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Shivam Bhagat Grade: 9
        

 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

Answers : (3)

AKASH GOYAL AskiitiansExpert-IITD
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


AskiitiansExpert-IITD


 


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6 years ago
vikas askiitian expert
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
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