Class 12th

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

${\left(t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}_x={\left(t_{mean}\right)}_y$
$\frac{\mathcal{l}n2}{{\lambda }_x}=\frac{1}{{\lambda }_y}$

${\lambda }_x=\left(\mathcal{l}n2\right){\lambda }_y$              

Given N_x = N_y = N₀

So activity A = $\lambda N$

$As{\lambda }_x<{\lambda }_y\Rightarrow A_x<A_y$           

So y will decay faster than x.

$f\left(x\right)=\left|x^2-2x-3\right|e^{\left|9x^2-12x+4\right|}$

$\Rightarrow f\left(x\right)=\{\begin{array}{l}\left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},x<-1\\ -\left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},-1\le x<3\\ \left(x^2-2x-3\right)e^{{\left(3x-2\right)}^2},x\ge 3\end{array}$

$f\text{'}\left(-1^{-}\right)\ne f\left(-1^{+}\right),$

$f\text{'}\left(3^{-}\right)\ne f\left(3^{+}\right)$

Number of non differential points is 2 at x = -1, 3.

$B_1=\frac{{\mu }_{0}l}{4\pi y}\left[sin{\theta }_1+sin90°\right]-------\left(i\right)$

B due to OX wire

$B_2=\frac{{\mu }_{0}l}{4\pi x}\left[sin{\theta }_2+sin90°\right]$       

As in the diagram direction of B₁ and B₂ are in downward direction

So B = B₁ + B₂

$B=\frac{{\mu }_{0}l}{4\pi y}\left(sin{\theta }_1+1\right)+\frac{{\mu }_{0}l}{4\pi x}\left(sin{\theta }_2+1\right)$

$B=\frac{{\mu }_{0}l}{4\pi xy}\left[\left(x+y\right)+\sqrt[]{x^2+y^2}\right]$  

In give diagram Diode 1 and 2 are in forward bias with R = $30\Omega$ and Diode 3 is reverse bias with R = infinite l₁ current is flowing through 20 $\Omega$  

So $\frac{l_1}{2}and\frac{l_1}{2}$ current will flow through Diode 1 and 2. As resistance is same applying KCl in ABCD Loop $-\frac{l_1}{2}*130-\frac{l_1}{2}*130-l_1*20+200=0$

-100l₁ + 200 = 0

l₁ = 2

$\left(p\wedge \sim q\right)\to \left(p\vee q\right)$ is tautology

$*=\wedge ,?=\vee$