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14/09/2021 10:18 am
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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 \(\frac{S_n}{S_{n+1}}\) is:
(1) \(\frac{2n-1}{2n}\)
(2) \(\frac{2n-1}{2n+1}\)
(3) \(\frac{2n+1}{2n-1}\)
(4) \(\frac{2n}{2n-1}\)
1 Answer
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14/09/2021 10:25 am
Correct answer: (2) \(\frac{2n-1}{2n+1}\)
Explanation:
Sn = Distance in nth sec. i.e. t = n – 1 to t = n
Sn+1 = Distance in (n + 1)th sec.
i.e. t = n to t = n + 1
So as we know
Sn = \(\frac{a}{2}\)(2n-1) a = acceleration
\(\frac{S_n}{S_{n+1}}\) = \(\frac{\frac{a}{2}(2n-1)}{\frac{a}{2}(2(n+1)-1)}\)
= \(\frac{2n-1}{2n+1}\)
\(\frac{S_n}{S_{n+1}}\) = \(\frac{2n-1}{2n+1}\)