To solve this problem, we need to understand the relationship between momentum and kinetic energy. Let's break it down step-by-step:
1. **Formula for Momentum and Kinetic Energy**: The momentum (\(p\)) of a body is given by the formula:
\(p = mv\)
where \(m\) is the mass of the body and \(v\) is its velocity.
The kinetic energy (\(KE\)) of a body is given by the formula:
\(\text{KE} = \frac{1}{2}mv^2\)
2. **Relating Momentum and Kinetic Energy**: We can express kinetic energy in terms of momentum. By substituting \(v = \frac{p}{m}\) (from the momentum formula) into the kinetic energy equation, we have:
\(\text{KE} = \frac{1}{2}m\left(\frac{p}{m}\right)^2 = \frac{p^2}{2m}\)
3. **Calculate the Increase in Kinetic Energy**: Given that the momentum of the body increases by 20%, the new momentum is:
\(p_{\text{new}} = p + 0.2p = 1.2p\)
The new kinetic energy based on the increased momentum is:
\(\text{KE}_{\text{new}} = \frac{{(1.2p)}^2}{2m} = \frac{1.44p^2}{2m}\)
4. **Determine the Percentage Increase in Kinetic Energy**: The initial kinetic energy was \(\frac{p^2}{2m}\). The percentage increase in kinetic energy can be calculated as:
\(\text{Percentage Increase} = \left(\frac{\text{KE}_{\text{new}} - \text{KE}}{\text{KE}}\right) \times 100%\)
\(= \left(\frac{\frac{1.44p^2}{2m} - \frac{p^2}{2m}}{\frac{p^2}{2m}}\right) \times 100%\)
\(= \left(\frac{1.44p^2 - p^2}{p^2}\right) \times 100%\)
\(= \left(0.44\right) \times 100%\)
\(= 44%\)
Hence, when the momentum of a body is increased by 20%, its kinetic energy increases by 44%.
\(K=\frac{p^2}{2m}\)
\(K=\frac{(1.2p)^2}{2m}\)
\(⇒\frac{K−K}{K}=(1.2)^2−1=0.44\)
⇒ Kinetic energy increases by 44%
So, the correct option is (C): 44%
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,

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 : \(A_2 \rightleftharpoons 2A\)
Kinetic energy of an object is the measure of the work it does as a result of its motion. Kinetic energy is the type of energy that an object or particle has as a result of its movement. When an object is subjected to a net force, it accelerates and gains kinetic energy as a result. Kinetic energy is a property of a moving object or particle defined by both its mass and its velocity. Any combination of motions is possible, including translation (moving along a route from one spot to another), rotation around an axis, vibration, and any combination of motions.
