Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

A body falls from a height $h=200 \text{ m}$ (at New Delhi). The ratio of distance travelled in each $2 \text{ sec}$ during $t = 0$ to $t = 6$ second of the journey is:

1 : 4 : 9

1 : 2 : 4

1 : 3 : 5

1 : 2 : 3

Step-by-Step Solution

Analyze the Motion: The body falls from rest ( $u=0$ ) under constant acceleration due to gravity ( $a=g$ ). The problem asks for the ratio of distances covered in successive equal time intervals of $\tau = 2 \text{ s}$ .
Galileo's Law of Odd Numbers: For a body falling from rest, the distances traversed during equal intervals of time stand to one another in the same ratio as the odd numbers beginning with unity .

Ratio = $1 : 3 : 5 : 7 : \dots$

Verification via Calculation:

Distance in 1st interval ( $0-2\text{ s}$ ): $h_1 = \frac{1}{2}g(2)^2 = 2g$ .
Distance in 2nd interval ( $2-4\text{ s}$ ): Total distance in $4\text{ s}$ ( $8g$ ) minus distance in $2\text{ s}$ ( $2g$ ) = $6g$ .
Distance in 3rd interval ( $4-6\text{ s}$ ): Total distance in $6\text{ s}$ ( $18g$ ) minus distance in $4\text{ s}$ ( $8g$ ) = $10g$ .
Ratio: $2g : 6g : 10g = 1 : 3 : 5$ .

Check Validity: The total distance fallen in $6\text{ s}$ is approximately $180 \text{ m}$ (taking $g=10 \text{ m/s}^2$ ), which is less than the initial height of $200 \text{ m}$ , so the body is still in flight.

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