Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```         A small block slides without friction down an inclined starting form rest. Let Sn be the distance travelled from time
t = n-1 to t = n. Then Sn/Sn+1 is -
(A)   2n-1/2n
(B)   2n+1/2n+1
(C)   2n-1/2n+1
(D)   2n/2n+1```
6 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

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.
Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

```
6 years ago
510 Points
```										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
```
6 years ago
navkirat
23 Points
```										Sir it is inclined plane so how do u took a=g as it is not horizontal plane So akash sir plz correct it
```
4 months ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »
• Complete Physics Course - Class 12
• OFFERED PRICE: Rs. 2,756
• View Details
• Complete Physics Course - Class 11
• OFFERED PRICE: Rs. 2,968
• View Details