Physics

Suppose acceleration of wedge is a and acceleration of block w. r.t. wedge is a₁ then N cos60° = Ma = 16 a -> N = 32 a

For block w.r.t. wedge

N + 8a sin 30° = 8g cos 30°

N = 8g cos 30° - 8a sin 30°

->32a = 8g cos 30° - 8a sin 30°

->a = $\frac{\sqrt[]{3}}{9}g$  

Now for 8 kg,

$8gsin30°+8acos30°=m{a}_1$  

$a_1=g*\frac{1}{2}+\frac{\sqrt[]{3}}{9}g*\frac{\sqrt[]{3}}{2}$  

$=\frac{2}{3}g$  

The Schrodinger wave equation is the foundation of quantum mechanics. Without it, we couldn't understand how atoms work or predict the behaviour of electrons in materials.

De Broglie wavelength

$\lambda =\frac{h}{mV}=\frac{h}{\sqrt[]{2mE}}E\to K.E.$              

$E=\frac{3}{2}$  KT for gas

$So\lambda =\frac{h}{\sqrt[]{3mKT}}=\frac{6.6*10^{-34}}{\sqrt[]{3*9*10^{-31}*1.38*10^{-23}*300}}$

$\begin{array}{l}\lambda =6.26*10^{-9}m\\ \lambda =6.26nm\end{array}$                

De Broglie wavelength

$\lambda =\frac{h}{mV}=\frac{h}{\sqrt[]{2mE}}E\to K.E.$              

$E=\frac{3}{2}$  KT for gas

$So\lambda =\frac{h}{\sqrt[]{3mKT}}=\frac{6.6*10^{-34}}{\sqrt[]{3*9*10^{-31}*1.38*10^{-23}*300}}$

$\begin{array}{l}\lambda =6.26*10^{-9}m\\ \lambda =6.26nm\end{array}$                

For circular motion F_{net   $=\frac{M{V}^2}{R}$}

_{$\sqrt[]{2F}+F\text{'}=\frac{M{V}^2}{R}$}

$\sqrt[]{2}\frac{GMM}{{\left(\sqrt[]{2}R\right)}^2}+\frac{GMM}{{\left(2R\right)}^2}=\frac{M{V}^2}{R}$                

$\frac{GM}{R}\left[\frac{1}{\sqrt[]{2}}+\frac{1}{4}\right]=V^2$              

$V=\frac{1}{2}\sqrt[]{\frac{GM}{R}\left(2\sqrt[]{2}+1\right)}$

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.

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

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

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

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