Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

From the top of a tower, a particle is thrown vertically downwards with a velocity of $10 \text{ m/s}$ . The ratio of the distances, covered by it in the 3rd and 2nd seconds of the motion is (Take $g=10 \text{ m/s}^2$ ):

5 : 7

7 : 5

3 : 6

6 : 3

Step-by-Step Solution

Identify the Formula: The distance covered by a particle in the $n$ -th second of uniformly accelerated motion is given by the formula: $S_n = u + \frac{a}{2}(2n - 1)$ where $u$ is initial velocity, $a$ is acceleration, and $n$ is the specific second .
Identify Given Values:

Initial velocity ( $u$ ) = $10 \text{ m/s}$ (downwards, taking down as positive).
Acceleration ( $a$ ) = $g = 10 \text{ m/s}^2$ .

Calculate Distance in 3rd Second ( $S_3$ ): Substitute $n=3$ : $S_3 = 10 + \frac{10}{2}(2 \times 3 - 1)$ $S_3 = 10 + 5(6 - 1) = 10 + 25 = 35 \text{ m}$
Calculate Distance in 2nd Second ( $S_2$ ): Substitute $n=2$ : $S_2 = 10 + \frac{10}{2}(2 \times 2 - 1)$ $S_2 = 10 + 5(4 - 1) = 10 + 15 = 25 \text{ m}$
Calculate Ratio: $\text{Ratio} = \frac{S_3}{S_2} = \frac{35}{25} = \frac{7}{5}$

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