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

Question

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

AKASH GOYAL AskiitiansExpert-IITD · Answer

Dear Shivam S=ut+0.5at 2 = 0.5at 2 (since u=0) At t=n-1 S= 0.5a(n-1) 2 At t=n S'=0.5a(n) 2 S n = S'-S =0.5 [n 2 - (n-1) 2 ] = 0.5a (n-n+1)*(n+n-1) S n = 0.5a*(2n-1) put n+1 in place of n in expression of S n S n+1 = 0.5a (2(n+1)-1)= 0.5a (2n+1) S n /S n+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..

vikas askiitian expert · Answer

a is the accleration of block then

S = 1/2 a t² (using this eq)

S_n = 1/2 a [n²-(n-1)²]

= 1/2 a [2n-1] ...........1

S_n+1 = 1/2a[(n+1)²-n²]

=1/2 a [2n+1] .........2

dividing eq 1 & 2

S_n/S_n+1 = 2n-1/2n+1

navkirat · Answer

Sir it is inclined plane so how do u took a=g as it is not horizontal plane So akash sir plz correct it

Kushagra Madhukar · Answer

Dear student, Please find the solution to your problem. S = ut+0.5at 2 = 0.5at 2 (since u=0) At t = n – 1 S = 0.5a(n – 1) 2 At t = n S' = 0.5a(n) 2 S n = S' – S = 0.5 [n 2 – (n – 1) 2 ] = 0.5a (n – n + 1)*(n + n – 1) S n = 0.5a*(2n – 1) put n+1 in place of n in expression of S n S n+1 = 0.5a (2(n + 1) – 1) = 0.5a (2n+1) S n /S n+1 = 2n – 1/2n + 1 Hence option (C) is correct Thanks and regards, Kushagra

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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

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

4 Answers

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