Class 12th

This is a long answer type question as classified in NCERT Exemplar

Let d be the diameter of the disc. The spot shall be invisible if the incident rays from the dot at O to the surface at d / 2 at the critical angle.

Let I be the angle of incidence. Using relationship between refractive index and critical angle,

SinC= 1/ $\mu$

$\frac{d/2}{h}$ = tani

$\frac{d}{2}=htani$

D= $\frac{2h}{\text{exception 25:}}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- as debroglie wavelength is $\lambda =\frac{h}{mv}=\frac{h}{p}$

P=h/ $\lambda$

P₁/p₂= ${\lambda }_2/{\lambda }_1$

If wavelength are equal then ratio is 1:1 so P₁=P₂

E= 1/2mv²= p²/2m

E inversely proportional to m

So $\frac{E_1}{E_2}$ = $\frac{m_2}{m_1}$ <1

E₁2

This is a multiple choice answer as classied in NCERT Exemplar

momentum per unit time per unit area = intensity/ speed of wave

 = I/c= radiation pressure (p)

Momentum is always double when a light gets reflected back as in that case the momentum which is positive to one side added to momentum which is negative to other side so momentum is always double

So it becomes 2I/c

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-(c,d)

Explanation- $\lambda =\frac{h}{mv}$

V= h/m $\mathit{\lambda }$

V1= $\lambda =10nm=10*{10}^{-9}m={10}^{-8}$

$\frac{6.6*{10}^{-34}}{9*{10}^{-31}*{10}^{-8}}={10}^{5}m/s$

V1= $\lambda =1/10nm={10}^{-10}$

$\frac{6.6*{10}^{-34}}{9*{10}^{-31}*{10}^{-10}}={10}^{7}m/s$

V1= $\lambda =1/10000nm={10}^{-13}$

$\frac{6.6*{10}^{-34}}{9*{10}^{-31}*{10}^{-13}}={10}^{10}m/s$

V1= $\lambda =1/1000000nm={10}^{-15}$

 

This is a Long Answers Type Questions as classified in NCERT Exemplar

Sol. 

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation- $\lambda =\frac{h}{mv}$

Force F= -eE=-e [E_oi]=eE_oi

So a=f/m=eE_oi/m

Velocity at xaxis=V_oi

Velocity at yaxis= eE_otj/m

Net velocity v= $\sqrt[]{v_x^2+v_y^2}=\sqrt[]{v_o^2+{\frac{\mathrm{e}\mathrm{E}\mathrm{o}\mathrm{t}}{\mathrm{m}}}^2}$

$\lambda =\frac{h}{mv}=h/m\sqrt[]{v_o^2+{\frac{\mathrm{e}\mathrm{E}\mathrm{o}\mathrm{t}}{\mathrm{m}}}^2}$

$\lambda ={\lambda }_o/\sqrt[]{1+\frac{e^2{E}_o^2{t}^2}{m^2{v}_o^2}}$

This is a Long Answers Type Questions as classified in NCERT Exemplar

Sol. 

(b, d) Here, in option (b) charge is moving in a circular orbit.

In circular motion, the direction of the motion of charge is changing continuously, thus it is an accelerated motion and this option is correct.

Also, we know that a charge starts accelerating when it falls in an electric field.

This is a Long Answers Type Questions as classified in NCERT Exemplar

Sol. 

$\begin{array}{l}LetP\left(n\right):{\left(AB\right)}^n=A^n{B}^n\\ Step1:Putn=1,P\left(1\right):AB=ABwhichistrueforn=1\\ Step2:Putn=k,P\left(k\right):{\left(AB\right)}^k=A^k{B}^{k}Letitbetrueforanyk\in N\\ Step3:Putn=k+1,P\left(k+1\right):{\left(AB\right)}^{k+1}=A^{k+1}{B}^{k+1}\\ L.H.S.{\left(AB\right)}^{k+1}={\left(AB\right)}^k.AB\\ =A^k{B}^k.AB\left[Fromstep2\right]\\ =A^{k+1}{A}^{k+1}R.H.S.\\ Hence,ifP\left(n\right)istrueforP\left(k\right)thenitistrueforP\left(k+1\right).\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(a, c, d) as we know that  $\lambda =\frac{c}{\upsilon }$ = $\frac{3*{10}^8}{{10}^9}$  = 0.3m and it is fall in region of radio wave.