UPSEE 2019 Solved Paper

Key Idea For a uniformly accelerated body, the displacement in nth second is given by the expression, sn=u+

1

2

a(2n−1). A body placed on a smooth inclined plane is shown in figure below,Distance travelled in nth second sn=u+

1

2

(g‌sin‌θ)(2n−1)
Here, initial velocity, u=0
⇒ sn=

g‌sin‌θ

2

(2n−1) ......(i)
Similarly, in (n+1)th second
s(n+1)=

g‌sin‌θ

2

[2(n+1)−1]
⇒ s(n+1)=

g‌sin‌θ

2

(2n+1) ......(ii)
From Eqs. (i) and (ii), we get the ratio of distances,

s(n+1)

sn

=

2n+1

2n−1

∴ The correct option is (c).