A. Person A has reaction time of $0.20 \text{ s}$.
B. Person B has reaction time of $0.22 \text{ s}$.
C. Person C has reaction time of $0.18 \text{ s}$.
D. Person D has reaction time of $0.19 \text{ s}$.
E. Person E has reaction time of $0.21 \text{ s}$.
What is the correct order of the distance travelled by the ruler for each person ?
When a ruler falls vertically from rest, the distance it travels ($d$) is determined by the time it is in the air. This motion is governed by the laws of kinematics under constant acceleration due to gravity ($g$).
The distance fallen by an object starting from rest is given by the equation:
$d = \frac{1}{2}gt^2$
where $g$ is the acceleration due to gravity and $t$ is the time of fall. In this problem, the time of fall is the reaction time of each person catching the ruler.
Since $g$ is constant ($9.8 \text{ m s}^{-2}$), the distance traveled ($d$) is directly proportional to the square of the reaction time ($t^2$). For positive reaction times:
The given reaction times are:
Ordering these times from largest to smallest:
Since the distance travelled is directly proportional to the reaction time (or its square), the order of distances will match the order of reaction times from longest to shortest.
Thus, the order of distances travelled is:
Distance for B > Distance for E > Distance for A > Distance for D > Distance for C
This corresponds to the order: B > E > A > D > C.
A thin wire of length 'L' and linear mass density 'm' is bent into a circular ring (in x-y plane) with centre 'C' as shown in figure. The moment of inertia of the ring about an axis yy' (tangent in the plane) will be :
Bob B of mass $m$ at rest is hanging vertically from the ceiling via a massless string of length $10\text{ m}$, as shown in the figure. Point mass A of mass $m$ travelling horizontally with speed $10\text{ ms}^{-1}$ hits bob B elastically. The bob B rises $h$ meter after the collision. Taking the acceleration due to gravity $g = 10\text{ ms}^{-2}$ and neglecting the size of the bob, the value of $h$ is :
A thin horizontal disc is rotating about a vertical axis passing through its fixed centre O. Its angular momentum is $L_A$ and $L_B$ computed about points A and B, respectively, with $OB = 2 \times OA$. The value of $\frac{L_A}{L_B}$ is :
Two circular discs of radius each $10 \text{ cm}$ are joined at their centres by a rod of length $30 \text{ cm}$ and mass $600 \text{ gm}$ as shown in figure.
If the mass of each disc is $600 \text{ gm}$ and applied torque between two discs is $43 \times 10^5 \text{ dyne.cm}$, the angular acceleration of the discs about the given axis $AB$ is________$\text{rad/s}^2$.
