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$\begin{array}{l}{\displaystyle {\int }_0^{1}x{e}^{x^2}dx=Isay},\\ Putting,x^2=t\Rightarrow 2xdx=dt\to xdx=\frac{dt}{2}\\ ={\displaystyle {\int }_0^1{e}^t\frac{dt}{2}}=\frac{1}{2}{\displaystyle {\int }_0^1{e}^{t}dt}\end{array}$

When,

x = 0, t = 0

x = 1, t = 1

$\begin{array}{l}=\frac{1}{2}{\left[e^t\right]}_0^1\\ =\frac{1}{2}\left[e^1-e^0\right]=\frac{\left(e-1\right)}{2}\end{array}$

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

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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’ = $\frac{M\times N}{M+N}$ =M/2=918m

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

Dividing eqn

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

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

$\begin{array}{l}{\displaystyle {\int }_2^3\frac{xdx}{x^2+1}}=\frac{1}{2}{\displaystyle {\int }_2^3\frac{2x}{x^2+1}}dx\\ =\frac{1}{2}{\left[log\left|x^2+1\right|\right]}_2^3\end{array}$

$\begin{array}{l}=\frac{1}{2}log\left|3^2+1\right|-\frac{1}{2}log\left|2^2+1\right|\\ =\frac{1}{2}\left[log\left|9+1\right|-log\left|4+1\right|\right]\\ =\frac{1}{2}log\frac{10}{5}\\ =\frac{1}{2}log2.\end{array}$

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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.

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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

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

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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

$\begin{array}{l}{\displaystyle {\int }_2^3\frac{dx}{x^2-1}}={\left[\frac{1}{2\times 1}log\frac{x-1}{x+1}\right]}_2^3\\ =\frac{1}{2}log\left|\frac{3-1}{3+1}\right|-\frac{1}{2}log\left|\frac{2-1}{2+1}\right|\end{array}$

$\begin{array}{l}=\frac{1}{2}log\frac{2}{4}-\frac{1}{2}log\frac{1}{3}\\ =\frac{1}{2}\left[log\frac{1}{2}-log\frac{1}{3}\right]\\ =\frac{1}{2}log\frac{1/2}{1/3}\\ =\frac{1}{2}log\frac{3}{2.}\end{array}$

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$\begin{array}{l}{\displaystyle 1_{}^{}\frac{dx}{1+x^2}}={\left[ta{n}^{-1}x\right]}_0^1\\ =ta{n}^{-1}\left(1\right)-ta{n}^{-1}\left(0\right)\\ =\; \pi /4-0\\ =\; \pi /4\end{array}$

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