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)

Step-by-Step Explanation:

To find the ratio of the distance travelled in consecutive time intervals, we use the kinematics equation for the distance travelled by a body in the n^th second of its motion.

The distance travelled by an object starting with an initial velocity $u$ and moving with constant acceleration $a$ during the n^th second (from $t=n-1$ to $t=n$) is given by the formula:

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

Since the block starts from rest at time $t=0$, its initial velocity is zero:

$u=0$

Substituting $u=0$ into the equation, we get the distance travelled in the n^th second as:

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

Similarly, the distance travelled in the next consecutive second, which is the (n + 1)^th second (from $t=n$ to $t=n+1$), is obtained by replacing $n$ with $n+1$ in the formula:

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

Simplifying the term inside the bracket:

$2(n+1)-1=2n+2-1=2n+1$

Therefore, the distance travelled in the (n + 1)^th second is:

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

Now, we find the ratio of $S_n$ to $S_{n+1}$:

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

Dividing out the common acceleration term $\frac{a}{2}$, we obtain:

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

Thus, the required ratio is $\frac{2n-1}{2n+1}$.