Class 12th

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

Explanation- in this type of situation centripetal force is balanced by electrostatic force

So mv²/r= -ke²/r²

According to bohr postuates

Mvr = h when n=1

On solving n  $\frac{h^2}{m^2{r}^2}\frac{1}{r}={ke}^2/r^2$

So after solving r= 0.51A⁰

So potential energy  $\frac{{ke}^2}{r}$ = -27.2eV , KE= mv²/2

= $\frac{1}{2}m\frac{h^2}{r^2}=\frac{h}{2m{r}^2}=+13.6eV$

But if R

If R>>r the electron moves inside the sphere with radius r'

Charge inside r'⁴=e $\frac{{r\text{'}}^3}{R^3}$

So r' = $\frac{h^2}{m}\frac{k}{e^2}\frac{R^3}{{r\text{'}}^3}$

So r'⁴= 0.51A⁰R³

= 510(A⁰)⁴

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

Explanation- total energy of electron

E= $\frac{\mu Z^2{e}^4}{8{\epsilon }_o^2{h}^2}\frac{1}{n^2}$

The frequency hv= $\frac{\mu e^4}{8{\epsilon }_o^2{h}^2}\left(1-\frac{1}{4}\right)=\frac{\mu e^4}{8{\epsilon }_o^2{h}^2}\frac{3}{4}$

$?\lambda$ = ${\lambda }_D-{\lambda }_H$

$100*\frac{?\lambda }{{\lambda }_H}$ = $\frac{{\lambda }_D-{\lambda }_H}{{\lambda }_H}*100$ = $\frac{{\mu }_D-{\mu }_H}{{\mu }_H}*100$

= $\frac{\frac{m_{e{M}_D}}{m_e+M_H}-\frac{m_{e{M}_D}}{m_e-M_H}}{\frac{m_{e{M}_D}}{m_e+M_H}}*100$

When m_e<H<

after solving we get 2.714 $*{10}^{-2}\%$

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

Explanation- as we know total energy in stationary orbit is

E_n=- $\frac{{-me}^4}{8n^2{{\epsilon }_0}^2{h}^2}$  where sign have usual meaning.

According to bohr third postulate h $\nu =E_{f-}{E}_i$

$\nu =$ - $\frac{{me}^4}{8{{\epsilon }_0}^2{h}^3}(\frac{1}{n_f^1}-\frac{1}{n_i^2})$

$\lambda \propto \frac{1}{\propto }$

Where $\propto$  is the reduced mass

Reduced mass for H=_H=;m_e(1-m_e/M)

D= _D; m_e(1-m_e/2M)

 =m_e(1-m_e/2M)(1+m_e/2M)

If for hydrogen deuterium, the wavelength

$\frac{{\lambda }_D}{{\lambda }_H}=\frac{{\propto }_H}{{\propto }_D}$ = (1+ $\frac{m_e}{2M}$ )^-1= (1- $\frac{1}{2*1840}$ )

${\lambda }_D={\lambda }_H*0.99973$  so lines emitted are 1217.7A⁰,1027.7A⁰,974.04A⁰

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) For house hold supplies, AC currents are used which are having zero average value over a cycle.

The line is having some resistance so power factor cos φ = R/Z≠0

so, φ not equal to π /2 ⇒ φ < /2

i.e., phase lies between 0 and π /2.

This is a multiple choice answer as classified in NCERT Exemplar

(c, d) When the AC voltage is applied to the capacitor, the plate connected to the positive terminal will be at higher potential and the plate connected to the negative terminal will be at lower potential.

The plate with positive charge will be at higher potential and the plate with negative charge will be at lower potential. So, we can say that the charge is in phase with the applied voltage.

P= E_rmsI_rmscos $\varnothing$

$\varnothing$ =90

So power, P = 0

This is a multiple choice answer as classified in NCERT Exemplar

According to the question power transferred Is P = I² Z cos?

as we know cos? = R/Z

R>0 and Z>0

cos? >0  so P>0

This is a multiple choice answer as classified in NCERT Exemplar

We have to transmit energy (power) over large distances at high alternating voltages, so current flowing through the wires will below because for a given power (P).

P= E_rmsI_rms, I_rms  is low when E_rms is high.

Power loss= I ² _rms R= low

This is a multiple choice answer as classified in NCERT Exemplar

(c, d) According to the question, the current increases on increasing the frequency of supply. Hence, the reactance of the circuit must be decreases as increasing frequency.

For a capacitive circuit,  capacitive reactance Xc= $\frac{1}{2\pi \vartheta \mathrm{C}}$

For an CR circuit Z= $\sqrt[]{R^2+\left(X_c^2\right)}$ when frequency increases X decreases.

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) inductive reactance XL= 2 $\pi \vartheta$ L

 

For an LCR circuit Z= $\sqrt[]{R^2+(X_L-X_C)}$ 2

As frequency (ν) increases, Z decreases and at certain value of frequency know as resonant frequency (ν₀), impedance Z is minimum that is Z_min= R current varies inversely with impedance and at Z_min current is maximum.

This is a multiple choice answer as classified in NCERT Exemplar

Secondary voltage V_s= 24V

Power associated with secondary P_s = 12W

I_s= $\frac{P_S}{V_S}$ = 12/24= 0.5A

I₀= Is $\sqrt[]{2}$ = 1/ $\sqrt[]{2}$ A