A small block slides down on a smooth inclined plane, starting from rest at time . Let be the distance travelled by the block in the interval to . Then, the ratio is [2021]

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Solution

Q.
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 [2021]

1 $\frac{2 n}{2 n - 1}$

2 $\frac{2 n - 1}{2 n}$

3 $\frac{2 n - 1}{2 n + 1}$

4 $\frac{2 n + 1}{2 n - 1}$

Ans.
(3)

The acceleration is $a = g \sin θ$

Initial velocity, $u = 0$

Distance travelled in $n^{th}$ second $S_{n^{t h}} = u + \frac{1}{2} a (2 n - 1)$

$S_{n^{t h}} = \frac{1}{2} a (2 n - 1)$ ...(i)

$S_{{(n + 1)}^{t h}} = \frac{1}{2} a [2 (n + 1) - 1]$ ...(ii)

On dividing eqn (i) by (ii), we get

$∴$ $\frac{S_{n^{t h}}}{S_{{(n + 1)}^{t h}}} = \frac{(2 n - 1)}{(2 n + 1)}$

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