Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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

\frac{2n+1}{2n-1}

\frac{2n}{2n-1}

\frac{2n-1}{2n}

\frac{2n-1}{2n+1}

Step-by-Step Solution

Identify the Formula: The distance traveled by a body undergoing uniformly accelerated motion in the $n$ -th second (interval from $t = n-1$ to $t = n$ ) is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$ where $u$ is the initial velocity and $a$ is the acceleration.
Apply Conditions: The block starts from rest, so $u = 0$ . The acceleration $a$ is constant ( $g \sin \theta$ on an incline). Thus: $S_n = \frac{a}{2}(2n - 1)$
Calculate for (n+1)-th Interval: For the next interval ( $t = n$ to $t = n+1$ ), we substitute $n$ with $(n+1)$ : $S_{n+1} = \frac{a}{2}[2(n + 1) - 1] = \frac{a}{2}(2n + 2 - 1) = \frac{a}{2}(2n + 1)$
Calculate Ratio: $\frac{S_n}{S_{n+1}} = \frac{\frac{a}{2}(2n - 1)}{\frac{a}{2}(2n + 1)} = \frac{2n - 1}{2n + 1}$
Context: This is a generalization of Galileo's law of odd numbers found in NCERT .

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