Class 11th

According to modified Ampere's law

$?B.dI={\mu }_0\left(I_C+I_D\right)$

For Loop  $L_1{I}_C\ne 0$ and $I_D=0$

For Loop  $L_2{I}_C=0$ and $I_D\ne 0$

Due to  $KCL{I}_C=I_D$

$T=m\widehat{?}{\omega }^2$

$T^{\prime }=m:{(2\omega )}^2$

$T^{\prime }=4T$

(Material)                      (Susceptibility ( $\chi$  )

Diamagnetic                 (II) $0>\chi \ge -1$

Ferromagnetic  (III) $\chi ?1$

Paramagnetic                (IV) $0<\chi <\epsilon$

Non-magnetic               (I) $\chi =0$

Divided into 10 parts

$R=\frac{\rho l}{A}$

$R^{\prime }=\frac{\rho l}{10A}=\frac{R}{10}$

$R_S=5*\frac{R}{10} [series]$

$R_S=50$

$R_P=\frac{R}{50}\left[ parallel \right]$

$R_{eq }=R_S+R_P$

$=52\Omega$

Energy difference $\Delta E=\frac{hc}{\lambda }$

$\therefore \lambda \propto \frac{1}{\Delta E}$

${(\Delta E)}_{6-2}>{(\Delta E)}_{5-2}>{(\Delta E)}_{4-2}>{(\Delta E)}_{3-2}$

${\lambda }_{6-2}<{\lambda }_{5-2}<{\lambda }_{4-2}<{\lambda }_{3-2}$

$A-III,\; B-IV,\; C-II,\; D-I$

$Y_1=\overline{A·A}$

$=\overline{A}$

$Y_2=\overline{B+B}$

$=\overline{B}$

$Y=\overline{Y_1+Y_2}$

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

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

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

$$For SHM, kinetic and potential energies are equal when displacement $x=\pm \frac{A}{\sqrt{2}}$ , where A is the amplitude. At this point, total energy is evenly shared between motion and position-dependent restoring forces.

At central point on screen, path difference is zero for all wavelength. So, central bright fringe is white and other fringes depend on wavelength as $\beta =\frac{\lambda D}{d}$ .

Therefore, other fringes will be coloured.

In SHM, the restoring force, F is directly proportional to the negative of displacement. That is, F=- kx. This follows Hooke's law and ensures motion is oscillatory about the mean position with angular frequency,  

$\omega =\sqrt{\frac{k}{m}}$

$A=90^{?}$

In prism,  $r_1+c=A$

$r_1=90^{?}-c$                  …(i)

$sinc=\frac{1}{\mu }\Rightarrow cosc=\frac{\sqrt[]{{\mu }^2-1}}{\mu }$

$\Rightarrow$ Apply Snell's law, on incidence surface

$1·sin30^{?}=\mu sin\left(r_1\right)\Rightarrow 1*\frac{1}{2}=\mu *sin\left(90^{?}-c\right)$

$\frac{1}{2}=\mu *\frac{\sqrt[]{{\mu }^2-1}}{\mu }$

On squaring  $\frac{1}{4}={\mu }^2-1$

$\Rightarrow {\mu }^2=\frac{5}{4}\Rightarrow \mu =\frac{\sqrt[]{5}}{2}$