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A small block slides without friction down an inclined plane starting from rest. Let So be the distance travelled from time t=n-1 to t=n, then Sn/Sn+1 is

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Best answer

Hi Hari,

please se the answer below. Hope it helps you

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

So the correct option is (C)

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