Physics Ncert Solutions Class 12th

${\lambda }_{air}=\frac{c}{f_c}$

Fermi level of p-type semiconductors will go downward

Using truth table of logic gates

$Y=\overline{\left(A\overline{B}\right)+\left(\overline{A}B\right)}=\overline{\left(\left(A\overline{B}\right)\right)}\overline{\left(\left(\overline{A}B\right)\right)}$

$Y=\left(\overline{A}+\overline{B}\right)\left(\overline{A}+\overline{B}\right)=\left(\overline{A}+B\right)\left(A+\overline{B}\right)=AB+\overline{A}\overline{B}$

This is an interesting question! Even though a proton has a positive charge, the net positive charge in a conducting material is always due to the removal of free electrons. 

This happens due to the availability of only free electrons in all conductors. Since protons are present in the nucleus, they can not roam freely in the conductor, as electrons travel freely in the conductor due to their presence in the outer shell.

$Y_1=\overline{A·A}$

$=\overline{A}$

$Y_2=\overline{B+B}$

$=\overline{B}$

$Y=\overline{Y_1+Y_2}$

$=\overline{\overline{A}+\overline{B}}$

$=\overline{\overline{A}}·\overline{\overline{B}}$

$=A·B is similar to output of AND Gate$

$T^{\prime }=2\pi \sqrt[]{\frac{{?}^{\prime }}{g}}$ where $t^{\prime }=\frac{\mathcal{l}}{2}$

$T=2\pi \sqrt[]{\frac{t}{g}}$

$T^{\prime }=\frac{x}{2}T$

$2\pi \sqrt[]{\frac{t}{2g}}=\frac{x}{2}2\pi \sqrt[]{\frac{t}{g}}$

$\frac{1}{\sqrt[]{2}}=\frac{x}{2}\Rightarrow x=\sqrt[]{2}$

According to transformer ratio,

$\frac{V_S}{V_P}=\frac{N_S}{N_P}=2:1$

A particle moving with uniform speed in a circular path maintains varying velocity and varying acceleration. It is because direction of both velocity as well as acceleration will change continuously.

$A_{max}=A_c+A_m$

$A_{min}=A_c-A_m$

$\frac{A_{min}}{A_{max}}=\frac{A_c-A_m}{A_c+A_m}=\frac{250-150}{250+150}$

$\Rightarrow \frac{100}{400}=\frac{50}{200}$  

$l=\frac{120-60}{4000}=0.015A$

Thus l₂ = l - L_L

= 0.015 – 0.006

= 0.009 = 9mA