Physics

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- k'_max= 2k_max

K'_max= hc/ $\lambda -?$

2K_max= hc/ ${\lambda }^{\text{'}}-{?}_0$

2 (1230/600- $?$ )= (1230/400- $?$ )

So $?$ =1.02eV

This is a Short Answer Type Questions as classified in NCERT Exemplar

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- during photoelectric emission the momentum is transferred to metal. metal absorbs absorb the photon and its momentum is transferred, and excited electron emitted.

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