Class 12th

Force on q₁ :

$F_{q_1}=q\left(V*B\right)=4\pi \left(0.5c\widehat{i}\right)*\frac{B_0}{\sqrt[]{2}}\left(cos\left(kz-\omega t\right)\widehat{i}+cos\left(kz-wt\right)\widehat{j}\right)$

$=4\pi \left(0.5c\right)\left(\frac{B_0}{\sqrt[]{2}}\right)cos\left(k\frac{\pi }{k}\right)\widehat{k}$      

For on q₂:

= $2\pi \left(0.5c\right)\left(\frac{B_0}{\sqrt[]{2}}\right)cos\left(k\frac{3\pi }{k}\right)\widehat{k}$

So ratio will be 2 : 1

$?a_r=e^{i\frac{2r\pi }{9}}$

$\therefore a_1,a_2,a_3,...........$ are in G.P.

$\left|\begin{array}{ccc}\hfill a_1\hfill & \hfill a_{1}r\hfill & \hfill a_1{r}^2\hfill \\ \hfill a_1{r}^3\hfill & \hfill a_1{r}^4\hfill & \hfill a_1{r}^5\hfill \\ \hfill a_1{r}^6\hfill & \hfill a_1{r}^7\hfill & \hfill a_1{r}^8\hfill \end{array}\right|={a_1}^3{r}^9\left|\begin{array}{ccc}\hfill 1\hfill & \hfill r\hfill & \hfill r^2\hfill \\ \hfill 1\hfill & \hfill r\hfill & \hfill r^2\hfill \\ \hfill 1\hfill & \hfill r\hfill & \hfill r^2\hfill \end{array}\right|=0$

$\left(i\right)a_2{a}_6-a_4{a}_8=a_{1}r.a_1{r}^5-a_1{r}^6{a}_1{r}^7={a_1}^2{r}^3-{a_1}^2{r}^{10}\ne 0$

$\left(ii\right)a_9=a_1{r}^8\ne 0$

$\left(iii\right)a_1.a_1{r}^8-a_1{r}^2.a_1{r}^6=0$

$\left(iv\right)a_5\ne 0$

Magnetization (M) is directly proportional to magnetizing field and magnetic susceptibility does not depend on temperature so option 3 is correct.

Using truth table

A            B            V₀

0            0

0            1            1

1            0            1

1            1

V = 10 $\left(1-e^{-t/RC}\right)$  

2 = 20 $\left(1-e^{-t/RC}\right)$  

$\frac{1}{10}=1-e^{-t/RC}$

$e^{t/RC}=\frac{10}{9}$

$\frac{t}{RC}=ln\left(\frac{10}{9}\right)=0.105$

$C=\frac{t}{R*0.105}=\frac{10^{-6}}{10*0.105}=0.95\mu F$              

$\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi co{s}^{4}x\right)}{x^4}=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi -\pi co{s}^{4}x\right)}{x^4}=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi si{n}^{2}x\left(1+co{s}^{2}x\right)\right)}{x^4}$

$=\underset{x\to 0}{lim}\frac{si{n}^2\left(\pi si{n}^{2}x\left(1+co{s}^{2}x\right)\right)}{{\pi }^{2}si{n}^{4}x{\left(1+co{s}^{2}x\right)}^2}*\frac{{\pi }^{2}si{n}^{4}x{\left(1+co{s}^{2}x\right)}^2}{x^4}=4{\pi }^2\underset{x\to 0}{lim}\frac{si{n}^{4}x}{x^4}=4{\pi }^2$

Since $5\Omega$ is connected across conductor so we can remove it.

$R_{eq}=1\Omega$

As observer is at O So height of water observed by observer

  $\Rightarrow \frac{H}{{\mu }_{\omega }}=\frac{H}{\left(4/3\right)}=\frac{3H}{4}$

given diagram (17.5 - H) is height of observer

$So\frac{3H}{4}=17.5-H$          

$\frac{7H}{4}=17.5$            

7H = 70

H = 10

Thermal resistance of spherical shell = $\frac{r_2-r_1}{4\pi K{r}_1{r}_2}$  

Rate of heat flow = $\frac{\Delta T}{R}=\frac{\left({\theta }_2-{\theta }_1\right)}{\left(\frac{r_2-r_1}{4\pi K{r}_1{r}_2}\right)}=\frac{4\pi K{r}_1{r}_2\left({\theta }_2-{\theta }_1\right)}{\left(r_2-r_1\right)}$