A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let S_n be the distance travelled by the block in the interval…

Question

A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let S_n be the distance travelled by the block in the interval t=n-1 to t=n. Then, the ratio s_ns_(n+1) is:

MARKS Content Team · Accepted Answer

Correct answer: (C) (2n-1)/(2n+1). As we know that the displacement covered by particle in n^thsecond is given by, ∵Sn^th=u+2n-12abut u=0∴s_n=(2n-1)/2a …1 s_(n+1)=(2(n+1)-1)/2a …(2) Now, equation (1) divide by equation (2) s_ns_(n+1)=(2n-1)/(2n+1)

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

Solution

Relevant Questions

A small block slides down on a smooth inclined plane, starting from rest at time t=0. Let Sn be the distance travelled by the block in the interval t=n-1 to t=n. Then, the ratio snsn+1 is:

Solution

Relevant Questions

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