Physics

V = 10 $\left(1-e^{-t/RC}\right)$  

2 = 20 $\left(1-e^{-t/RC}\right)$  

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

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

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

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

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

In SHM sum of kinetic and potential energy will be constant and average kinetic energy & average potential energy in one time will be remains same.

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.

Heisenberg uncertainty principle :

$\begin{array}{l}\Rightarrow \Delta x\Delta p\ge ?\\ \Rightarrow \Delta x\left(m\Delta v\right)\ge ?\end{array}$

$\Rightarrow \Delta x\left(m\sqrt[]{\frac{3RT}{m}}\right)\ge ?$

$\Rightarrow \Delta x\propto \frac{1}{\sqrt[]{m}}$

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

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

Acceleration due to gravity at r distance above the surface = $\frac{GM}{{\left(R+r\right)}^2}$  

Acceleration due to gravity at r distance below the surface = $\frac{GM}{R^3}\left(R-r\right)$

So, ratio = $\frac{\left(R-r\right){\left(R+r\right)}^2}{R^3}=\frac{\left(R-r\right)\left(R^2+r^2+2Rr\right)}{R^3}=1+\frac{r}{R}-\frac{r^2}{R^2}-\frac{r^3}{R^3}$