In above diagram, ma=pseudo force due to acceleration of plank,
$f=$ force of static friction, $m g=$ weight of block and $N=$ normal reaction.
Here, we must note that due to motion of plank, normal reaction force does not passes through centre of mass.
If line of action of normal force at a distance $x$ from centre, then for equilibrium of block, net torque about centre of mass must be zero.
$\Rightarrow \quad \tau_{\substack{\text { Normal } \\ \text { reaction }}}=\tau_{\text {Friction }}$
$m g$ and $m a$ does not produces any torque about centre of mass as their line of action passes through centre of mass.
$\Rightarrow N \cdot x=f \cdot l / 2$
$\Rightarrow m g x=m a \frac{l}{2}$
${[\because N=m g \text { and } f=m a]}$
$\Rightarrow x=\left(\frac{a}{g}\right) \frac{l}{2}$
$\text { or } x=\frac{1}{10} \cdot \frac{l}{2} \quad\left[\because a=\frac{g}{10}\right]$
$\text { or } x=\frac{l}{20}$
So, coordinates of point from which reaction passes are $\left(-\frac{l}{20}, 0\right)$.
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If two springs $S_1$ and $S_2$ of force constants $k_1$ and $k_2$, respectively, are stretched by the same force, it is found that more work is done on spring $S_1$ than on spring $S_2$.
Statement $-1$: If stretched by the same amount, work done on $S_1$, will be more than that on $S_2$
Statement $-2$ : $k_1 < k_2$.
