UPSEE 2019 Solved Paper

Key Idea For a uniformly accelerated body, the displacement in nth second is given by the expression, $s_n = u + \frac{1}{2} a(2n-1) .$ A body placed on a smooth inclined plane is shown in figure below,Distance travelled in $n$th second $s_n = u + \frac{1}{2}(g \sin \theta)(2n-1)$ Here, initial velocity, $u = 0$ ⇒ $s_n = \frac{g \sin \theta}{2}(2n-1)$ ......(i) Similarly, in $(n+1) t h$ second $s_{(n+1)} = \frac{g \sin \theta}{2}[2(n+1)-1]$ ⇒ $s_{(n+1)} = \frac{g \sin \theta}{2}(2n+1)$ ......(ii) From Eqs. (i) and (ii), we get the ratio of distances, $\frac{s_{(n+1)}}{s_n} = \frac{2n+1}{2n-1}$ $\therefore$ The correct option is (c).