth wedge whose vertical section is a curve $y = \frac{x^2}{\sqrt{3}}$ as shown in the figure,where $x$ represents the horizontal direction and $y$ represents the vertical direction. When released from a point where $y = \frac{1}{4\sqrt{3}}$,what will be its acceleration? $(g = 10 \ m/s^2)$

Find the reading of the spring balance as shown in the figure. Assume that mass $M$ is in equilibrium and the block of mass $2.5 \text{ kg}$ is also in equilibrium. All surfaces are smooth. (Take $g = 10 \text{ m/s}^2$) (in $\text{ N}$)