A small block slides without friction down an inclined plane starting from rest. Let sn be the distance travelled…..

Q. A small block slides without friction down an inclined plane starting from rest. Let sn be the distance travelled from t = n-1 to t = n. Then \displaystyle \frac{S_n}{S_{n+1}} is:

(a) \displaystyle \frac{2n-1}{2n}

(b) \displaystyle \frac{2n+1}{2n-1}

(c) \displaystyle \frac{2n-1}{2n+1}

(d) \displaystyle \frac{2n}{2n+1}

Ans: (c)
Sol:
\displaystyle S_n = 0+\frac{a}{2}(2n-1)

\displaystyle S_{n+1} = 0+ \frac{a}{2}(2(n+1)-1)

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

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

Author: Rajesh Jha

QuantumStudy.com

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