Class 12th

The last sentence talks about investment in cryptocurrencies while the first 3 sentences talk about Dogecoin.

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

V = 10 $\left(1-e^{-t/RC}\right)$  

2 = 20 $\left(1-e^{-t/RC}\right)$  

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

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

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

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

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

${\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.

21 is a link here. As the 1^st sentence talks about the other 50% investment. No other option has this link.

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

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

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

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

3 can never start the sentence. 2 seems to be introducing a fact. 2413 makes a logical sequence.

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

While the first 3 sentences talk about lack of sleep and its repercussions, only the last sentence talks about cats and their sleeping time.