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.

To find the ratio of the distance travelled in the interval from $t=n-1$ to $t=n$ to the distance travelled in the next interval, we can use the formula for the distance travelled by an object undergoing constant acceleration in the $n$-th second.

The formula for the distance travelled in the $n$-th second is given by:
$S_n=u+\frac{a}{2}(2n-1)$
where:
- $u$ is the initial velocity,
- $a$ is the constant acceleration down the smooth inclined plane,
- $n$ is the specific second or time interval.

Since the block starts from rest at $t=0$, we have $u=0$. Substituting this value into the equation gives:
$S_n=\frac{a}{2}(2n-1)$

Next, we determine the distance travelled in the next interval, which is the $(n+1)$-th second (from $t=n$ to $t=n+1$). We find this by replacing $n$ with $n+1$ in our equation:
$S_{n+1}=\frac{a}{2}\left(2\right(n+1)-1)$
Simplifying the term inside the parentheses:
$2(n+1)-1=2n+2-1=2n+1$
So, the expression for $S_{n+1}$ becomes:
$S_{n+1}=\frac{a}{2}(2n+1)$

Now, we take the ratio of $S_n$ to $S_{n+1}$:
$\frac{S_n}{S_{n+1}}=\frac{\frac{a}{2}(2n-1)}{\frac{a}{2}(2n+1)}$
Cancelling the common factor $\frac{a}{2}$ from both the numerator and the denominator, we get:
$\frac{S_n}{S_{n+1}}=\frac{2n-1}{2n+1}$