A ladder rests against a frictionless vertical wall, with its upper end $6\mathrm{m}$ above the ground and lower end $4\mathrm{m}$ away from the wall. The weight of ladder is $500\mathrm{N}$ and centre of
gravity $=\left(\frac{1}{3}\right)rd$distance from the lower end. Wall's reaction will be, in $\mathrm{N}$

Two blocks of masses $M_1=4 \mathrm{kg}$ and $M_2=6 \mathrm{kg}$ are connected by a string of negligible mass passing over a frictionless pulley as shown in the figure below. The coefficient of friction between block $M_1$ and the horizontal surface is $0.4$. When the system is released, the masses $M_1$ and $M_2$ start accelerating. What additional mass $m$ should be placed over $M_1$, so that the masses $\left(M_1+m\right)$ slide with a uniform speed?

Assertion: A car moving on a horizontal road may be thrown out of the road in taking a turn due to lack of proper centripetal force.

Reason: If a particle moves in a circle, describing equal angles in equal intervals of time. Then the velocity vector changes its magnitude.