Question

A small block slides down on a smooth inclined plane, starting from rest at time $t=0.$ Let $S_n$ be the distance travelled by the block in the interval, $t=n-1$ to $t=n$, then, the ratio $\frac{S_n}{S_{n+1}}$ is

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

Answer 1

The correct option is C $\frac{2n-1}{2n+1}$

Suppose, $\theta$ is inclination of inclined plane.

So, the acceleration along inclined plane $a=g\sin \theta$

Initial speed, $u=0$.

By equation of uniformly accelerated motion, distance travelled by object during $n^{th}$ second,

$S_n=u+\frac{a}{2}(2n-1)$

$S_n=0+\frac{g\sin \theta }{2}(2n-1)=\frac{g\sin \theta }{2}(2n-1)...\left(i\right)$

Distance travelled during $(n+1{)}^{th}$ second.

$S_{n+1}=0+\frac{g\sin \theta }{2}\left[2\right(n+1)-1]=\frac{g\sin \theta }{2}(2n+1)...\left(ii\right)$

Using $\left(1\right)$ and $\left(2\right)$,

$\frac{S_n}{S_{n+1}}=\frac{(2n-1)}{(2n+1)}$

Hence, $\left(c\right)$ is the correct answer.