A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sₙ be the distance travelled by the block in the interval t=n−1 to t=n. Then, the ratio Sₙ/(Sₙ₊₁) is :

The correct option is 2ₙ-1/2ₙ+1.

Let the small block have a constant acceleration of $a$ as it slides down the smooth inclined plane. We are given that the block starts from rest at time $t=0$, meaning its initial velocity is $u=0$.

The distance traveled by a uniformly accelerating object in the $n^{\text{th}}$ second (the interval from $t=n-1$ to $t=n$) can be calculated using the standard kinematic formula:

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

By substituting the initial velocity $u=0$ into this equation, we get the distance traveled during the $n^{\text{th}}$ interval:

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

Next, we apply the same logic to find the distance traveled by the block in the very next time interval, which is the ${(n+1)}^{\text{th}}$ second (from $t=n$ to $t=n+1$). We simply replace $n$ with $n+1$ in our formula:

$S_{n+1}=\frac{a}{2}\left(2\right(n+1)-1)$

Simplifying the expression inside the parentheses, we have:

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

Finally, we need to find the ratio of the distance traveled in the $n^{\text{th}}$ interval to the distance traveled in the ${(n+1)}^{\text{th}}$ interval. We do this by dividing $S_n$ by $S_{n+1}$:

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

The common acceleration term $\frac{a}{2}$ cancels out perfectly from both the numerator and the denominator, leaving us with the final ratio:

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

This matches our correct option perfectly.