
In a symmetrical graph, the average value of speed corresponds to the midpoint, which is also the most probable speed.
However, when calculating the root mean square (rms) speed, we compute the average of the squares of the speeds and then take the square root. Since higher speeds contribute more to the squared value, the result will be higher than the average speed value.
So the Correct Option is (B): $1: 1: 1.224$
Find the time required to complete a reaction 90% if the reaction is completed 50% in 15 minutes.
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 
It is branch of physics that defines motion with respect to space and time is known as kinematics.
Inverse Kinematics: Inverse Kinematics do the reverse of kinematics.
There are four basic kinematics equations:

Another branch of kinematics equations which deals with the rotational motion of anybody.
