Physics

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- the debroglie wavelength of the particle can be varying cyclically between two values ${\lambda }_1$ and ${\lambda }_2$ , if particle is moving in an elliptical orbit with origin as its focus.

Let v₁, v₂, be the speed of particle at A and B respectively and origin is at focus O. If ${\lambda }_1$ and ${\lambda }_2$ are the de-Broglie wavelengths associated with particle while moving at A and B

respectively. Then,

${\lambda }_1$ =h/mv₁

${\lambda }_2$  =h/mv₂

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{v_2}{v_1}$

${\lambda }_1$ > ${\lambda }_2$

So v₂>v₁

By law of conservation of angular momentum, the particle moves faster when it is closer to

focus.

From figure, we note that origin O is closed to P than A

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- energy spent to convert into water = mass $*$ latent heat

= mL= 1000g $*$ 80cal/g

= 80000cal

Energy of phtons used= nT $*$ E=nT

So nTh $*v$ =mL

T= mL/nhv

T $\propto$ 1/n where v is constant

T $\propto 1/v$ when n is fixed

T $\propto$ 1/nv

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- ${\lambda }_e=\frac{h}{m_e{v}_e}$ = $\frac{h}{m_e\left(\frac{c}{100}\right)}=\frac{100h}{m_{e}c}$

E_e=1/2m_ev_e²

M_ev_e= $\sqrt[]{2E_e{m}_e}$

${\lambda }_e=\frac{h}{m_e{v}_e}=\frac{h}{\sqrt[]{2m_e{E}_e}}$

E_e= h²/2 $\lambda$ _e²m_e

For photon

E_p= hc/ ${\lambda }_p$ = hc/2 ${\lambda }_e$

$\frac{E_e}{E_p}=\frac{hc}{2{\lambda }_e}*\frac{2{\lambda }_e^2{m}_e}{h^2}$ = 100

$\frac{E_e}{E_p}=\frac{1}{100}={10}^{-2}$

P_e=m_ev_e= m_{e $*\frac{c}{100}$}

$\frac{P_e}{m_{e}c}=\frac{1}{100}={10}^{-2}$

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

Any ray entering at an angle I shall be guided along AC if the angle ray makes with the face AC (φ) is greater than the critical angle as per the principle of total internal reflection φ +r =900, therefore sinφ = cosr

Sinφ> $\frac{1}{\mu }$

Cosr> $\frac{1}{\mu }$

1-cos²r<1-1/ ${\mu }^2$

Sin²r<1-1/ ${\mu }^2$

Sin²r<1-1/ ${\mu }^2$

Sini = $\mu$ sinr

I= $\frac{\pi }{2}$

If that is greater than the critical then all other angle of incidence shall be more than the critical angle.

1< ${\mu }^2$ -1

$\mu$ > $\text{exception 25:}$

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

(i) As we know that power P_f=1/f=1/0.1+1/0.02=60D

By corrective lens the object distance at far pointP'_f=1/f'=1$\infty +$/1/0.02=50D

Total power P'_f=P_f+P_g

P_g=-10D

(ii) This power of accommodation is 4D for the normal eye then

4=P_n-P_f where P_n power of near point

So P_n= 64D

1/x_n+1/0.002=64 then x_n= 1/14=0.07m

(iii) P_n'=P_f'+4=54

After solving x_n'=4=0.25m

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

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

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