Physics

According to transformer ratio,

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

Excess force  $=T*2\pi R$

$=\frac{7}{100}*2*3.14*\frac{4.5}{100}$

$=197.82*10^{-4}$

$=19.8*10^{-3} N$

$=19.8mN$

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

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

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

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

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

$Y_1=\overline{A·A}$

$=\overline{A}$

$Y_2=\overline{B+B}$

$=\overline{B}$

$Y=\overline{Y_1+Y_2}$

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

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

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

(A) If $c$  is the velocity of light

so,   $E=h\nu$   (Energy of photon)

(B) Velocity of photon is equal to velocity of light i.e. c.

(C)  $\lambda =\frac{h}{p}$

$p=\frac{h}{\lambda }$

$p=\frac{hv}{c}$

(D) In photon-electron collision both total energy and total momentum are conserved.

$$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.

$V.C=MSD-VSD$     … (i)

given :  $\left(N+1\right)VSD=NMSD$

$VSD=\left(\frac{N}{N+1}\right)MSD$     … (ii)

From (1) and (2)

$V.C=\left(MSD\right)-\frac{N}{N+1}\left(MSD\right)$

$=MSD\left(1-\frac{N}{N+1}\right)=\frac{MSD}{N+1}$

$=\frac{0.01}{N+1}=\frac{1}{100\left(N+1\right)}$

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  $\frac{1}{4}={\mu }^2-1$

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