Physics

(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  $\frac{1}{4}={\mu }^2-1$

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

Uniform circular motion has constant speed. But SHM requires velocity and acceleration to vary sinusoidally with displacement. SHM is the projection of uniform circular motion. But it's not the motion itself, due to differing energy and force characteristics.

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.

$A_{max}=A_c+A_m$

$A_{min}=A_c-A_m$

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

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

By work energy theorem

Work done = change in K.E.

Work done by friction work done by spring

$=0-\frac{1}{2}m{V}^2$              

As 90% of K.E. is losed by friction so that

$-\frac{90}{100}\left(\frac{1}{2}m{V}^2\right)-\frac{1}{2}K{x}^2=-\frac{1}{2}m{V}^2$                

-K -> -16 * 10⁵

K = 16 * 10⁵