Question

A body slides down a frictionless inclined plane starting from rest. If $S_n$ and $S_{n+1}$ be the distance travelled by the body during nth and $(n + 1) {th}$ seconds, then the ratio $\frac{S_{n+1}}{S_n}$ is

• A. $\frac{2n - 1}{2n + 1}$
• B. $\frac{2n }{2n + 1}$
• C. $\frac{2n + 1}{2n - 1}$
• D. $\frac{2n }{2n - 1}$

Answer 1

Answer: (C)

$\frac{2n + 1}{2n - 1}$

Answer 2

Key Idea For a uniformly accelerated body, the displacement in nth second is given by the expression,
$s_{n} = u +\frac{1}{2} a \left(2n-1\right)$.A body placed on a smooth inclined plane is shown in figure below,

Distance travelled in $n$th second
$s_{n} = u +\frac{1}{2} \left(g \,sin\, \theta\right)\left(2n-1\right)$
Here, initial velocity, $u =\theta$
$\Rightarrow s_{n} =\frac{ g \,sin\, \theta }{2} \left(2n-1\right) \quad...\left(i\right)$
Similarly, in $\left( n+1\right)$ th second
$s_{\left(n+1\right)} =\frac{ g \,sin\, \theta}{2} \left[2\left(n+1\right)-1\right]$
$\Rightarrow s_{\left(n+1\right)} = \frac{g \,sin\, \theta}{2}\left(2n+1\right) \quad...\left(ii\right)$
From Eqs. $\left(i\right)$ and $\left(ii\right)$, we get the ratio of distances,
$\frac{s_{\left(n+1\right)}}{s_{n}} = \frac{2n+1}{2n-1}$