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 interva - Physics MCQ NEET 2021 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

The distance travelled in the n^{t h} second starting from rest (u = 0) is: S_{n} = u + \frac{a}{2} (2 n - 1) = \frac{a}{2} (2 n - 1) The distance travelled in the (n + 1)^{t h} second is: S_{n + 1} = \frac{a}{2} [2 (n + 1) - 1] = \frac{a}{2} (2 n + 2 - 1) = \frac{a}{2} (2 n + 1) The ratio is: \frac{S _{n}}{S _{n + 1}} = \frac{\frac{a}{2} ( 2 n - 1 )}{\frac{a}{2} ( 2 n + 1 )} = \frac{2 n - 1}{2 n + 1} (Note: Option labels in source material might vary, but the mathematical ratio is \frac{2 n - 1}{2 n + 1}).

Try it yourself before checking the explanation above.

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:

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

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

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

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

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