Back to Directory
NEET PHYSICSMedium

A small block slides down on a smooth inclined plane, starting from rest at time t=0t = 0. Let SnS_n be the distance travelled by the block in the interval t=n−1t = n - 1 to t=nt = n. Then, the ratio SnSn+1\frac{S_n}{S_{n+1}} is

1

2n−12n\frac{2n-1}{2n}

2

2n−12n+1\frac{2n-1}{2n+1}

3

2n+12n−1\frac{2n+1}{2n-1}

4

2n2n−1\frac{2n}{2n-1}

Step-by-Step Solution

Let θ\theta be the inclination of the inclined plane, acceleration a=gsin⁡θa = g \sin \theta. Sn=u+a2(2n−1)S_n = u + \frac{a}{2}(2n-1). With u=0u=0, Sn=gsin⁡θ2(2n−1)S_n = \frac{g \sin \theta}{2}(2n-1). Similarly, Sn+1=gsin⁡θ2(2(n+1)−1)=gsin⁡θ2(2n+1)S_{n+1} = \frac{g \sin \theta}{2}(2(n+1)-1) = \frac{g \sin \theta}{2}(2n+1). Thus, SnSn+1=2n−12n+1\frac{S_n}{S_{n+1}} = \frac{2n-1}{2n+1}.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started