Physics Ncert Solutions Class 12th

Using truth table

A            B            V₀

0            0

0            1            1

1            0            1

1            1

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

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

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

From observation, we can say that right of $-5\mu c$ charge, net electric force (or electric field) can be zero.

$So,\frac{k\left(5\mu c\right)q}{x^2}=\frac{k\left(20\mu c\right)q}{{\left(5+x\right)}^2}$

$\Rightarrow \frac{1}{x^2}=\frac{4}{{\left(5+x\right)}^2}$

$\Rightarrow \frac{5+x}{x}=2$

x = 5 cm

 0.02i * R_G = 0.98i * 5

[in parallel, pot. Diff is same]

or, R_G = 49 * 5

R_G = 245 $\Omega$

$\eta =\frac{1}{4}or25\mathrm{\%}$

$\Delta T_{sink}=-58°C\left(iedecreased\right)$

$\eta \text{'}=\frac{1}{2}or50\mathrm{\%}$

T_{sin k} =?

As we know, the efficiency for a reversible engine is given by :

$\eta =1-\frac{T_2}{T_1}\left[T_2\to sinktemperature\right]$

On comparing both, we get

$T_2=\frac{3}{4}{T}_1\&\left(T_2-58°C\right)=\frac{T_1}{2}$

$\Rightarrow T_2=174°C$

Histamine stimulate the secretion of HCl

$dM=\sqrt[]{2R{h}_T}+\sqrt[]{2R{h}_R}$

$?h_T+h_R=160$

[Given]

So, hR = 160 - hT

$d_M=\sqrt[]{2R{h}_T}+\sqrt[]{2R\left(160-h_T\right)}$

$\therefore d_M=\sqrt[]{2*6400000*80}+\sqrt[]{2*6400000*80}$

$=2*4*8*10^3$

dM = 64 km

(a) $Calamine:ZnC{O}_3$  

(b) Malachite : $CuC{O}_3.Cu{\left(OH\right)}_2$  

(c) Siderite : FeCO₃

(d) Sphalerite : ZnS

Maximum current through zener diode,

  $i=\frac{2}{10}=0.2A$             

When unregulated voltage is 14V,

Potential difference across Rs,

              V_S = 14 – 10 = 4V

$R_S=\frac{V_S}{i}=\frac{4}{0.2}=20\Omega$