Physics Ncert Solutions Class 12th

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

Explanation- nuclei ₂He⁴ and ₁He³ have the same mass number . the element having more number of proton having more repusion so less binding energy. So ₂He⁴ have more number of proton so more repulsion so less binding energy.

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

Explanation- B.E = (M_p+M_H-M_N)c²

B.E= (118.9058+1.0078252-119.902199)c²

B.E=0.0114362 c²

B.E= (M_p+M_H-M_N)c²

B.E= (119.902199+1.0078252-120.902822)c²

B.E= 0.0059912c²

  (ii) the existence of magic numbers indicates that the shell structure of nucleus is similar to the shell structure of an atom. This also explains peaks in binding energy curve.

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

Explanation- 

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

Explanation- P_n=P_p+P_e

P_p+P_e=0

P_e=P_p=P

E_p= ${(m}_p^2{c}^4+p_p^2{c}^2)$ ^1/2

E_e= ${(m}_e^2{c}^4+p_p^2{c}^2)$ ^1/2

= ${(m}_e^2{c}^4+p_e^2{c}^2)$^1/2

From conservation of energy

^{${(m}_p^2{c}^4+p_p^2{c}^2)$ 1/2}= ${(m}_e^2{c}^4+p_e^2{c}^2)$ ^1/2= m_nc²

m_pc²= 936MeV, m_nc²=938MeV, m_ec²=0.51MeV

since the energy difference n and p is small

m_pc²+ $\frac{p^2{c}^2}{2m_p^2{c}^4}=m_n{c}^2-pc$

pc= m_nc²-m_pc² = 938-936= 2MeV

E_p=( $m_p^2{c}^4+p^2{c}^2$ )^1/2= $\sqrt[]{{0.51}^2+2^2}$

= 2.06MeV

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

Explanation-E= $\frac{m{e}^4}{\pi {\epsilon }_0^2{h}^2}=13.6eV$

If proton and neutron had charge e each and were governed by the same electrostatic force, then in the above equation we would need relace m

So m' = $\frac{M*N}{M+N}$ =M/2=918m

Hence binding energy= 918me'⁴/8 ${\epsilon }_0^2{h}^2=2.2MeV$

Dividing eqn

918 ( $\frac{e\text{'}}{e})$ ⁴= $\frac{2.2MeV}{13.6eV}$ = $\frac{2.2*{10}^6}{13.6}$

$\frac{e\text{'}}{e}$= 3.64

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) When circular coil expands radially in a region of magnetic field such that the magnetic field is in the same plane as the circular coil or the magnetic field has a perpendicular (to the plane of the coil) component whose magnitude is decreasing suitably in such a way that the cross product of magnetic field and surface area of plane of coil remain constant at every instant.

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

Explanation- B.E= 2.2MeV

E-B=K_n+K_P= $\frac{p_n^2}{2m}+\frac{p_p^2}{2m}$

Conservation of momentum= P_n+P_P= E/C

E=B $p_n^2-p_p^2=0$

It only happen if P_n=P_p=0

Let E=B+X where X<

X= (E/C-P_P)/2m+ $\frac{P_0^2}{2m}$

2 $P_P^2-\frac{2E{P}_p}{c}$ + $\frac{E^2}{C^2}-2mX$ =0

Pp= $\frac{\frac{2E}{c}?\text{exception 25:}}{4}$

$\frac{4E^2}{C^2}=8(\frac{E^2}{c^2}-2mX)$

$16mX$ = 4E²/c²

$X=\frac{E^2}{4m{c}^2}=\frac{B^2}{4m{C}^2}$

This is a multiple choice answer as classified in NCERT Exemplar

(a, d) Mutual inductance of a coil increases when they came closer also the relation for the same will be given by

M₂₁= ${\mu }_0$ n₁n₂ r₁²l= M₁₂

M₂₁= M₁₂=M

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

Explanation- let ³⁸S have N₁ active nuclei and ³⁸Cl have N₂ active nuclei

$\frac{d{N}_1}{dt}={-\lambda }_1{N}_1$ + ${\lambda }_1{N}_1$

N₁=N_{0 $e^{-\lambda 1t}$}

$\frac{d{N}_2}{dt}={-\lambda }_1{N}_2{e}^{\left({-\lambda }_{1}t\right)}$ + ${\lambda }_2{N}_2$

Multiplying by $e^{\lambda 2t}dt$ and then integrating both sides we got

N2= $\frac{N_o{\lambda }_1}{{\lambda }_2-{\lambda }_1}(e^{-\lambda 1t}-e^{-\lambda 2t})$

After solving it we get time t= (log $\frac{{\lambda }_1}{{\lambda }_2}$ )/ ${\lambda }_1-{\lambda }_2$

t= $\frac{log\frac{2.48}{0.62}}{2.48-0.62}$ = $\frac{2.303*2*0.3010}{1.86}=0.745s$

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, c) Here, magnetic flux linked with the isolated coil change when the coil being in a time varying magnetic field, the coil moving in a constant magnetic field or in time varying magnetic field.