Step 1: Relation between kinetic energy and momentum - Kinetic energy \[ K = \frac{P^2}{2m}, \] where \(P\) is momentum and \(m\) is mass.
Step 2: Calculate new kinetic energy - Initial momentum \(P_1\), kinetic energy \[ K_1 = \frac{P_1^2}{2m}. \] New momentum \[ P_2 = P_1 + 0.5P_1 = 1.5P_1. \] New kinetic energy \[ K_2 = \frac{P_2^2}{2m} = \frac{(1.5P_1)^2}{2m} = \frac{2.25P_1^2}{2m} = 2.25K_1. \]
Step 3: Calculate percentage increase in kinetic energy - Percentage increase is given by: \[ \frac{K_2 - K_1}{K_1} \times 100 = \frac{2.25K_1 - K_1}{K_1} \times 100 = 125\%. \]
Final Answer: The percentage increase in kinetic energy is 125%.
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Knowing the initial position \( x_0 \) and initial momentum \( p_0 \) is enough to determine the position and momentum at any time \( t \) for a simple harmonic motion with a given angular frequency \( \omega \).
Reason (R): The amplitude and phase can be expressed in terms of \( x_0 \) and \( p_0 \).
In the light of the above statements, choose the correct answer from the options given below:
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)