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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
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
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 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. Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
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
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.
Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar respectively : Click here to download the toolbar..
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
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
Sir it is inclined plane so how do u took a=g as it is not horizontal plane So akash sir plz correct it
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