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 S_n/S_n + 1 is :

The correct option is 2n - 1 / 2n + 1.

Step-by-Step Explanation:

Let us analyze the motion of the block sliding down a smooth inclined plane.
Since the inclined plane is smooth, the acceleration of the block is constant. Let this constant acceleration be $a$.
The block starts from rest at time $t=0$, so its initial velocity $u=0$.

The distance travelled by an object in the $n^{\mathrm{th}}$ second (i.e., between the time interval $t=n-1$ to $t=n$) under constant acceleration is given by the formula:
$S_n=u+\frac{a}{2}(2n-1)$

Since the block starts from rest, we substitute $u=0$ into the equation:
$S_n=\frac{a}{2}(2n-1)$ --- (Equation 1)

Similarly, the distance travelled by the block in the next interval, which is the ${(n+1)}^{\mathrm{th}}$ interval (from $t=n$ to $t=n+1$), is represented as $S_{n+1}$.
By replacing $n$ with $n+1$ in the general formula:
$S_{n+1}=\frac{a}{2}\left[2\right(n+1)-1]$
$S_{n+1}=\frac{a}{2}(2n+2-1)$
$S_{n+1}=\frac{a}{2}(2n+1)$ --- (Equation 2)

Now, we need to find the ratio $\frac{S_n}{S_{n+1}}$. Dividing Equation 1 by Equation 2:
$\frac{S_n}{S_{n+1}}=\frac{\frac{a}{2}(2n-1)}{\frac{a}{2}(2n+1)}$

Cancelling the common factor $\frac{a}{2}$ from both the numerator and the denominator, we get:
$\frac{S_n}{S_{n+1}}=\frac{2n-1}{2n+1}$

Thus, the ratio of the distance travelled in the $n^{\mathrm{th}}$ interval to that in the ${(n+1)}^{\mathrm{th}}$ interval is indeed 2n - 1 / 2n + 1.