Class 11th

Internal forces in an isolated system do not affect the total momentum. But do note that the mutual forces between pairs of particles in the system can cause individual particles to change their momentum. Now, these internal forces are always equal and opposite, as you can recall from Newton's Third Law. 
Due to that, the individual momentum changes cancel out in pairs. What happens is that the total momentum of the system remains unchanged. That further allows the Second Law of Motion to be applied to a body or a system of particles. The internal forces sum to a force that is mathematically nulled out. 

An isolated system is one that has no external force acting on it. This means that for the total momentum to remain unchanged, there must be no net force originating from outside the system. This net external force should not be able to influence the motion of the isolated system, as per the law of conservation of momentum.

Case – I : When disk slides down

$t_1=\sqrt[]{\frac{2L}{gsin\theta }}..........\left(i\right)$              

Case – II : When disk rolls down

$t_2=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+{\beta }^2}}}=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+{\left(\frac{k}{R}\right)}^2}}}=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+{\left(\frac{R/\sqrt[]{2}}{R}\right)}^2}}}=\sqrt[]{\frac{2L}{\frac{gsin\theta }{1+\frac{1}{2}}}}=\sqrt[]{\frac{4L}{3gsin\theta }}......\left(ii\right)$             

$\Rightarrow \frac{t_1}{t_2}=\sqrt[]{\frac{2L}{gsin\theta }}*\sqrt[]{\frac{3gsin\theta }{4L}}=\sqrt[]{\frac{3}{2}}$              

Using conservation of linear momentum, we can write

$mu=M{v}_1\Rightarrow v_1=\frac{mu}{M}........\left(1\right)$           

Using conservation of angular momentum about centre of mass of Rod, we can write

$v_1+\frac{\mathcal{l}\omega }{2}=u\Rightarrow \frac{mu}{M}+\frac{3mu}{M}=u\Rightarrow \frac{m}{M}=\frac{1}{4}$     . (2)

Using definition of e, we can write

$v_1+\frac{\mathcal{l}\omega }{2}=u\Rightarrow \frac{mu}{M}+\frac{3mu}{M}=u\Rightarrow \frac{m}{M}=\frac{1}{4}$  

Initially wall will act as observer, so with the help of Doppler's effect, we can write

$f\text{'}=\frac{c}{c-v_0}{f}_0$  

Now wall will act as source of sound of frequency f', so With the help of Doppler's effect, we can write

$f\text{'}\text{'}=\frac{c+v_0}{c}{f}^{\text{'}}=\frac{c+v_0}{c-v_0}{f}_0$              

$\Rightarrow 500=\left(\frac{c+v_0}{c-v_0}\right)400\Rightarrow 5c-5v_0=4c+4v_0\Rightarrow v_0=\frac{c}{9}=\frac{330}{9}m/s=\frac{330}{9}*\frac{18}{5}km/h=132km/h$

Let ball starts its motion with horizontal velocity v₀, so with the help of conservation of mechanical energy, we can write

$\frac{1}{2}m{v}_0^2=\frac{1}{2}m{x}^2\Rightarrow v_0=x\sqrt[]{\frac{k}{m}}=0.05*\sqrt[]{\frac{100}{0.1}}=0.5*\sqrt[]{10}m/s$              

t = Time required to fall the ball =    $\sqrt[]{\frac{2h}{g}}=2*\sqrt[]{\frac{1}{10}}s$

^{$=d=v_0*t=0.5*\sqrt[]{10}*2*\sqrt[]{\frac{1}{10}}=1m$}              

Heat absorbed in cyclic process = Work done = 100? Joule

The equation of wave at any time t will be $y=\frac{1}{1+{\left(x?vt\right)}^2},$ so v * 1 = 2 -> v = 2m/s