What is ultraviolet catastrophe in laymans terms? From what I understand, ultraviolet catastrophe means that at that time scientists believed that for energy to be higher the frequency should be higher. But frequency is higher in case of ultraviolet light but ultraviolet light has less heat. 
Does this mean that even though ultraviolet light is having greater frequency it has less energy?
Also since violet color has most energy ultraviolet light has less energy than violet even though ultraviolet is more hotter than normal heat?
 A: The idea is related to thermal equilibrium of a black body. The classical theory predicts that this type of bodyy will emit light on all frequencies, with energy density increasing with frequency $B_\nu \sim\nu^2$. 
And you can actually see this is not case (hence the catastrophe part), a furnace is a good example of a black body, the classical model will conclude that there'll be light with an arbitrarily high energy density coming out of the furnace, which is not observed in reality
A: 
From what I understand, ultraviolet catstrophe means that at that time scientists believed that for energy to be higher the frequency should be higher. But frequency is higher in case of ultraviolet light but ultraviolet light has less heat.

This is a wrong understanding. More energy means also utimately more heat, once the electromagnetic frequencies are absorbed.
This the catatastropy:

As the other answer explains, in classical electromagnetic theory the radiation (radiated power density) from a heated body could not be fitted. It resulted with a climbing frequency spectrum,( ~1/wavelength) whereas the quantized formula fitted the data with the Planck radiation formula. It was called a catastrophy because classically it went to a singularity, very large frequencies.

Does this mean that even though ultraviolet light is having greater frequency it has less energy?

No , the energy of a photon is h*nu, as discovered by Planck.

Also since violet color has most energy ultraviolet light has less energy than violet even though ultraviolet is more hotter than normal heat?

No .
Ultra violet, as the name says , is beyond violet in frequency, thus with higher energy.
A: I think I have the answer. 
An ideal black body at thermal equilibrium emits radiation in all frequency ranges. It emits more energy as the frequency increases. The higher the frequency the higher the energy. Since ultraviolet light has higher frequency it has more energy. By calculating the total amount of radiated energy, it can be shown that a blackbody would release an infinite amount of energy. The maximum wavelength emitted by a black body radiator is infinite.
Lets take an example of furnace here heat is provided to the furnace which is a blackbody, which means it absorbs all the radiation and reflects none. Now when we calculate the heat emitted we can say that the amount of heat emitted in the furnace is infinite. So since energy is infinite it violates the law of conservation of energy since input heat was not infinite but output emission was infinite. This is ultraviolet catastrophe.
I am politely asking whether my reasoning is correct?
A: Another way of explaining is this: The predictions suggested that an ideal black body at thermal equilibrium would emit energy proportional to the frequency squared. Now since an object if it is more hotter, it has more energy. since ultraviolet light has higher frequency it should be hotter(more energy) and an object burning with high heat should emit light at the ultraviolet end of the spectrum. This does not happen so this is ultraviolet catastrophe. Am I right?
