Physics Ncert Solutions Class 12th

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

Explanation-

$?p=\frac{h}{?x}=\frac{h}{2\pi ?x}$

= $\frac{6.64*{10}^{-34}}{2*22/7*{10}^{-9}}=1.05*{10}^{-25}$ kgm/s

E= p²/2m= $\frac{{(1.05*{10}^{-25})}^2}{2*9.1*{10}^{-31}}$ J= 3.8 $*{10}^{-2}$ eV

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

Let us consider a portion of a ray between x and x+dx inside the liquid. Let the angle of incidence at x be θ and let it enter the thin column at height y. Because of the bending it shall emerge at x+dx  with an angle θ+dθ and at a height y+dy . From Snell's law,

$\mu ysin\theta =\mu (y+dy)sin(\theta +d\theta )$

$\mu ysin\theta \left(\mu y+\frac{d\mu }{dy}dy\right)sin\theta cosd\theta +cos\theta si$ nd $\theta$

$\mu ycos\theta d\theta$ $\frac{d\mu }{dy}dysin\theta$

d $\theta -\frac{d\mu }{dy}dytan\theta$

tan $\theta =\frac{dx}{dy}$

$\theta =-\frac{1d\mu d}{\mu dy}$

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

$\mu$ > $\text{exception 25:}$