Class 11th

$COP=\frac{Q_t}{W}=\frac{T_{low}}{T_{high}-T_{low}}=\frac{250}{300-250}=5$

$\frac{Q_L}{W}=5$

$W=\frac{500*4.2}{5}420J$

Elongation in the wire is given as

$?I=\frac{Fl}{AY}$

$T_{max}=\frac{m{v}^2}{l}+mg$

$T_{max}=3mg$

$?l_{max}=\frac{3mgl}{\pi r^{2}Y}$

From equation of uniformly accelerated motion

$v=u+at$

$7.5=15+5a$

$a=-1.5m/s^2$

$nowF=ma$

$\left|\stackrel{?}{F}\right|=120*1.5$

$=180N$

In NEET exam, the weightage of Equilibrium is around 3% to 6% and in JEE Main exam, the weightage is 4% to 6.6%.

There are two conditions needed for equilibrium. The net force and net torque acting upon an object must be zero.

There are four types of equilibrium constants: Formation constants, stability constants, association constants, and binding constants.

It consists of concepts such as Le Chatelier's Principle, equilibrium constant, and factors affecting equilibrium. It is not considered an easy chapter.

A reversible reaction where rate of the forward reaction equals to the rate of the backward reaction is called the equilibrium. It results in constant reactants and products concentrations over time.