Class 12th

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

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

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

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

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

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

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

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

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

This is a 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.

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