Physics

Action-reaction forces act on different objects. That's why they don't or cannot cancel out.

For instance, when you push a wall, you can observe two things. 

• Your hand pushes the wall (action)

• The wall pushes your hand (reaction)

These are equal and opposite. But a close scientific examination will tell you that they are indeed acting on different things. To find your hand's motion, only consider the force on your hand (wall pushing back). The force from your hand affects the wall's motion, which is not yours.

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.

$\mu =\frac{A_m}{A_c}=\frac{V_1-V_2}{V_1+V_2}=\frac{200-120}{200+120}=\frac{1}{4}\Rightarrow$ Modulation index

Here V_C (t) = 160 sin $\left(2\pi *10^{6}t\right)Vand{V}_m\left(t\right)=A_{m}sin\left(2\pi *10^{3}t\right)V,so$  

$A_m=\frac{A_c}{4}=\frac{160}{4}=40V$              

Case – I : When disk slides down

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

Case – II : When disk rolls down

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

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

Here $?=45°$ and current leads voltage, so $X_C>X_L$

$tan?=\frac{X}{R}=tan45°=\text{?}1\Rightarrow X=X_C-X_L=R$

$\Rightarrow X_C=R+L\omega \Rightarrow X_C=1+30*10^{-3}*300=10$

$\Rightarrow \frac{1}{C\omega }=10\Rightarrow C=\frac{1}{10*\omega }=\frac{1}{10*300}=\frac{1}{3}*10^{-3}F$  

For microscopic lens

$m=1+\frac{D}{f}\Rightarrow 6=1+\frac{25}{f}\Rightarrow f=5cm$

Now in second case, this lens acts as objective of compound microscope, so according to question

    $m=m_O*m_e=\frac{LD}{f_o{f}_e}\Rightarrow f_e=\frac{LD}{m{f}_o}=\frac{60*25}{12*5}=25cm$