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```
A small block slides without friction down an inclined starting form rest. Let S n be the distance travelled from time t = n-1 to t = n. Then S n /S n+1 is - (A) 2n-1/2n (B) 2n+1/2n+1 (C) 2n-1/2n+1 (D) 2n/2n+1

```
9 years ago 419 Points
```							Dear Shivam
S=ut+0.5at2= 0.5at2   (since u=0)
At t=n-1
S= 0.5a(n-1)2
At t=n
S'=0.5a(n)2
Sn= S'-S =0.5 [n2 - (n-1)2] = 0.5a (n-n+1)*(n+n-1)
Sn= 0.5a*(2n-1)
put n+1 in place of n in expression of Sn
Sn+1= 0.5a (2(n+1)-1)= 0.5a (2n+1)
Sn/Sn+1 = 2n-1/2n+1
option (C)

All the best.
AKASH GOYAL

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```
9 years ago
```							a is the accleration of block then
S = 1/2 a t2                         (using this eq)
Sn = 1/2 a [n2-(n-1)2]
= 1/2 a [2n-1]             ...........1
Sn+1 = 1/2a[(n+1)2-n2]
=1/2 a [2n+1]          .........2
dividing eq 1 & 2
Sn/Sn+1 = 2n-1/2n+1
```
9 years ago
```							Sir it is inclined plane so how do u took a=g as it is not horizontal plane So akash sir plz correct it
```
3 years ago 605 Points
```							Dear student,Please find the solution to your problem. S = ut+0.5at2 = 0.5at2   (since u=0)At t = n – 1S = 0.5a(n – 1)2At t = nS' = 0.5a(n)2Sn = S' – S = 0.5 [n2 – (n – 1)2] = 0.5a (n – n + 1)*(n + n – 1)Sn= 0.5a*(2n – 1)put n+1 in place of n in expression of SnSn+1 = 0.5a (2(n + 1) – 1) = 0.5a (2n+1)Sn/Sn+1 = 2n – 1/2n + 1Hence option (C) is correct Thanks and regards,Kushagra
```
2 months ago
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