A small body starts its motion on a friction surface from rest on an inclined plane. If $S_{n}$ is the distance travelled in tme $t = n - 1$ to $t = n$ then the ratio of $S_{n}$ and $S_{n + 1}$ will be:

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

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

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

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\frac{2 n}{2 n + 1}

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Solution

The correct option is C $\frac{2 n - 1}{2 n + 1}$
A small body starts its motion on a friction surface from rest on and inclined plane
If $S_{n}$ is the distance travelled in time $T = n - 1$ to $t = n$
The ratio of $S_{n}$ and $S_{n + 1}$ will be
Solution
Distance travelled upto $n + 1$ sec
$= D (n + 1) = \frac{1}{2} g sin θ (n + 1)^{2}$
$S_{n} =$ Distance travelled from $t = n - 1 s e c$
To $n s e c$
$= \frac{1}{2} g sin θ n_{(}^{2} n - 1)^{2} = \frac{1}{2} g sin θ (2 n - 1)$
$S_{n + 1} =$ distance travelled from
$t = n s e c$ to $n + 1 s e c$
$\frac{1}{2} g sin θ (2 n + 1) r a t i o = \frac{S_{n}}{S_{n + 1}} = \frac{2 n - 1}{2 n + 1}$

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