Class 12th

$\frac{N{}_p}{N_s}=\frac{V_p}{V_s}\Rightarrow N_p=\frac{V_p}{V_s}*N_s=\frac{220}{12}*24=440$

For line of intersection of two planes

put z = $\lambda$ then

$x+2y=6-\lambda \&y=4-2\lambda \Rightarrow x+2\left(4-2\lambda \right)=6-\lambda$           

-> $x=3\lambda -2$

Now $\overrightarrow{a}.\overrightarrow{PQ}=0gives9\lambda -15+4\lambda -4+\lambda -1=0$  

$So,P\left(\frac{16}{7},\frac{8}{7},\frac{10}{7}\right)=P\left(\alpha ,\beta ,\gamma \right)$

$\Rightarrow 21\left(\alpha +\beta +\gamma \right)=21*\frac{34}{7}=102$           

$v=\frac{p}{m}\Rightarrow v_p:v_d:v_{\alpha }=1:\frac{1}{2}:\frac{1}{4}=4:2:1$

$F_{mag}=qvB\Rightarrow F_P:F_d:F_{\alpha }=1*4:1*2:2*1=2:1:1$

In the medical entrance test NEET, this chapter has a moderate weightage. You can expect around 2-3 questions of this chapter that contributes to the 3-5% of the total marks in the Physics section.

Given x + 2y – 3z = a

2x + 6y – 11z = b

x – 2y + 7z = c

Here    $\Delta =\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 2\hfill & \hfill 6\hfill & \hfill -11\hfill \\ \hfill 1\hfill & \hfill -2\hfill & \hfill 7\hfill \end{array}\right|\begin{array}{cc}\hfill \begin{array}{l}=\left(42-22\right)-2\left(14+11\right)-3\left(-4-6\right)\\ \end{array}\hfill & \hfill =20-50+30=0\hfill \end{array}$

For infinite solution 

20a – 8b – 4c = 0 Þ 5a = 2b + c

Class 12 physics chapter 1 Electric Charges and Fields is an important chapter for the CBSE Board exam. In unit 1 of the CBSE Board exam of class 12 Physics, there are questions from three chapters including - Electric Charges and Fields, Electrostatics, and Electrostatic Potential and Capacitance. This unit carries a total 16 marks. 

Given $\frac{dy}{dx}=\frac{x{y}^2+y}{x}=y^2+\frac{y}{x}$  

OR   $\frac{dy}{dx}-\frac{y}{x}=y^{2}OR\frac{1}{y^2}\frac{dy}{dx}-\frac{1}{x}.\frac{1}{y}=1.........\left(i\right)$          

->   Since curve intersect x + 2y = 4 at x = -2 then y = 3 so

From (ii) $\frac{-2}{3}=-2+cORc=2-\frac{2}{3}=\frac{4}{3}$  

put x = 3, then $\frac{3}{y}=\frac{-9}{2}+\frac{4}{3}=\frac{-19}{6}$  

$\Rightarrow y=\frac{-18}{19}$  

Magnetic field on the axis of a circular coil at distance x from centre, B = $\frac{{\mu }_{0}Ni{r}^2}{2{\left(r^2+x^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

$\frac{{\left(r^2+{\left(0.2\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\left(r^2+{\left(0.05\right)}^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}=8$ $\frac{r^2+{\left(0.2\right)}^2}{r^2+{\left(0.05\right)}^2}=4\Rightarrow r^2=0.01\Rightarrow r=0.1m.$  

Number of half lives of Y = 3

Number of half lives of X = 6 [As half life of X is half of that of Y].

$\frac{N_1}{2^6}=\frac{N_2}{2^3}\Rightarrow \frac{N_1}{N_2}=8$

Given f (k) = $\{\begin{array}{l}k+1,kisodd\\ k,kiseven\end{array}$

  $?g:A\to A$           such that g (f (x) = f (x)

Case I : If x is even then g (x) = x . (i)

Case II : If x is odd then g (x + 1) = x + 1 . (ii)

From (i) & (ii), g (x) = x, when x is even

So total no. of functions = 10⁵ * 1 = 10⁵