Class 12th

we know,   $C=\frac{1}{\sqrt[]{{\mu }_0\in o}}=3*10^8,Given,{\in }_r=1$

$v=\frac{1}{\sqrt[]{{\mu }_0{\in }_r{\in }_o{\mu }_r}}=2*10^8$            

$\Rightarrow \frac{3*10^8}{2*10^8}=\sqrt[]{{\in }_r{\mu }_r}$               

$={\left(1.5\right)}^2={\in }_r{\mu }_r$              

${\mu }_r=2.25$               

Number of solution of the equation |cos x| = sin x for $x\in \left[-4\pi ,4\pi \right]$ $$ will be equal to 4 times the number of solutions of the same equation for $x\in \left[0,2\pi \right]$ $$

Graphs of y = |cos x| and y = sin x are as shown below.

Hence, two solutions of given equation in [0, 2]

$$ $\Rightarrow$  total of 8 solutions in [4, 4]

V = 210 sin 3000 t

$\omega$ = 3000

$X_L=\omega L=30$                

$X_C=\frac{1}{\omega c}=\frac{40}{3}$                

$tan?=\frac{V_L-V_C}{V_R}=\frac{X_L-X_C}{R}$                

$\begin{array}{l}tan?=0.17\\ ?=ta{n}^{-1}\left(0.17\right)\end{array}$                

For $x^2+\alpha x=\beta >0\forall x\in R$  to hold, we should have   ${\alpha }^2-4\beta <0$

If $\alpha =1,\beta$  can be 1, 2, 3, 4, 5, 6 i.e., 6 choices

If a  = 4, b can be 5 or 6 i.e., 2 choices

hence total favourable outcomes

= 6 + 5 + 4 + 2 + 0 + 0 = 17

Required probability =  $\frac{17}{36}$

If n is number of trials, p is probability of success and q is probability of unsuccess then,

Mean = np and variance = npq.

Here np + npq = 24         …. (i)

np. npq = 128                   …. (ii)

and q = 1 – p                    …… (iii)

From eq. (i), (ii) and (iii) :

$=\left(32+\frac{32*31}{2}\right).\frac{1}{2^{32}}$      

$=\frac{33}{2^{28}}$

  $B=\eta {\mu }_{0}l$

 if l ->2l

$\eta \to \raisebox{1ex}{$n$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$                

$B^1=\frac{n}{2}{\mu }_{0}2l$

$B^1=B$                

  $?\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=0$ …….(i)

then

$\overrightarrow{a}+\overrightarrow{c}=-\overrightarrow{b}$

then

$\left(\overrightarrow{a}+\overrightarrow{c}\right)*\overrightarrow{b}=-\overrightarrow{b}*\overline{b}$

$\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}$  $\therefore \overrightarrow{a}*\overrightarrow{b}+\overrightarrow{c}*\overrightarrow{b}=\overrightarrow{0}$ …….(ii)

Now (S1) : $\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\stackrel{}{}\text{exception 25:}$

$\left|\overrightarrow{a}*\overrightarrow{b}+\overrightarrow{c}*\overrightarrow{b}\right|-\left|\overrightarrow{c}\right|=6\left(2\sqrt[]{2}-1\right)$

$\therefore cos\left(\angle ACB\right)=\sqrt[]{\frac{2}{3}}$

$\therefore \angle ACB=co{s}^{-1}\sqrt[]{\frac{2}{3}}$

$$ $\therefore S\left(2\right)$  is correct

Kindly go through the solution

 $\begin{array}{l}T\text{?}\to Absolutetemperature\\ x\to susceptibilities\end{array}\}x\propto \frac{1}{T}$

Let a, b, c be direction ratios of plane containing lines

$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$

and

$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$

Equation of plane P is : 1 (x – 3) 1 (y + 4) + 2 (z – 7) = 0

$\Rightarrow x-y+2z-21=0$

Distance from point (2, 5, 11) is

$d=\frac{\left|2+5+22-2\right|}{\sqrt[]{6}}$

$\therefore d^2=\frac{32}{3}$

Given

E₁ = E₂ = E

$I=\frac{2E}{R=r_1+r_2}$   - (i)

Potential drop across second cell is

  $V_A-E_2+I{r}_2=v_B$

According to question V_A – V_B = 0

E₂ -lr₂ = 0

$\Rightarrow R=r_2-r_1$